Hey thanks everyone for your support as always! Check out my quantum mechanics playlist here for more videos: th-cam.com/play/PLOlz9q28K2e4Yn2ZqbYI__dYqw5nQ9DST.html Also, as always, let me know what other physics topics you'd like me to cover in future videos :)
Please explain how the operator works i.e how it generates useful values and what they mean. Please explain the implications of the collapse of the wavefunction. Love ur content!! Also was wondering how entropy and wavefunction are related on a quantum level. If there are many possible configurations of probabilities then the entropy () in the future before measurement is high. But after measurement, does the collapse of the wavefunction cause entropy to become lower? Does this not violate 2nd law of thermodynamics in some special way...just like how the speed of light is the fastest thing only IN space but there are theoretical exeptions. Please clarify. Thanks for the content 😀.
Hi parth , after watching your video series, I have subtle lots of doubts, related to some old video topics as well, how could I ask all the questions please let me know, I really want to clear those doubts,
Maybe this is what you were saving for a future video, but the operator being complex (specifically Hermitian) is required so that its eigenvalues are guaranteed to be real. So even though the operator is complex, what's measured in experiment is always real:) Complex "observables" confused me so much in undergrad until that connection was made lol. Nice job as always!
Hermitians have one one more advantage that they can be applied to any of the Dirac brackets for deriving expectation values as they have real eigenvalues.
Maybe another detail to is mention that it's not the operator itself that has a complex form, but its representation in position space. In momentum space, for example, it's the position operator that takes a complex form. I think the more fundamental concept is the commutation relations between the two operators, since they are independent of the representation being used.
I found your overview to be thorough and it was very thoughtfully presented. I do appreciate when explainers take time to “unpack the math° and strip away the sometimes abstruse notations to show “what’s really going on” at the level that one could then sit down and actually perform the calculation with Calculus II level knowledge. Thank-you.
I've watched a lot of quantum physics related content. I've been engaged in quantum physics courses. I had never seen an explanation that was so easy to follow but still bringing all the info I needed to not feel like something was missing.
You are a talented teacher, you speak so elegantly and manage to make this stuff sound so simple. You just earned yourself a subscriber and I will definitely be recommending you to my friends!
When you apply an operator to an eigenstate, you end up multiplying the eigenstate by the eigenvalue, this is not the same as what happens when you make a measurement: when you make a measurement the general state collapses randomly to an eigenstate with a probability given by |c|^2 where c is the coefficient of that eigenstate in the original state - applying an operator and making a measurement are not the same thing :) Operators give us a spectrum of possible eigenvalues for any given observable, and they also provide us with a means of calculating an expectation value for any given observable
The state doesn't collapse. The individual quantum system doesn't have a state. It has energy, momentum and angular momentum and when you are doing a measurement, then you are taking a certain amount of energy, momentum and angular momentum out of the system. The state thing tracks the properties of the quantum mechanical ensemble and allows you to calculate the probability distributions for how much energy, momentum and angular momentum you take out of the system.
Small mistake?! You are too generous :). Quantum momentum operator being a spatial derivative, versus the classical momentum as a time derivative - underlines the huge difference between classical and quantum reality and deserves attention.
You can say I suppose that In QM momentum is a term in the exponent, to get the momentum out of the exponent you have to take the differential with respect to distance and multiply with i*hbar to remove unwanted factors. The wave function will not change as it is an exponential, therefore the wave function is on both sides of the equation.
Yes, and the reason why that is so is because of the translation symmetry of spacetime. At the end of the day the "correct" equations (Dirac, not Schroedinger) are representations of the Poincare group. The generators here are the translations and rotations and the general group elements can be written as a complex exponential. If we would teach trigonometric functions like it's the 21st century instead of the 3rd century BC (Euclid), then all of this would be a totally trivial matter even for the average high school student.
I was a EE student 20 years ago. The imaginary part i is most certainly the phase parameter of the particle or device. Capacitors and inductors also have the i component as their parameter when you want to calculate the circuit's behavior in a phasor diagram.
There are no particles in nature. There are only quanta of energy. The linearity of the theory comes from the physical independence of members of the quantum mechanical ensemble. The scalar product form and the necessity for rotations in a unitary framework arise out of Kolmogorov and the complex representation is just one of several possible (there is a quaternion version of quantum mechanics if you are interested). That the complex numbers show up can also be motivated with temporal and spatial homogeneity due to Lie-group symmetries. It's the same reason why they show up in the phasor diagram. At the end of the day this is all geometry.
6:17 This is giving me flashbacks to Fourier series problems in my signals and systems class last year. Fourier series (which are how we write complicated wave functions as sums of simpler wave functions) are awesome as a concept and super useful, but computing them by hand is a *ginormous* pain in the butt. Thanks heavens for Matlab and wolfram alpha.
Momentum in quantum mechanics is, like many other stuff in physics, not demostrated formally, instead is calculated using an example like the solution of the wave function. This is what I like the most about your video.
2:00 It is, as I remember it, theoretically impossible to have totally precise measurements of observables that have a continuum of possible outcomes, like position usually does. So even after measuring the particle, we only get a somewhat certain result back, and the wave function collapses to a more localized, but not single-positioned wave function.
6:51 Teacher: Are you familiar with differentiation? Students: No. Teacher: Well, it’s a mathematical process by which you find the gradient. Students: What’s a gradient? Teacher: Well, it’s what you get by differentiation…
This was particularly frustrating to me because there's a much better high-level description of derivatives that seems like it would give a much more intuitive idea of why the momentum operator involves taking a derivative. Derivatives tell you about how a function changes, and partial derivatives tell you about how a particular component of the function changes. Thus, by taking the partial derivative of the wave function with respect to position, you're finding out how the position component changes, i.e. what its momentum is.
It's easy, really. Measuring a quantum system means giving it energy, which changes it. If you hit an electron even with a photon, you impart energy. You could use a low energy photon with longer wavelength and the measurement is more "blurry". Or use a high energy photon (shorter wavelength) but then you are imparting energy to the system. You simply cannot measure a quantum system without interfering with it = hence the change.
@@Dinnye01 Sure, giving energy to a system can change it and we usually work with small systems in quantum mechanics, but that's also true of classical systems and neither the wave function collapse nor the uncertainty principles (an "intrinsic blurriness) depend on that - I interpreted the question to be more along the lines of "how/why does the wave function collapse happen" (to which the answer is idk : p )
@@user-sl6gn1ss8p No, FullMetal's answer is correct and complete. You need at least one photon to observe something, but that one photon changes the system. If you're asking why measuring something about a bowling ball doesn't cause it to start rolling because of the energy transfer from the photons then you're not ready for quantum mechanics.
