Ch 8: Why is probability equal to amplitude squared? | Maths of Quantum Mechanics
ฝัง
- เผยแพร่เมื่อ 5 ต.ค. 2024
- Hello!
This is the eighth chapter in my series "Maths of Quantum Mechanics." In this episode, we'll dive into how we calculate probabilities in quantum mechanics. We'll derive the Born rule for calculating probability as amplitude squared, then put together our fundamental framework for how we mathematically model a quantum particle.
If you have any questions or comments, shoot me an email at:
quantumsensechannel@gmail.com
Thanks!
Animations:
All animations created by me within Python, using Manim. To learn more about Manim and to support the community, visit here:
Link: www.manim.comm...
Music:
--------------------------------------------------------------
♪ Galatea by HOME
Link : svnsetwaves.ba...
--------------------------------------------------------------
I honestly cannot be more grateful for this series. Huge props man! Congrats.
This channel has already seen massive growth, wouldn't be surprised if you reached 100k subs by the end of 2023!
Jeeesh, Your not lying. XD
I remember when it was around 2000-3000 subs only, and now it is 16000+. Those are some stonks right there. 📈
@@shadow15kryans23
A quantum leap...
(Incidentally, that's a stupidity of language that I hate. A quantum leap is the smallest possible difference. The leap, or rather the climb, to 100,00 viewers and up is not the smallest at all: Congratulations!)
But it’s almost the end of 2024, and it’s 64K. Quantum itself isn’t as popular as other lower level stuff. How many k doesn’t matter. This channel is a goldmine!
It is very interesting how this subject is imparted in different fields. For instance, I study ChemE and we skipped many steps in order to reach the Schrödinger equation in like two hours. Since our interest is to describe atomic orbitals, the rest of the semester was just that. I'm really satisfied with your explanation videos so far!
Indeed, It was also like that when I studied Chem Eng in college.
I study physics - it was actually quite similar for us. For us, the Schrödinger equation was basically just shown to us, the significance of it was explained, and we went from there.
I suppose since it can't be proven (it's an axiom of the theory) there isn't much point in getting hung up over the motivation behind it.
@@wallabooboo4082 Except that humans don't understand concepts through rote axioms and logic, so in order to actually fully learn something, you *need* the motivation
I am studying quantum computing right now, and your explanations about quantum state (particle) and observable (physical quantities) are so much better than any other textbook. I truly appreciate your year and a half of effort. Thank you so much!
I agree with your point that most textbooks present most of these properties as axioms. Studying through textbooks feels like I am just pushing through math, rather than understanding the true physics. Do you mind sharing how you went about studying this field and how you got these intuitions? Your videos help me a lot, but I also want to train myself to also gain these intuitions as well.
yes please! it would be so helpful if i could start getting the intuition for the underlying physics in quantum mechanics by myself, rather then having to look for explanations or videos online. i feel so lucky that this series exists, but it is (naturally) limited, and i would love to be able to expand on some specific topics/ideas in a more theoretical manner, instead of just mindlessly computing the mathematics.
This series has very high pedagogical value. It will likely become a must-watch for all budding physicists and anyone interested in this topic. Kudos to the author and much gratitude from everyone else!
This series is amazing! In three years of lectures I got only a very murky idea of why the maths of QM is the way it is. Keep up the good work!
Simply brilliant. This series clearly shows where the mathematics of quantum mechanics comes from, not only from mathematical treatment, but also from physical intuition. I'm speechless. Thank you and congratulations for these videos!
Today I just happened to land on this video series by chance. Now I see I have commented a year ago too. Let me say again this video series is the best. If you unveil the rigorous mathematical jargons/constructs, and a few bizarre quantum phenomena (like wave-particle duality, tunneling effect, and entanglement), QM is just as normal classical physics. Just digest (believe; take by faith) the few weird QM phenomena, and the rest is just School physics. But, only this sort of teaching is possible to convince that. Thank you again. Fantastic!