@@user-sl6gn1ss8p To answer your question the wave function collapses because the wave function represents the probabilities associated to the particle. When you know something for sure (because you measured it) it is not a wave of probabilities anymore, it is a definite result. Like flipping a coin, in the air the result is unknown but as soon as I snatch it to see heads or tails, it is definitely one of the other. That's why the wave function collapses to a specific value.
Very nicely done. Thanks. It would be helpful to get your brief take on why and how the plane wave works -- aside from the mathematical convenience of its partials fitting the needs of the Schrödinger Equation (as you hint at here). Thanks again. As in your other vids, your explanation here is crystal.
It works because the physical vacuum has translation symmetry. The plane wave is basically just the explicit solution of the Lie-group of translations. The spherical harmonic functions are the solutions to the rotations. You are being introduced to quantum mechanics in the wrong order. Instead of being shown the differential equation, it would be much better if we would tell you how to work with continuous symmetry groups, instead, because those are universal for all equations that have these symmetries. The only change we get from e.g. the hydrogen functions, is the radial part of the solution. Why? Because the 1/r potential doesn't violate the rotational symmetry, hence it can be factored out right away. We never ever have to think about it... assuming we already understand how symmetries, equations and solutions of equations come together.
this is the clear explanation i've been waiting for. well, one of them. i've seen lots of descriptions of squaring the wave function to get a probability, but not a concrete connection between the probability of a thing and the wave function of a thing. i'm still digesting this, but when you introduced the system & wavefunction, i'd have liked to see a few examples of different systems before you focused on a particle's momentum, so i had a better picture of psi's general role in describing a system's evolution ... psi describes this variable in this system when these variables are held constant. very very good though i always wondered why they switched notation for partial derivatives. if you have extra independant variables, dy/dx could only make sense if you treat the other variables as constant ... you're specifically talking about sliding infinitessimally in the x direction only, whether you have one or many iv's
In my opinion, all particles obey this equation: c² = Ψ^i* g_ij Ψ^j Where Ψ^r is the amplitude, that in time dt, a particle jumps from point 'A' to neighborpoint A+dr. And g_ij is metrictensor. For lightbending in gravity, pathintegral would help. Just my intuition.
I don’t know if you said this, but it depends on the space. If you are in momentum space, then the momentum operator is just momentum itself. Then, the position operator is the partial derivative of the wave function w.r.t momentum, multiplied by the constant -i*hbar. The operations are essentially the opposite of what is done in position space.
So drilling the topic a little bit further the question would be why there is a p in wave function in such a form but I guess this is a big topic maybe for another video. Thanks for the great explanation, you are doing amazing job.
Great video!! Have just taken a course on modern physics and this expended my knowlage even further. I would really like to understand though, why the probability function is the square of the wave function? (Why square? Why not some other manipulation on the wave function...). Thanks!
As far as i understand it the probality is defined as |ψ|^2 because you need it to be a real number (0\leq x \leq 1) and multiplying a complex number with its complex conjugate is pretty much the easiest way to manipulate a complex number so that the result is garanteed to be a real number.
For me, it took the path integral formulation to finally understand why the wavefunction is complex and probability is its modulus squared. If you shoot a photon at a wall with two slits, the photon will delocalize, take several paths, and along each path it will accumulate some phase shift - in much the same way that a classical electromagnetic wave changes phase as it travels. In fact, for high-intensity light source, the experiment can be well described with classical EM. There, the intensity can be computed by taking the modulus squared of the complex amplitude of the wave. If you then decrease the intensity to the point of emitting a single photon at a time, the intensity becomes the probability of detection. And the complex amplitude (phasor) becomes the wavefunction.
I think understanding the momentum operator boils down to realizing how exp(position operator) generates momentum by applying a phase differently depending on the position….
Plane waves are not eigenstates of the momentum operator, since they are not normalizable, i.e. they are no admissible physical states. Hermiticity is not a consequence of real values obtained by measurements, since you are free to multiply your measured values with i where i²=-1 and call them imaginary; it is a consequence of the orthogonality of states obtained by the reduction of states during a measurement. The wave function does not give the probability of finding a particle somewhere; this concept is no longer valid in relativistic quantum mechanics.
What you show in the eigenvalue equation isn't a wavefunction. |psi> is just a state, the wavefunction is given by the projection of |psi> onto real space: psi(x)=
4:53 you can't show e^ix like that in figure...it is misleading... instead you should write it as cosx and should mention it is combination of e^ix and e^-ix.
well it is defined to give the probability that's why it gives the "Probability". Now that's a silly answer I know let me further explain it a bit. So squaring the wavefunction turns out to satisfy all the axioms for what a function needs to be in order for it to be a "Probability distribution" and it also turned out that making The square of the wavefunction to be the probability distribution allows the machinery of QM very mathematically general. Physicist Max Born figured out this nature of the square of the wavefunction in mathematical terms and hence found it sufficient enough to postulate this into quantum mechanics. Later mathematician Von Neuman while generalizing the mathematical background of quantum mechanics via introducing Hilbert Spaces also showed how choosing the square of the Wavefunction makes sense in a general mathematical sense. For more details look at L^{2} functions in Hilbert space.
@@dhritimanroyghatak2408 so the answer is basically: because if we do it that way for no particular reason at all it turns out to work remarkably well?
@@philip8498 well I did mention some of the reasons altho they were mostly mathematical does not disregard their capacity to model physical reality. Maybe I did a poor job let me try again. 1) So fist of all for a function to be a probability distribution it must fulfill the various properties of a probability distribution. All everyday probability distribution may they be in the domain of economics to biology to generalized statistics obey and follow those properties. It turned out that square of wavefunction also follows these properties hence it makes sense to define it so. 2) Now there's more. It turned out once u do define it to be the probability density u end up with an algorithm to measure and assign probabilities to various physical quantities within QM such as position, momentum, intensity etc which can be experimentally measured by repetitive measurements and they matched the theoretically calculated value. So defining the square of the wavefunction as a probability distribution not just only satisfies all the mathematical properties of a probability distribution but also gives strong experimental validation. 3) Also in any wavemechanics weather it be sound, water, EM, anything the square of the amplitude has always has its importance as being proportional to the intensity. Hence giving square of the amplitude in QM(which is a wave mechanics) a purpose made it complete and general with respect to wave mechanics. All these r very strong reasons to postulate the statistical interpretation in QM and that is what the founding fathers of QM aptly did.
@@dhritimanroyghatak2408 First off, you didn't tackle at all the Why just the How. Which does not answer my question I am asking why does squaring the wave function, a mathematical representation of fundamental units of nature, give the probability of that units location. If I squared the function that represents you would I get your approximate location? Makes no sense at all What you have described is more along the lines of "its true because it was an accident that we defined to be true" and when we force it to be true, that is the only time/method that gives the measurements similar enough to force that model to be "true". Self fulfilling prophecy much? I'm siding with @Philip on this Seems more like a con artist a long time ago tried to convince someone of something rather than a method of explanation
Relationship between the real Simple Harmonic Oscillator and the QFT (Dirac) Oscillator/ladder operator in the complex plane, and the extension of that into the hyperplane
Do you have a video/video recommendation, about how to practically measure a quantum state and how to interpret the result with the eigenequation. You mentioned such a video at 08:07. Thank you in advice.
In the 'actual world' of course all dimensions are involved. To make life simpler while learning, we can restrict attention to one of the space dimensions without losing any physics, but gaining explanatory clarity. Later on, generalize to - as many dimensions as you like! But the additional complication will come from the geometry, not the physics. You can explain all the physics in terms of the wave function's behavior in (or along) a single dimension.
@@ParthGChannel would I be correct in saying that the momentum operator could then be represented in vector form using the up-side-down triangle? (I'm not sure if it is div, grad or curl)... Or would that be a classical way of thinking about it
Fantastic video (as always)! One question: is the quantic momentum measured in Kg x m/s? If so, does this formulation still apply in the case of particles without mass? 🤔
Hi Parth, excellent video. As I watched your video, I thought about how particle accelerators will yield the positions of particles after collisions, and how these are used to determine their masses. My challenge and request is to link the concepts in this video, particularly the landscape of eigenvalues of the wave function, to the practical outputs of such accelerators. Thank you for your videos.
@@yashdadhwal3034 But what's the problem with imaginary numbers? Imaginary numbers are actual numbers which we need to use to understand the reality of the universe we live in. The name is a misnomer, they are not imaginary but real. First we hesitated about negative numbers now about imaginary numbers.
@@nisargbhavsar25 the thing is it would really be interesting to see equation of quantum mechanics if one derive them from a wave function which is combination of sine and cos rather including e raise to power i
@@nisargbhavsar25 I always try to figure out why imaginary numbers appear in quantum mechanics the real meaning but I came to find out it is just schrodinger took a plane wave solution of wave function and derive everything from it which I really doesn't like I mean it's like a postulate you can't question in a theory because all assumptions are made over it
Because the harmonic oscillator keeps exchanging momentum with its "support" potential. That is already so in classical mechanics. We are "fixing the spring" to an absolute coordinate system, so the motion does not conserve momentum and kinetic energy. It's one of those implicit high school/undergrad physics approximations that robs you of the chance to develop a solid intuition into the real behavior of systems early on. Instead you are being given a shortcut into a toy treatment of these problems that at the end of the day is false. The actual hydrogen solution, where the potential is between two quanta, a heavy nucleus and a light electron, also contains the plane wave continuum for the momentum of the center of mass motion. The Eigenspectrum of that is real. There is another effect, which in reality is far more important: the coupling to the electromagnetic field makes all but the ground state unstable, i.e. technically only the ground state energy is real. All other states have imaginary admixtures that come from the possible decays into all other allowed states (not just single photon but also multi-photon transitions AND virtual terms). This introduces natural line broadening that can be seen in atomic spectroscopy.
-i.h = h/i; just noting how the minus sign appeared. Also noting that for a compactly supported wave function, you need to sum an infinite number of the 'sine waves' which then leads folks into thinking that the wave function had infinite extent (because each 'sine wave' does), and forgetting the wave-particle duality aspect that such 'particles' are localised, rather than being a point. Just as any real wave function isn't infinite. Finally, the probability vs measurement, can be compared to a shuffled card deck. You know the probabilities for any hand, but until it is dealt, you won't know the hand you're given!
oh this is so insightful, though indirectly. i was always taught and read in books that the wave function can be expressed as A×exp(i×(kx + ωt)). _No one_ showed it to me this way as exp(i÷h×(px + Et)). now i understand _why_ the momentum operator is such. 🤦
This should make sense because if h-bar goes to 0 then there is no spacing between energy levels anymore and they are continuous, which is what they appear to be to us on the macro scale.
Yes. If h-bar goes to zero, then the uncertainty principle does not prevent both position and momentum distributions to be sharply peaked, i.e. classical.
How accurate can we measure the speed of light? If a pair of entangled photons are aimed through a detector, do they always get to the detector at the proper speed of light for the distance traveled? If the wave form gives a probability, then they should many of the times not collapse at the same time at the detector or get detected at the same time yes? Can someone reference the experiment where this has been tested or has that not been an experiment performed?
The waveform gives a probability for the momentum, not for the velocity. A photon's momentum is independent of the velocity (c) and only dependent on its wavelength.
part of the trouble comes from the fact that a quantum object's speed is not well defined. c as the speed of light is a notion from classical relativity, and so it has been measured with classical methods. subsequently, c has been given a defined value and does not truly refer to "the" speed of light, but rather as a constant pertaining to the geometry of spacetime, which is still important as you can tell from the fact that we do not observe macroscopic objects travelling at superluminal speeds.
@@itsalongday Just to clarify, the question is aimed at referencing the difference traveled to the detector should always be the speed of light. But in wave form the time they are detected should be different as the probability implies when they are detected one should be detected before the other which would mean one photon is not traveling at the speed of light.
@@Errenium yes c is the speed of causality, but in this reference, when in the same medium, 2 entangle pairs of photons should hit at the same time, but probability means they should not. I guess in my OP, the speed of light doesn't need to be measured, just ensured that both entangle pairs are moving through the same medium as the actual speed doesn't matter but that their distance does.
Why is there a unique choice of eigenvalue lambda for the operator? Or is lambda some kind of probabilistic object (e.g. we get any particular lambda based on some probability measure supported on spectrum of the momentum operator)?
hcut*K=P This expression is extremely useful in every modern physics topics be it condensed matter physics or Quantum mechanics as such. Anyway before going to quantum mechanics make sure you have complex analysis,linear algebra,basic calculus,waves etc packed inside your bag.
In a sense QM is an axiomatic theory justified by "it works if we do it that way" And part of the axioms is to say that physical systems are described by wavefunctions, and physical observables like charge and number of particles, energy etc. are defined to be represented by operators acting on these wavefunctions So yes, you are basically right. A bit disappointing, but this is also the magic-mystery of physics yet to be solved imo :D
It's always just taken as a given that we use some weird square root of probability. Not probability itself, which would make some sense. It'd be nice if there was any sort of rationale or intuition behind it.