Smart, very cleaver~ it’s not my first time being amazed by how concrete the logic of math is and being amazed by people who come up with them! Nevertheless, being amazed by you for explaining it so intuitively and easy to understand. Thank you. Fabulous ~
Undoubtedly this series is the best. Who says QM is not intuitive? Congratulations
This is one of the most accessible treatments I have seen. I struggled with QM in the 80s and you make it seem so clear.
One thing I’ve picked up from your material, and others such as Binney from Oxford, is that teaching using Dirac notation is the way to go. When I was at Cambridge, the mandatory second year course made no mention of it, yet it’s such a huge aid to understanding. Only once students had specialised, in the third year, was the notation introduced, so most students would not be exposed to it.
One last thing, I’ve been trying to figure this out for some time, before seeing your course, and I wonder if you have anything you can say in reply.
The Hilbert space is spanned by the eigenstates of a particular observable. And bases have the same size. Ergo each observable should have the same number of eigenstates. Yet - and you presented this yourself in this video - position and momentum will often have continuous (uncountably infinitely many) eigenstates denoted by psi(x) or phi(p) and yet energy will often have discrete eigenstates (finite, or countably infinite) E1, E2, E3 etc.
But any basis of the Hilbert space should have the same cardinality of any other basis of that Hilbert space. So if we can find an uncountable basis of x eigenstates or p eigenstates, then surely there should also be an uncountable basis of E eigenstates. But there isn’t!!
Great series so far. Following with interest.
Hello! Thank you so much for the kind words, it does mean a lot.
In reference to your question about cardinality: someone below asked a very similar question. Like I said to them, I frankly don’t have the formal maths training to answer this with a lot of rigor, but I can give you some insight.
It seems to bother you that, for example, the energy basis and position basis are countable and uncountable, respectively. And reasonably so, it’s almost as if the position basis has “more states” than the energy basis, right? But infinity is a very tricky thing, and equating infinities is often a losing game.
As an example, let’s take a function f(x) that has a well defined Taylor series. f(x) is defined by an uncountable set of values for the input of real numbers. Now let’s Taylor expand it: f(x) can now be defined by each coefficient of the Taylor series, which is a countable set. So…we have that an uncountable set of values f(x) carries the same information as a countable set of coefficients. Yet, it’s the same function.
In truth, I haven’t thought too much about this. But I think the crux of the paradox is that we are talking about “well behaved” functions, ie functions with a Taylor series. This constraint is so strong, that you are able to condense the information of the function down to a discrete set. Again, just thoughts off the top of my brain, I would have to dig deep into the math to confirm this.
So in QM, the coefficient wavefunction psi(x) can’t be whatever it wants, it also has to be well behaved. Hence, I think this allows for a bridge between the uncountable and the countable, much like we found for Taylor series.
Hopefully this answered some of your questions!
-QuantumSense
@@quantumsensechannel I wonder it's on the same level as Hilbert Hotel's paradox. Even if tecnically we have infinities, their relations between themselves are different. But as i've seen informally on math courses, you can pretend that and infinite countable set can be strongly approximated to a subset of an uncountable set. Also, don't know the full argument, just believe on the premise.
Thank you, though I think something else is going on: from Wikipedia…
As a consequence of Zorn's lemma, every Hilbert space admits an orthonormal basis; furthermore, any two orthonormal bases of the same space have the same cardinality, called the Hilbert dimension of the space.[69] For instance, since l2(B) has an orthonormal basis indexed by B, its Hilbert dimension is the cardinality of B (which may be a finite integer, or a countable or uncountable cardinal number).
@@gavintillman1884 Strictly speaking, the problem is that the position and momentum operators do not have eigenstates. They are examples of operators that fall outside the basic framework presented in this series, and the mathematics for handling these is just different. Instead, you work with something called the "spectrum" of these operators, and this is facilitated by the spectral theorem in functional analysis. This is ignored in most introductory treatments of quantum mechanics, though, because you almost never actually care about operators with a continuous spectrum like the position or momentum operators in that introductory setting: all you care about is the Hamiltonian, and a select few operators, all of which most definitely have a discrete spectrum. For the vast majority of applications, this alone is good enough.
Thankfully things have changed in Cambridge. We mostly use Dirac notation in second-year quantum, and I am so grateful for it.
Thank you so much for these clear and intuitive introductory videos to quantum mechanics! Your focus on intuitive yet extensive descriptions of the mathematics behind quantum mechanics are not only enlightening but encouraging to someone who’s often seen the subject shrouded in mysticism. Again, this is some of the clearest work I’ve seen! Remain awesome! ;)
Amazing content, builds intuition to the subject the same way the essence of calculus playlist does, so grateful
This is a complex subject but your series makes it seems so intuitive
Fantastic series. Extremely clear and methodical and, in addition, the pacing of the narration allowed me to absorb what you were saying with out the next concept overlapping and clobbering it.
Really a spectacular series of quantum mechanics,with deep sense of mathematics.
It is just wonderful. Linking Born's rule and Gleason's theorem to probability of measuring an eigen value. Grateful.
Amazing video! This was the best one yet! Such a clear explanation of a really cool derivation :)
what a motivation! good job man ,best explanation about max born idea, great
Instantly subscribe, after I've just seen the first video of your playlist in my youtube feed. Love structurized courses, like yours. Hope you'll keep going!
This was the most mind-boggling of them all, thank you so much
Great video! I have no background in physics, but I absolutely love the beauty of this framework. Thank you for making it accessible to a layman like myself. I think the visual at 5:51 is a neat way to show how taking |c| as a probability gives rise to problems. It also gives an early insight into the eventual solution: as c1 and c2 are projections of a vector with length k on two orthogonal vectors, we know through Pythagoras that c1^2 + c2^2 = k^2, irrespective of the orthogonal vectors we choose. So, as long as we do not change k, c1^2 + c2^2 will be a constant for any observable and we can treat each squared projection as a probability (after standardizing).
Thank you so much for this! Over the years it has bothered me immensely that i could not prove the born rule, and now i can finally put this one to rest. The closest i got was realizing that taking the amplitude of the inner product was like summing the projections of a normalised vector on each orthonormal basis, but this sum of projections would change if i rotated my orthonormal basis (or equivalently rotated my vector, but the implication still is that the sum of probabilities of all mutually exclusive events would not always equal 1), and because of Pythagoras theorem i figured that summing the square of the projection of the vector onto each orthonormal baisis was the only way i could think of to always make the sum of probabilities equal 1 regardless of the orientation of the orthonormal baisis. However I did not know how to prove that this was the only way to do it, the Pythagoras stuff was only a backwards way of proving that the born rule satisfied normality, and although i did suspect it was the only way to have the sum of probabilities equal 1 irrespective of the orientation of the orthonormal baisis i did not know how to prove it or attack such a proof. I was told that the born rule was an assumption, and we didnt know why it worked, but I always suspected that there had to be a solid mathematical reason for it to be the only reasonable way to represent probabilities of outcomes when using the probability of outcomes of mutually exclusive events to describe states.
Well done. I am neither a physicist nor a (mathematician) but my passion and understanding to both leads me to keep learning and watching such channels/resources. And I must confess that your channel has simplified tremendously the complex view of quantum mechanics (to some extents physics) that I have never seen elsewhere. Most of the quantum concepts and theorems/axioms just fell from the sky as such. Basically there is no foundation from where they come.
It even looks like to me that quantum mechanics was "invented" by magicians or appeared out of the blue.
Congrats and keep it up.
I throw you a challenge though: one of the concepts that I could grasp from QM is that any particle can be at several places at the same time (just like I the double slit experiment). Could you pls go through this as I cannot go over this since it seems also to contradict both Classical / Relativistic mechanics.
In case I have misspelled, pls correct me.
Well done again! I shout out from Luanda - Angola 😅
Love this series, man
I'm just hooked with your this series ❤️🤩🤩 . Love your work 🌸 keep going 💖
Simply brilliant video! Bravo!
Based on the sigma/integral difference at 1:30, I am assuming that the energy eigenbasis is a countably infinite set, but the position (and momentum) eigenbasis is uncountably infinite (having the cardinality of the continuum). How can this be possible for the same single vector space?