Nothing ever collapses and there are no particles. A "detection" is an irreversible energy transfer from the free field to the detector. There are systems for which all of these energy transfers are independent, i.e. one can't find any correlations between them. For systems for which this is the case we can build an ensemble theory, in which we are replacing the single system made from a source, the free field and the detector, with an ensemble of infinite copies that contain exactly one quantum of energy each. Quantum mechanics is the theory of that ensemble. It's not the theory of the single system, which you would notice very quickly if you tried to replace a thermal source with a laser in optical experiments and you would do correlations between individual quanta for times scales that are shorter than the coherence time of the laser. But since you are not taking laser physics at the same time as you are taking introduction to quantum mechanics nobody giving the QM lectures cares about telling you that the precondition for "single particle QM" is statistical independence of individual detection events and this is NOT guaranteed for every experiment. You actually have to test that simplifying assumption experimentally before you blindly apply QM. OK, so let's say the statistical independence assumption holds, then we can build an ensemble theory, which mathematically leads to Kolmogorov's axioms. The main one there is that P(x) + P(x̄) =1 (the sum of the probabilities of an event set and its complement is one). This is a partition of unity and it can be fulfilled with the algebraic solution p + (1-p) =1 (that's ordinary probability theory) and with the geometric solution sin^2(phi)+cos^2(phi)=1 (that's quantum mechanics). The latter leads to the usual definition of vectors, scalar products and linear vector spaces and in the infinite dimensional limit this becomes the Hilbert space of non-relativistic QM. Translation symmetry of spacetime in those vector spaces leads to wave-like solutions FOR THE ENSEMBLE. And if nobody ever told you this before, then you need to get your student loans back, because you got a really crappy physics education. ;-)
PARTH, I like your videos very much. Tristan Needham in VISUAL COMPLEX ANALYSIS, page 217 gives a definition of i, which I think is important for what you discuss in this video (please look it up). For a complex number z=x+iy, a change in x result in change in y, Tristan defines i as the ratio of change in y (effect) to the change in x (cause). I am sure this should give an insight into QM, as you refer to ih or i/h, implying cause is at right angle to effect. Empiricists thought cause and effect are independent of each other. Complex Analysis proves they are related by i. Does this give you an added insight into QM?
No, it does not. The reason for the use of complex numbers in QM is that position and momentum distributions are connected by Fourier transformation that in turn is a change of basis transformation with an infinite two dimensional basis. The Fourier transformation is also the model of fundamental uncertainty principle between position and momentum that determine the state of motion of a classical particle. Also, in QM those definite values of position and momentum become fuzzy distributions.
@@sonarbangla8711 The unitary Fourier transformation transforms the same information of the state of motion between position and momentum basis. That is also the reason for uncertainty principle. The quantum fuzziness comes from the fact that we need to consider all the possible (typically infinite) eigenvalues of the operator (physical observable) in question.
@@sonarbangla8711 There is a trade off with information. If you measure position with high accuracy, then the position distribution indeed becomes a delta function, BUT at the same time the momentum becomes super fuzzy due to the uncertainty principle.
Every point z = x+iy in the complex plane becomes -i*z = y - ix, using i^2=-1. If you picture that as coordinates (x,y), it's a 270 degree rotation about the origin, to (y,-x).
@@bentoomey15 Thanks, but I meant, does the wave function then look kind of like a sinusoidal wave rotated 270 degrees, in effect looking like the graph of x = sin y?
@@paulthompson9668 Ah, sorry. Yes, you are right, that's what happens. To be a little extra, the wavefunction actually "fills" the entire complex plane, and the sine wave shown is just a slice of what the given wavefunction looks like along the real axis. That whole function gets rotated, and the slice that looks like sine is now on the imaginary axis. The whole thing is really happening in 4 real dimensions (because the wavefunction outputs complex numbers), so the best we can do is simplified visualizations when things are real-valued.
Maybe I'm wrong, but I'm also confused about the eigenvalue equation (from 3 minutes) you say "we apply an operator to our wave function..." .But it appears to me that the equation shown multiplies a vector - a ket (not a wavefunction) by a Hermitian matrix (not an operator). The result is the same but the method isn't.
Actually it’s hermitian, meaning it only has real eigenvalues, that’s like the first thing a physics undergrad proves In their QM class. I know you’re probably gonna say this but still, this is clearly click bait to any physics undergrad.
Hey thanks everyone for your support as always! Check out my quantum mechanics playlist here for more videos: th-cam.com/play/PLOlz9q28K2e4Yn2ZqbYI__dYqw5nQ9DST.html
Also, as always, let me know what other physics topics you'd like me to cover in future videos :)
Please explain how the operator works i.e how it generates useful values and what they mean. Please explain the implications of the collapse of the wavefunction. Love ur content!!
Also was wondering how entropy and wavefunction are related on a quantum level. If there are many possible configurations of probabilities then the entropy () in the future before measurement is high. But after measurement, does the collapse of the wavefunction cause entropy to become lower? Does this not violate 2nd law of thermodynamics in some special way...just like how the speed of light is the fastest thing only IN space but there are theoretical exeptions. Please clarify. Thanks for the content 😀.
Hi
parth , after watching your video series, I have subtle lots of doubts, related to some old video topics as well, how could I ask all the questions please let me know, I really want to clear those doubts,
how about a noncommutative explanation from Heisenberg? thanks
Maybe this is what you were saving for a future video, but the operator being complex (specifically Hermitian) is required so that its eigenvalues are guaranteed to be real. So even though the operator is complex, what's measured in experiment is always real:) Complex "observables" confused me so much in undergrad until that connection was made lol. Nice job as always!
Oh hey Andrew :)
Hermitians have one one more advantage that they can be applied to any of the Dirac brackets for deriving expectation values as they have real eigenvalues.
Dude thanks for tagging this on, super useful!!
Nice to find you here, papa Andrew
Maybe another detail to is mention that it's not the operator itself that has a complex form, but its representation in position space. In momentum space, for example, it's the position operator that takes a complex form. I think the more fundamental concept is the commutation relations between the two operators, since they are independent of the representation being used.
I found your overview to be thorough and it was very thoughtfully presented. I do appreciate when explainers take time to “unpack the math° and strip away the sometimes abstruse notations to show “what’s really going on” at the level that one could then sit down and actually perform the calculation with Calculus II level knowledge. Thank-you.
I've watched a lot of quantum physics related content. I've been engaged in quantum physics courses. I had never seen an explanation that was so easy to follow but still bringing all the info I needed to not feel like something was missing.
You are a talented teacher, you speak so elegantly and manage to make this stuff sound so simple. You just earned yourself a subscriber and I will definitely be recommending you to my friends!
When you apply an operator to an eigenstate, you end up multiplying the eigenstate by the eigenvalue, this is not the same as what happens when you make a measurement: when you make a measurement the general state collapses randomly to an eigenstate with a probability given by |c|^2 where c is the coefficient of that eigenstate in the original state - applying an operator and making a measurement are not the same thing :)
Operators give us a spectrum of possible eigenvalues for any given observable, and they also provide us with a means of calculating an expectation value for any given observable
The state doesn't collapse. The individual quantum system doesn't have a state. It has energy, momentum and angular momentum and when you are doing a measurement, then you are taking a certain amount of energy, momentum and angular momentum out of the system. The state thing tracks the properties of the quantum mechanical ensemble and allows you to calculate the probability distributions for how much energy, momentum and angular momentum you take out of the system.