The possible coefficient combinations (countably many coefficients) for the energy basis can be represented as a function from integers to reals. The possible coefficient combinations (uncountably many coefficients) for the position basis can be represented as a function from reals to reals.
Since this is a single vector space, this gives a one-to-one mapping between “functions from integers to reals” and “functions from reals to reals”. I don’t know how cardinal arithmetic works, but surely, such a mapping can’t exist, right?
Hello! Thank you for watching.
Although I don’t think I have the absolute mathematical rigor to fully answer your question, I can offer one insight.
You’re right that this seems impossible to map continuous to discrete, but note that this is something you’ve probably done without even noticing it. Specifically though Taylor series. Let’s look at the space of real functions with Taylor series (ie, well behaved functions). We can take such a function, and expand it in a Taylor series. Once we do that, note that all the information of the function is mapped into the Taylor coefficients, which are countable. So for every well behaved function, we have a map to a discrete set of coefficients (which is a function from integers to reals like you say).
So Taylor series represent the exact map that you claim can’t exist! Now the big stipulation here is “well behaved”. I don’t remember what the actual conditions are, but this well-behavedness is constraining enough that it allows us to map a continuous function down to a discrete set. And the wavefunctions in QM need to satisfy such well behavedness, which is where we can also get such a map from continuous to discrete.
Again, I lack the formal mathematical training to answer this question with more rigor, but hopefully you see that such an idea isn’t so foreign in mathematics.
-QuantumSense
I think the trick resides in the normalization of the continous functions like position and momentum, they are really uncountable, but have limits, position from 0 to L (not in infinity) and 0 to m*c for momentum (for massive particles). If, normalized, those quantitatives are brought to an interval from 0 to 1, and that specific interval is quote-unquote cardinally equivalent to the full integer set
@@linuxp00 Not true. Any interval of the real numbers, no matter how small, is uncountably infinite.
The original proof for the uncountability of the real numbers was performed on just the [0,1] interval.
@@Raphael4722 what I meant was that even if we treat the possibilities for position and momentum as discrete, from -infinity to +infinity. Usually, when dealing with finite intervals we apply some kind of squishing (normalization), from infinite discrete set of possibilities to a finite continous set of possibilities. The term from 0 to 1, is not literally that the length of the interval must be necessarily unitary, you can multiply both bounds by any number, like L, 0 stays fixed and 1 is resized to L.
I don't know how to thank you but have one from me. Never thought quantum could be so intuitive.
Given a norm, the length of a vector is the same under every basis, and since we must also satisfy that the probability adds to 1, assuming that length = 1, it seems convenient to define the probability function as ci*·ci such that it adds up to = ||ψ||^2 = 1.
Less formal, but perhaps easier to understand.
Truly excellent video!! This series is invaluable ❤ keep up the great work! ^^
This is cool! I got the idea, though I don’t think I’m able to do it myself. It’s convincing.
I owe you a lot . Your vedioes are the work of art . Thanks a lot
I have watched all of your videos and simply don't have words for how amazing they are... Magnificently clear explanation of ideas while emphasizing intuition. Many sources (Books and videos) gloss over and just list the postulates and throw ideas as if they have fallen from the sky. I really appreciate your incredible effort and wish you the very best for your channel.
I'm wondering where you acquire this physical intuition about 'Quantum Ideas'? I will appreciate any recommendation about books or some other sources emphasizing physical intuition... Thank you.
This one's a bit heavy compared to previous ones. I didn't manage to completely follow it, but I feel satisfaction in knowing that an explanation exists for "amplitude squared" and one day, there's a reasonable chance that I'll fully understand it. Meanwhile... onto ch9.
Thank you for explanation ..plz tell me why did you expand ket phi on two basis one is continous (for momentum) and the other in discrete basis ( for energy) and you said it is the same ket phi .. in other way can we associate the same ket phi to more than one observable
I just now realized that too I think that’s how the framework is designed in qm: The ket phi is your state vector, which can be represented as a linear combination of a certain observable’s eigenbasis depending on which observable is being described. When we want to look at momentum per se instead of position we switch to the momenta eigenbasis, but we still have the same state vector.
It’s just like if you have an arrow vector and you change basis the arrow is still the same you’re just describing it a different way.
Hope that makes sense, if anyone has an alternate interpretation or thinks I’m confused somewhere pls let me know - toodles ;)
This. Is. Just. Magnificent.
Oh my god that mathematical argument for the Born rule is gorgeous
Actually, if I'm not mistaken, the equation at 12:05 already implies that the function f _needs_ to be even, since the equation must be true for both positive and negative choices.
The calculus in the derivation of the Born rule is 🔥🔥🔥
In telecommunication engineering; this was a no-brainer. We use Fourier transforms to analyze waves to find their frequency response.
Frequency response is just another name for the probability of the wave to exist at that a giver or a particular frequency
That is why engineers use the terms power density and probability density interchangeably
This is the ONLY way to analyze single+ noise to design filters in DSP
Once I saw the derivation(s) for the Schrodinger wave equation; I was shocked.
Really interesting motivation for the Born rule, did you learn it like this in a lecture or did you come up with it yourself? Have never seen this before...
Hello! Thank you for watching.
I discovered this approach to deriving the Born rule, but I do not know if it is completely novel (there are lots of physicists out there, and maybe someone has done this before). I have seen some pdfs online that try something similar, but not in the way I did in the video.
-QuantumSense
Question around 1:30: how come the same quantum state can be expressed by both discrete superposition and continuous superposition? Former is infinite dimension but countable and latter is infinite dimension but uncountable. So is the dimension of the Hilbert space countable or uncountable?
This was brilliant. Filled a void in my mind. After seeing it I wonder if it has to do with measure theory. It may have a connection.
At first blush, it would have seemed easier to first try the special case when we have just two variables c_1 and c_2; but we would then have been prevented from using separation of variables because we would have had just one independent variable and so been stuck with the equation 1 over c_1 times f'(c_1) equal 1 over |c_2| times f'(c_2), with |c_2| equal to sqrt(k^2 - c_1^2). So there would have been no possibility of playing on independence to get zero on the right-hand side when differentiating with respect to c_1.
This shows it takes some experience with differential equations to know when heuristic simplification might lose valuable information.
I thought of a picture that makes the Born Rule kind of easy to see, building off this video. I'll try to describe it to get it clear in my head (and because I think it may be a useful picture for others also), but anyone let me know if I missed anything.
Like explained here, you start with constructing every possible measurement (like x in space) as an eigenbasis in Hilbert space, the value being the likelihood of being found there. Then the quantum state vector is a size 1 vector that meanders in that space, rotating about the origin. We can start by imagining it perfectly overlapping one eigenbasis (A), so the probability is 1 (100%), which means the measurement will definitely find it there. But as we move the state vector away from A in a 2D plane towards only one other eigenbasis B, like the "next point over" , the amplitude is the projection of the state vector on to the 2 axes. (Tricky to imagine for a continuous set, but I still think position is more concrete to imagine as to "moving one point over", and the bit of fudging doesn't change the basic point; but if one doesn't like it, just substitute a discrete category.) As it takes amplitude from A, it gives it to B, but not where it adds up to "1". Because the state vector is always size 1, and A and B are orthoginal, the state vector will move in a circular arc from A to B, and the projection will be cos(angle of the state vector to A or B). Now we can check this at any angle, but to start, if the state vector is exactly halfway, then the projection will be cos(45 deg) to both A and B, which is 1/sqrt(2) = 0.7.... If you move it 30 from A, then the projections are cos(30d) = sqrt(3)/2 = 0.8... for A and cos(60d) = 1/2 for B.
But this isn't probability yet, because again adding them all up doesn't add up to 1, but it's over. If you re-scaled the values so they added up to one, the projection lengths would get smaller, and it turns out to be the square that exactly does that scaling. But there's an easier way to picture this as well. What really makes the space shared between A & B relate to probability is each one's value relative to the other (and all the other vectors, but keep them 0 for now). What's important for probability isn't the absolute value of A & B. What's important is the size of A relative to B. And on that, e.g., for the halfway point where the abs value is cos(45 deg), A and B are exactly the same size. So each one is half the size of the total length (A+B). Then it's straightforward to see that the size of one (1/sqrt(2)) relates to 1/2 probability by just squaring it. You have to show a little more to show that that situation is going to still hold as you move the state vector towards B degree by degree, but I think it's easy (or easier) to see from this picture now. Well that's how I've pictured the source of Born's Rule in an intuitive way.