Still can't believe his channel isn't called Parth to Knowledge...
I missed a trick!
Fart to knowledge
Very clear explanation
Bravo. so many well-worded explanations on this channel. too bad this didn't exist 20 years ago
Sir, this is a first-class educational video and you explain with straight and precise language extremely difficult concepts. Thank you!
also at 8:20 there is a small mistake, you have written p = -ihd/dt, the derivative is supposed to be wrt x!
Oops! Great spot :)
Small mistake?! You are too generous :). Quantum momentum operator being a spatial derivative, versus the classical momentum as a time derivative - underlines the huge difference between classical and quantum reality and deserves attention.
Thank you! I'm studying at the university now and finaly got it! 🙌
You can say I suppose that In QM momentum is a term in the exponent, to get the momentum out of the exponent you have to take the differential with respect to distance and multiply with i*hbar to remove unwanted factors. The wave function will not change as it is an exponential, therefore the wave function is on both sides of the equation.
Yes, and the reason why that is so is because of the translation symmetry of spacetime. At the end of the day the "correct" equations (Dirac, not Schroedinger) are representations of the Poincare group. The generators here are the translations and rotations and the general group elements can be written as a complex exponential. If we would teach trigonometric functions like it's the 21st century instead of the 3rd century BC (Euclid), then all of this would be a totally trivial matter even for the average high school student.
I was a EE student 20 years ago. The imaginary part i is most certainly the phase parameter of the particle or device. Capacitors and inductors also have the i component as their parameter when you want to calculate the circuit's behavior in a phasor diagram.
There are no particles in nature. There are only quanta of energy. The linearity of the theory comes from the physical independence of members of the quantum mechanical ensemble. The scalar product form and the necessity for rotations in a unitary framework arise out of Kolmogorov and the complex representation is just one of several possible (there is a quaternion version of quantum mechanics if you are interested). That the complex numbers show up can also be motivated with temporal and spatial homogeneity due to Lie-group symmetries. It's the same reason why they show up in the phasor diagram. At the end of the day this is all geometry.
6:17
This is giving me flashbacks to Fourier series problems in my signals and systems class last year. Fourier series (which are how we write complicated wave functions as sums of simpler wave functions) are awesome as a concept and super useful, but computing them by hand is a *ginormous* pain in the butt. Thanks heavens for Matlab and wolfram alpha.
Momentum in quantum mechanics is, like many other stuff in physics, not demostrated formally, instead is calculated using an example like the solution of the wave function. This is what I like the most about your video.
In the Heisenberg picture, the momentum is the mass times the velocity operator (time derivative of the position operator).
2:00 It is, as I remember it, theoretically impossible to have totally precise measurements of observables that have a continuum of possible outcomes, like position usually does. So even after measuring the particle, we only get a somewhat certain result back, and the wave function collapses to a more localized, but not single-positioned wave function.
wow thats great thanks! Its amazing how easy QM can be sometimes, abstract and complex but not necessarily too complicated.
You are the best one who can describe anything to anyone
So good! I wish you would be giving lectures at my Uni
6:51
Teacher: Are you familiar with differentiation?
Students: No.
Teacher: Well, it’s a mathematical process by which you find the gradient.
Students: What’s a gradient?
Teacher: Well, it’s what you get by differentiation…
Are you by any chance channeling
Søren Kierkegaard? There's a similarity of ironic humor!
This was particularly frustrating to me because there's a much better high-level description of derivatives that seems like it would give a much more intuitive idea of why the momentum operator involves taking a derivative. Derivatives tell you about how a function changes, and partial derivatives tell you about how a particular component of the function changes. Thus, by taking the partial derivative of the wave function with respect to position, you're finding out how the position component changes, i.e. what its momentum is.
Thank you for your time and effort.
Definitely want to hear more about how measuring the wave function concretely changes the wave function.
It's easy, really. Measuring a quantum system means giving it energy, which changes it.
If you hit an electron even with a photon, you impart energy. You could use a low energy photon with longer wavelength and the measurement is more "blurry". Or use a high energy photon (shorter wavelength) but then you are imparting energy to the system. You simply cannot measure a quantum system without interfering with it = hence the change.
That said, I'm interested in the math side of it too, so good topic idea!
@@Dinnye01 Sure, giving energy to a system can change it and we usually work with small systems in quantum mechanics, but that's also true of classical systems and neither the wave function collapse nor the uncertainty principles (an "intrinsic blurriness) depend on that - I interpreted the question to be more along the lines of "how/why does the wave function collapse happen" (to which the answer is idk : p )
@@user-sl6gn1ss8p No, FullMetal's answer is correct and complete. You need at least one photon to observe something, but that one photon changes the system. If you're asking why measuring something about a bowling ball doesn't cause it to start rolling because of the energy transfer from the photons then you're not ready for quantum mechanics.
@@user-sl6gn1ss8p To answer your question the wave function collapses because the wave function represents the probabilities associated to the particle. When you know something for sure (because you measured it) it is not a wave of probabilities anymore, it is a definite result. Like flipping a coin, in the air the result is unknown but as soon as I snatch it to see heads or tails, it is definitely one of the other. That's why the wave function collapses to a specific value.
This is really good, thank you
my god it is so crystal clear now thank you so much bhaiya.... thats like a champ teacher!!!!!!! thankyou so much again
Very nicely done. Thanks. It would be helpful to get your brief take on why and how the plane wave works -- aside from the mathematical convenience of its partials fitting the needs of the Schrödinger Equation (as you hint at here). Thanks again. As in your other vids, your explanation here is crystal.
It works because the physical vacuum has translation symmetry. The plane wave is basically just the explicit solution of the Lie-group of translations. The spherical harmonic functions are the solutions to the rotations. You are being introduced to quantum mechanics in the wrong order. Instead of being shown the differential equation, it would be much better if we would tell you how to work with continuous symmetry groups, instead, because those are universal for all equations that have these symmetries. The only change we get from e.g. the hydrogen functions, is the radial part of the solution. Why? Because the 1/r potential doesn't violate the rotational symmetry, hence it can be factored out right away. We never ever have to think about it... assuming we already understand how symmetries, equations and solutions of equations come together.
this is the clear explanation i've been waiting for. well, one of them. i've seen lots of descriptions of squaring the wave function to get a probability, but not a concrete connection between the probability of a thing and the wave function of a thing. i'm still digesting this, but when you introduced the system & wavefunction, i'd have liked to see a few examples of different systems before you focused on a particle's momentum, so i had a better picture of psi's general role in describing a system's evolution ... psi describes this variable in this system when these variables are held constant. very very good though
i always wondered why they switched notation for partial derivatives. if you have extra independant variables, dy/dx could only make sense if you treat the other variables as constant ... you're specifically talking about sliding infinitessimally in the x direction only, whether you have one or many iv's
Well explained
In my opinion, all particles obey this equation: c² = Ψ^i* g_ij Ψ^j
Where Ψ^r is the amplitude, that in time dt, a particle jumps from point 'A' to neighborpoint A+dr.