17:39 I mean, we assumed that the quantum states should be absolute, and shouldn't depend on the choice of the observation, because that's more aligned with our idea of a physical particle, but this is an assumption (of particles being real, not depending on how it's observed) and it might be argued that there is no absolute length of our quantum state prior to the measurement.
So, it kind of makes sense that this is stated as an axiom, as it comes from our physical perception of the world. (If how we've derived is the only way to do it, this is the first time I've seen this)
I've just researched a bit about this and this assumption is called non-contextuality. And I've learnt that Gleason's theorem implies that non-contextual hidden variables can't exist, and contextual hidden variables are not desirable by hidden variable theorists.
This part is getting philosophical real quick :D For anyone that's interested, that's kind of the topic of Nobel Physics Price 2022, and it was explained beautifully by PBS Space Time's videos.
man.... Thank you!!
Absolutely beautiful.
I do have a small criticism regarding the form, i.e. the use of LaTeX.
1. When using labels it's preferable to write them as normal script rather than the italic one used for maths. So instead of $p_{total}$, one should use $p_{\text{total}}$.
2. For differentials it's common to use normal script to, i.e. the d should be upright. This can be easily achieved using either the physics package, or the commath package. In the latter case, $\dif x$ will produce a properly formatted dx, while for ordinary and partial derivatives one can use $\od[n]{x}{f}$ and $\pd[n]{x}{f}$, respectively, where n is an optional argument for the order of the derivative, x is the variable to derive by and f the value to derive. Thus, for example, $\od[5]{t}{p(t)}$ would typeset the 5th derivative of the function p(t).
Small question regarding the motivation for the probability, it seems like we are reaching the desired square value because we want to make the norm of the wave function invariant. And that's fine, but it's not clear why we want that, it doesn't seem like the norm corresponds to a anything physical at all (in fact it seems to matter so little many courses I've seen normalize wave functions implicitly without mentioning it). So it seems strange to have that as a foundational assumption when it isn't really justified by the physics.
Great video, thanks 😊
Are there books that explain this series (ie. Math behind Qm)?
Best video by far
I have a question. I didn't quite understand how exactly different definite states are different bases of the same hilbert space - does that mean that one can express momentum as a linear combination of energy vectors? And how can they even be bases if they have different dimensions? How can the Hilbert space contain both an energy basis, which is discrete and might even be finite, as well as a position basis, which is continuous?
thank you very much....
I stopped the video to try and guess why myself, and I arrived at:
For multiple reasons, we want unitary transformations to maintain a probability of 1 if we started there.
Because of the definition of a unitary transformation (one that preserves norm or equivalently inner product), this leads us naturally to the idea that the norm of a vector might be its probability.
For convenience, we want the probability function to distribute over addition.
For example, if we have
|x> = a|0> + b|1>
Then we want
P(|x>) = P(a|0>) + P(b|1>)
However, the norm doesn't do that.
|| |x> || = sqrt( || a|0> ||^2 + || b|1> ||^2 )
But a simple variation on it does (if we're working with an orthogonal basis).
|| |x> ||^2 = || a|0> ||^2 + || b|1> ||^2
So, we get that the probability of a vector is equal to the norm of that vector squared.
P(|x>) = || |x> ||^2
The vectors that make up an orthonormal basis are unit vectors, so the scalar out front entirely determines their norm.
P(a|0>) = P(|a|^2)
thanks for the derivation
When you do get to the Schrodinger's eqn, could you also brush on the master eqn, i.e the Linbladian? Much appreciated!
Can you have detailed examples for lecture 7 and 8
Very well done.
Brilliant ❤
Hi, i’m trying to show the expetation value formula for the quantum state, and i have two questions:
-I do not understand when the absolute value of the coefficient appears
-Every operator that gives us the mesure of a quantum state is diagonalizable?