And g_ij is metrictensor. For lightbending in gravity, pathintegral would help. Just my intuition.
I don’t know if you said this, but it depends on the space. If you are in momentum space, then the momentum operator is just momentum itself. Then, the position operator is the partial derivative of the wave function w.r.t momentum, multiplied by the constant -i*hbar. The operations are essentially the opposite of what is done in position space.
Hey cool explanation in simplest form...
So drilling the topic a little bit further the question would be why there is a p in wave function in such a form but I guess this is a big topic maybe for another video. Thanks for the great explanation, you are doing amazing job.
Great video! You explain physics so well! It would be nice if you made a video about the higgs mechanism
PLEASE PLEASE make a video on the DIRAC EQUATION
Great lesson, Parth G. Just a comment: last formula is really correct (derivative over X, not over time)? Thank you for all you excellent support.
Great video!! Have just taken a course on modern physics and this expended my knowlage even further.
I would really like to understand though, why the probability function is the square of the wave function? (Why square? Why not some other manipulation on the wave function...). Thanks!
As far as i understand it the probality is defined as |ψ|^2 because you need it to be a real number (0\leq x \leq 1) and multiplying a complex number with its complex conjugate is pretty much the easiest way to manipulate a complex number so that the result is garanteed to be a real number.
For me, it took the path integral formulation to finally understand why the wavefunction is complex and probability is its modulus squared. If you shoot a photon at a wall with two slits, the photon will delocalize, take several paths, and along each path it will accumulate some phase shift - in much the same way that a classical electromagnetic wave changes phase as it travels. In fact, for high-intensity light source, the experiment can be well described with classical EM. There, the intensity can be computed by taking the modulus squared of the complex amplitude of the wave. If you then decrease the intensity to the point of emitting a single photon at a time, the intensity becomes the probability of detection. And the complex amplitude (phasor) becomes the wavefunction.
5:04 Such a simple explanation! :-) I like it.
I think understanding the momentum operator boils down to realizing how exp(position operator) generates momentum by applying a phase differently depending on the position….
you deserve a Like
Plane waves are not eigenstates of the momentum operator, since they are not normalizable, i.e. they are no admissible physical states. Hermiticity is not a consequence of real values obtained by measurements, since you are free to multiply your measured values with i where i²=-1 and call them imaginary; it is a consequence of the orthogonality of states obtained by the reduction of states during a measurement. The wave function does not give the probability of finding a particle somewhere; this concept is no longer valid in relativistic quantum mechanics.
I want to see a video on the Parth Integral formulation of quantum mechanics.
I've seen a video or two on it, but it'd be cool if you did a series on the mathematics of perturbation theory
What do you think about the definition of momentum in bohmian mechanics ?
What you show in the eigenvalue equation isn't a wavefunction. |psi> is just a state, the wavefunction is given by the projection of |psi> onto real space: psi(x)=
4:53 you can't show e^ix like that in figure...it is misleading... instead you should write it as cosx and should mention it is combination of e^ix and e^-ix.
I was scratching my head too lol 🤔 good catch 😇
Awesome video. One question though, at 6:45 you mention resources below to get more info on partial differentiation and i don't see it in the notes.
Khan academy legit starting point 🤔
Can you please explain why the wave function squared gives "Probability"?
well it is defined to give the probability that's why it gives the "Probability". Now that's a silly answer I know let me further explain it a bit. So squaring the wavefunction turns out to satisfy all the axioms for what a function needs to be in order for it to be a "Probability distribution" and it also turned out that making The square of the wavefunction to be the probability distribution allows the machinery of QM very mathematically general. Physicist Max Born figured out this nature of the square of the wavefunction in mathematical terms and hence found it sufficient enough to postulate this into quantum mechanics.
Later mathematician Von Neuman while generalizing the mathematical background of quantum mechanics via introducing Hilbert Spaces also showed how choosing the square of the Wavefunction makes sense in a general mathematical sense.
For more details look at L^{2} functions in Hilbert space.
It comes from wave intensity.
@@dhritimanroyghatak2408 so the answer is basically: because if we do it that way for no particular reason at all it turns out to work remarkably well?
@@philip8498 well I did mention some of the reasons altho they were mostly mathematical does not disregard their capacity to model physical reality. Maybe I did a poor job let me try again.
1) So fist of all for a function to be a probability distribution it must fulfill the various properties of a probability distribution. All everyday probability distribution may they be in the domain of economics to biology to generalized statistics obey and follow those properties.
It turned out that square of wavefunction also follows these properties hence it makes sense to define it so.
2) Now there's more. It turned out once u do define it to be the probability density u end up with an algorithm to measure and assign probabilities to various physical quantities within QM such as position, momentum, intensity etc which can be experimentally measured by repetitive measurements and they matched the theoretically calculated value.
So defining the square of the wavefunction as a probability distribution not just only satisfies all the mathematical properties of a probability distribution but also gives strong experimental validation.
3) Also in any wavemechanics weather it be sound, water, EM, anything the square of the amplitude has always has its importance as being proportional to the intensity. Hence giving square of the amplitude in QM(which is a wave mechanics) a purpose made it complete and general with respect to wave mechanics.
All these r very strong reasons to postulate the statistical interpretation in QM and that is what the founding fathers of QM aptly did.
@@dhritimanroyghatak2408 First off, you didn't tackle at all the Why just the How. Which does not answer my question
I am asking why does squaring the wave function, a mathematical representation of fundamental units of nature, give the probability of that units location. If I squared the function that represents you would I get your approximate location? Makes no sense at all
What you have described is more along the lines of "its true because it was an accident that we defined to be true" and when we force it to be true, that is the only time/method that gives the measurements similar enough to force that model to be "true". Self fulfilling prophecy much? I'm siding with @Philip on this
Seems more like a con artist a long time ago tried to convince someone of something rather than a method of explanation
Plz make dirac equation video
8:28 should it be del x instead of del t?
Brother any video on thermodynamics??
Perhaps it would help in the last step just to explain that 1/i = - i.
Relationship between the real Simple Harmonic Oscillator and the QFT (Dirac) Oscillator/ladder operator in the complex plane, and the extension of that into the hyperplane
Wow 😍😍 love you sir
Do you have a video/video recommendation, about how to practically measure a quantum state and how to interpret the result with the eigenequation. You mentioned such a video at 08:07. Thank you in advice.