Thanks
My god, it's beautiful
I have a question about the proof of f(c_i). Why do we take n-constants c_i? Shouldn't there be infinite ones?
Hi I'm having trouble derivating the problem shown at 20:13
Should the operator E(hat) be written as ΣEi|Ei⟩⟨Ei| ?
This is the only expression I could think of, I know Σ|Ei⟩⟨Ei| is the identity operator, but can any operator other than the identity operator just be written as the eigen value times the identity operator eg. ΣEi|Ei⟩⟨Ei|？
1:30 Here is one thing has always bothered me in this subject. The dimension of {E} should be strictly larger than {x} because there are countably many energy eigenstates but uncountably many position eigenstates, so how can both bases span the same vector space when one is bigger than the other?
Never heard of the Gleason's theorem 👌
you are the best
would you upload the lecture notes as well? it would be a great help!
If we would be able to measure the outcome of every possible existing particle would quantum mechanics then be deterministic?
amazing
How can one vector be represented in various bases which have different units? Shouldn't the basis have the same unit as the vector?
Anyone know how to do that last expansion with the expected value the way he intedned?
Never mind I just figured it out. The trick is to realise that if you know what a vector's representation is in the basis of the eigen vectors of a matrix (yeah I know it's actually an abstract operator), then you can just multiply the appropriate eigen values to each eigenvector coefficient to get the result. It's hard to explain it in a youtube comment.
The Ei in the sum he shows are just the eigen values of the energy operator. The pi is just |ci|^2, or amplitude squared.
good job
question - at 13:35, aren’t the right hand sides of the two equations not necessarily equal, since f’ on the top refers to the derivative w/ respect to c1, and f’ on the bottom refers to the derivative w/ respect to c2? what am I missing?
Hello! Thank you for watching.
And ah, this is a good question. Remember that f is really only a function of a single variable, I think that’s what makes it confusing. So the term that you’re talking about shows up because we are using the chain rule. So what does the “prime” mean there? It’s the derivative of f (and I can say THE derivative, because remember f is a function of a single variable, so f’(c) is the only derivative we can take), and then plug in sqrt(k^2 -…) as the argument. I mean, you could replace c with c_1 or c_2, but it doesn’t really matter, because you’re evaluating the derivative at the same point sqrt(k^2-…).
So in essence, df(c_1)\dc_1 evaluated at sqrt(k^2…) and df(c_2)/dc_2 evaluated at sqet(k^2…) would give you the exact same thing, it’s just using a different label. These different variable labels only matter in the context of the total constraint equation. Let me know if this didn’t clear it up.
-QuantumSense
Can anyone help me out with the exercise at the end? I know applying the energy operator the the psi ket should give sum(c_i E_i). But how does applying the psi bra give the probability? From what I got from the video the square of the coefficient should be the probability, but how does the psi bra square the coefficient?
I need some help too
1:35 Woahhhhhh! You are claiming that an Eigenbasis based on momentum is eqivalent to one based on position. That's one hell of a claim. I had thought that you were implying that depending upon the observable you could define a basis and hence a vector space ... not that all vector spaces were the same one. There seems to be a missing lense here. The 'lense' being assosciated with the measurable in question. This is deeply wrong. That apart and in the sense of glasnost, the rest of your presentation is quite exiting. 🙂 (21:23 is what I found missing at the begining)
Very elegant
Phenomenal
Seems to me, that despite your explanation, the Born Rule can be assumed simply because it makes the result both real and positive.
I.e., why can you not use the same methods to show the Born Rule to be |x|^k where k is positive and even?
Hello, thank you for watching.
Note that the absolute magnitude is a rule that makes the probability function real and positive, yet it gave us the problem with the changing lengths. And in fact, any abs(x)^k would give you changing lengths, except for k=2. You can try this out. Plug in k=4, and check if the lengths stay the same after changing basis, you’ll find that they don’t. So if we assume that we have a constant length in any basis, that constrains the power to be k=2 (which was the essence of our proof). Let me know if this didn’t clear it up.