How did you get the equation at 7:25
I am looking in the description block for resources about partial differentiation - and see none.
7:56 and 8:20
Sorry, I am confused. Partial derivative with respect to t or x?
Why is the momentum operator only dependent on the x-axis of position? Is there a specific reason why the y and z axis are ignored?
We are just ignoring it here for simplicity - but in reality, there is a y and z dependency as well, in the same way as x is seen here!
In the 'actual world' of course all dimensions are involved. To make life simpler while learning, we can restrict attention to one of the space dimensions without losing any physics, but gaining explanatory clarity. Later on, generalize to - as many dimensions as you like! But the additional complication will come from the geometry, not the physics. You can explain all the physics in terms of the wave function's behavior in (or along) a single dimension.
@@ParthGChannel would I be correct in saying that the momentum operator could then be represented in vector form using the up-side-down triangle? (I'm not sure if it is div, grad or curl)... Or would that be a classical way of thinking about it
@@shaungovender7805 indeed you would be correct to do so. so in general the momentum operator in 3D would be -i h grad (wavefunction).
Fantastic video (as always)! One question: is the quantic momentum measured in Kg x m/s? If so, does this formulation still apply in the case of particles without mass? 🤔
Hi Parth, excellent video. As I watched your video, I thought about how particle accelerators will yield the positions of particles after collisions, and how these are used to determine their masses. My challenge and request is to link the concepts in this video, particularly the landscape of eigenvalues of the wave function, to the practical outputs of such accelerators. Thank you for your videos.
Instead of using e raise to power i as wave function can't we use sin or cos i mean in that way we don't have to deal with i in equation maybe.
It is actually easier to represent waves in complex form because it's derivatives can be written in its own form easier than for sin and cos.
@@nisargbhavsar25 yeah but due to that i appears in equation
@@yashdadhwal3034 But what's the problem with imaginary numbers? Imaginary numbers are actual numbers which we need to use to understand the reality of the universe we live in. The name is a misnomer, they are not imaginary but real.
First we hesitated about negative numbers now about imaginary numbers.
@@nisargbhavsar25 the thing is it would really be interesting to see equation of quantum mechanics if one derive them from a wave function which is combination of sine and cos rather including e raise to power i
@@nisargbhavsar25 I always try to figure out why imaginary numbers appear in quantum mechanics the real meaning but I came to find out it is just schrodinger took a plane wave solution of wave function and derive everything from it which I really doesn't like I mean it's like a postulate you can't question in a theory because all assumptions are made over it
That's well and good for the plane wave; but why does the momentum EIGENSTATE for a quantum harmonic oscilator have a complex value?
Because the harmonic oscillator keeps exchanging momentum with its "support" potential. That is already so in classical mechanics. We are "fixing the spring" to an absolute coordinate system, so the motion does not conserve momentum and kinetic energy. It's one of those implicit high school/undergrad physics approximations that robs you of the chance to develop a solid intuition into the real behavior of systems early on. Instead you are being given a shortcut into a toy treatment of these problems that at the end of the day is false. The actual hydrogen solution, where the potential is between two quanta, a heavy nucleus and a light electron, also contains the plane wave continuum for the momentum of the center of mass motion. The Eigenspectrum of that is real.
There is another effect, which in reality is far more important: the coupling to the electromagnetic field makes all but the ground state unstable, i.e. technically only the ground state energy is real. All other states have imaginary admixtures that come from the possible decays into all other allowed states (not just single photon but also multi-photon transitions AND virtual terms). This introduces natural line broadening that can be seen in atomic spectroscopy.
Can you help us visualise the equation in the nest videp
-i.h = h/i; just noting how the minus sign appeared.
Also noting that for a compactly supported wave function, you need to sum an infinite number of the 'sine waves' which then leads folks into thinking that the wave function had infinite extent (because each 'sine wave' does), and forgetting the wave-particle duality aspect that such 'particles' are localised, rather than being a point. Just as any real wave function isn't infinite.
Finally, the probability vs measurement, can be compared to a shuffled card deck. You know the probabilities for any hand, but until it is dealt, you won't know the hand you're given!
Then, similar to the momentum operator we should be able to construct energy operator as well. But the doesn't give us the hamiltonian . Why ?
Wonderful
That question has been bothering me for a good year. But now i feel kinda stupid since the explanation is that easy
8:19 Wait, isn't Psi derived with respect to x?
oh this is so insightful, though indirectly. i was always taught and read in books that the wave function can be expressed as A×exp(i×(kx + ωt)). _No one_ showed it to me this way as exp(i÷h×(px + Et)). now i understand _why_ the momentum operator is such. 🤦
Does the quantum form converge or match the classical form in some limit?
quantum matches classical in the limit that h-bar goes to 0.
This should make sense because if h-bar goes to 0 then there is no spacing between energy levels anymore and they are continuous, which is what they appear to be to us on the macro scale.
Yes. If h-bar goes to zero, then the uncertainty principle does not prevent both position and momentum distributions to be sharply peaked, i.e. classical.
How accurate can we measure the speed of light? If a pair of entangled photons are aimed through a detector, do they always get to the detector at the proper speed of light for the distance traveled? If the wave form gives a probability, then they should many of the times not collapse at the same time at the detector or get detected at the same time yes? Can someone reference the experiment where this has been tested or has that not been an experiment performed?
The waveform gives a probability for the momentum, not for the velocity. A photon's momentum is independent of the velocity (c) and only dependent on its wavelength.
part of the trouble comes from the fact that a quantum object's speed is not well defined. c as the speed of light is a notion from classical relativity, and so it has been measured with classical methods. subsequently, c has been given a defined value and does not truly refer to "the" speed of light, but rather as a constant pertaining to the geometry of spacetime, which is still important as you can tell from the fact that we do not observe macroscopic objects travelling at superluminal speeds.
@@itsalongday Just to clarify, the question is aimed at referencing the difference traveled to the detector should always be the speed of light. But in wave form the time they are detected should be different as the probability implies when they are detected one should be detected before the other which would mean one photon is not traveling at the speed of light.
@@Errenium yes c is the speed of causality, but in this reference, when in the same medium, 2 entangle pairs of photons should hit at the same time, but probability means they should not. I guess in my OP, the speed of light doesn't need to be measured, just ensured that both entangle pairs are moving through the same medium as the actual speed doesn't matter but that their distance does.
Very good!👍
So what is p? Is it momentum of a quantum particle or maybe probability of momentum?
There is a wave function for momentum also, and that function implies all the possible classical values of p.
@@xjuhox ah that’s interesting thanks!
Why is there a unique choice of eigenvalue lambda for the operator? Or is lambda some kind of probabilistic object (e.g. we get any particular lambda based on some probability measure supported on spectrum of the momentum operator)?
Parth, I love you!
This video is gonna save IITB Freshers 😜
hcut*K=P
This expression is extremely useful in every modern physics topics be it condensed matter physics or Quantum mechanics as such.
Anyway before going to quantum mechanics make sure you have complex analysis,linear algebra,basic calculus,waves etc packed inside your bag.
Is there a way to derieve Q.M operators mathematically or is it just take it as you find it.
In a sense QM is an axiomatic theory justified by "it works if we do it that way"
And part of the axioms is to say that physical systems are described by wavefunctions, and physical observables like charge and number of particles, energy etc. are defined to be represented by operators acting on these wavefunctions
So yes, you are basically right.
A bit disappointing, but this is also the magic-mystery of physics yet to be solved imo :D
That explains why when I study I feel unmotivated, my momentum is imaginary...
are particles are states of spacetime and other fields or they are actually particles in spacetime?
So particles? uh hate it..
It's always just taken as a given that we use some weird square root of probability. Not probability itself, which would make some sense. It'd be nice if there was any sort of rationale or intuition behind it.
wonderful
Why is it a wave in the first place?
Assuming we detect a particle, it collapses, we let it go on its marry way. Why would it become wavy?
Nothing ever collapses and there are no particles. A "detection" is an irreversible energy transfer from the free field to the detector. There are systems for which all of these energy transfers are independent, i.e. one can't find any correlations between them. For systems for which this is the case we can build an ensemble theory, in which we are replacing the single system made from a source, the free field and the detector, with an ensemble of infinite copies that contain exactly one quantum of energy each. Quantum mechanics is the theory of that ensemble. It's not the theory of the single system, which you would notice very quickly if you tried to replace a thermal source with a laser in optical experiments and you would do correlations between individual quanta for times scales that are shorter than the coherence time of the laser. But since you are not taking laser physics at the same time as you are taking introduction to quantum mechanics nobody giving the QM lectures cares about telling you that the precondition for "single particle QM" is statistical independence of individual detection events and this is NOT guaranteed for every experiment. You actually have to test that simplifying assumption experimentally before you blindly apply QM.
OK, so let's say the statistical independence assumption holds, then we can build an ensemble theory, which mathematically leads to Kolmogorov's axioms. The main one there is that P(x) + P(x̄) =1 (the sum of the probabilities of an event set and its complement is one). This is a partition of unity and it can be fulfilled with the algebraic solution p + (1-p) =1 (that's ordinary probability theory) and with the geometric solution sin^2(phi)+cos^2(phi)=1 (that's quantum mechanics). The latter leads to the usual definition of vectors, scalar products and linear vector spaces and in the infinite dimensional limit this becomes the Hilbert space of non-relativistic QM. Translation symmetry of spacetime in those vector spaces leads to wave-like solutions FOR THE ENSEMBLE. And if nobody ever told you this before, then you need to get your student loans back, because you got a really crappy physics education. ;-)
4 18
There is little mistake at the end, there is a time operator
PARTH, I like your videos very much. Tristan Needham in VISUAL COMPLEX ANALYSIS, page 217 gives a definition of i, which I think is important for what you discuss in this video (please look it up). For a complex number z=x+iy, a change in x result in change in y, Tristan defines i as the ratio of change in y (effect) to the change in x (cause). I am sure this should give an insight into QM, as you refer to ih or i/h, implying cause is at right angle to effect. Empiricists thought cause and effect are independent of each other. Complex Analysis proves they are related by i. Does this give you an added insight into QM?
No, it does not. The reason for the use of complex numbers in QM is that position and momentum distributions are connected by Fourier transformation that in turn is a change of basis transformation with an infinite two dimensional basis. The Fourier transformation is also the model of fundamental uncertainty principle between position and momentum that determine the state of motion of a classical particle. Also, in QM those definite values of position and momentum become fuzzy distributions.
@@xjuhox What about z transformed to the w plane, doesn't eigen values get rid of the fuzziness?
@@sonarbangla8711 The unitary Fourier transformation transforms the same information of the state of motion between position and momentum basis. That is also the reason for uncertainty principle. The quantum fuzziness comes from the fact that we need to consider all the possible (typically infinite) eigenvalues of the operator (physical observable) in question.
@@xjuhox Does the Fourier transform lead to observable Dirac's delta function which takes care of fuzziness.
@@sonarbangla8711 There is a trade off with information. If you measure position with high accuracy, then the position distribution indeed becomes a delta function, BUT at the same time the momentum becomes super fuzzy due to the uncertainty principle.
Some tricky questions please.......
In what way does multiplying by -i rotate the wave function by 270 degrees on the complex plane?
Every point z = x+iy in the complex plane becomes -i*z = y - ix, using i^2=-1. If you picture that as coordinates (x,y), it's a 270 degree rotation about the origin, to (y,-x).
@@bentoomey15 Thanks, but I meant, does the wave function then look kind of like a sinusoidal wave rotated 270 degrees, in effect looking like the graph of x = sin y?
@@paulthompson9668 Ah, sorry. Yes, you are right, that's what happens.
To be a little extra, the wavefunction actually "fills" the entire complex plane, and the sine wave shown is just a slice of what the given wavefunction looks like along the real axis. That whole function gets rotated, and the slice that looks like sine is now on the imaginary axis.
The whole thing is really happening in 4 real dimensions (because the wavefunction outputs complex numbers), so the best we can do is simplified visualizations when things are real-valued.
@@bentoomey15 Arrgh, that was the *simplified* visualization?
Maybe I'm wrong, but I'm also confused about the eigenvalue equation (from 3 minutes) you say "we apply an operator to our wave function..." .But it appears to me that the equation shown multiplies a vector - a ket (not a wavefunction) by a Hermitian matrix (not an operator). The result is the same but the method isn't.
Nice
Actually it’s hermitian, meaning it only has real eigenvalues, that’s like the first thing a physics undergrad proves In their QM class. I know you’re probably gonna say this but still, this is clearly click bait to any physics undergrad.
8:05 it's -iħ(d/dx) but again in 8:30 it's -iħ(d/dt)....why are they different?
Actually it is -ih(d/dx) for momentum operator. 8:30 is a typo mistake
It's a typo
Why is Ψ something like exponential function in the first place?
It isn't. It can be an arbitrary normalized complex L2 function. We get exponential functions for states with constant momentum.
I understood that Hermicity needed it to be imaginary, so the eigenvalues would be real.
I really love the vedios but I need the math aspect ...please include the math ...
Marvelous…
A triumph of time management!
The wave equation I have seen is of the form e^i(kx-ωt) ,how does k become p and ω become E?
h bar. If it does not help, try a bar :)
Momentum can be written as
p = h-bar k
And energy
E = h-bar (omega)
Hope this helps!
Momentum operator is also just p the momentum
position op is partial derivative with p
For sure not.