-QuantumSense
@@quantumsensechannelthanks for your reply, I just found your channel and have already watched the first 10 eps! Very nice.
I was just confused where the ^2 came from at 7:50 because you seemed to assume it conserved length, then go on to show that it does, which lacks uniqueness. I remember you said there’s a deeper formal proof, and that you were just giving intuition, which is fair.
Looking back it seems to be the result of the Pythagorean theorem, and not much else?
Hello,
At 7:50, I have merely written down the condition that length must be the same in any basis. c_i^2 just comes from the length of our vector, it has nothing to do with the probability function. It’s just the Pythagorean theorem as you say. This is a condition that we are stating must be true, but it is not a statement on f(c), yet. We then used this condition to *prove* that the only function that satisfies this is the square function.
So we have the following logical outline: if we have a probability function that conserves the length, we proved that it must be the square function. We are not assuming that the square function conserves the length, we are instead proving it. Let me know if this still does not clear it up.
-QuantumSense
@@quantumsensechannel I didn't follow the proof all the way but I believe you. I only wonder why then certain views like QBism take the Born rule to be axiomatic, and try to derive as much of QM from that -- that is, if it's so easy to derive from within.
Hello,
There’s a few reasons for this. One is that we’re replacing one axiom with another: instead of having the born rule as an axiom, we now have the axiom that the quantum state must have the same length in any basis. Which axiom is better? It’s a matter of opinion.
But a more significant reason is one I mentioned in the video: my proof (as well as Gleason’s theorem) are only valid in dimensions of 3 and greater. Yet, the born rule still applies in 2D quantum vector spaces (think about spin 1/2 vector spaces where the states are spin up and spin down - we still have the born rule!). So we would still need to axiomatically declare that the born rule applies in any dimensional vector space, regardless of our proof.
So I’m not trying to replace the Born rule as an axiom, I just tried to show why we should believe that it is a very good axiom to have (and why it is somewhat intuitive).
-QuantumSense
How can we confidently say that derivatives of f(c) exist? It might be the case that it is not differentiable, and not even continuous I guess.
Its physics, everything is continuous and differentiable /j
OK... cool I think I get it... the chain rule thing was a bit tricky I thought there would be a 2 somewhere but it turns out the 1/2 gets canceled with a 2 to become unity... I got google to clarify with an example... but at th-cam.com/video/PZUZgOUOOIU/w-d-xo.html there's an f'() term that is kept in the equation, which seems like a double representation? don't we just need the term next to it? It doesn't really matter since we are only concerned with showing how the left side equates to the same thing regardless of the ci... TTYL...
Wait, but if the range depends on the previous range theyre not actually independent? Like, i get physics plays it fast and loose but this seems too fast and loose, the ranges are monotonic descreasing surely there should be a second DE to represent that?
Intuition says thats also why it still works in 2D too, because theyre still monotonic decreasing in width. Hmm.
Isnt quantum collapse basically just a result of the random walk filtration being fixed? Like, if the wave function is a high dimensional state space and time is a sigma algebra on that state space, then observation fixes you in one of the subbranches of the sigma algebra? Idk ive been doing a bunch of stochastic process stuff maybe im confused
And then the reason we dont see isolated particles wandering from their prepared state over time would be because they would have some kind of zero expectation over time, aka they'd be be martingales. I think. Maybe. Hmm
How can we generalize this argument for basis of infinite dimension?
I think it can't be infinite. That's the reason the vector is in the Hilbert space. You can refer to his initial videos on Hilbert space.
How did we find the coefficients in the angular momentum basis? 6:23
Hello, thank you for watching.
And we would do it in the same way we would in any other orthonormal basis, you would take the inner product with the corresponding basis vectors. So c_1 = (which would just be the dot product in R2), and so on and so forth. I didn’t tell you what the basis vectors were explicitly, but I can give you what they are if you’re interested, and you’ll see that it gives you the values you see in the video. Let me know if this didn’t clear it up.
-QuantumSense
Hi, yes it’d be very helpful in understanding if you could provide the exact values of the basis vectors
Hello,
The L basis vectors are 1/sqrt(5) [1,2] and 1/sqrt(5) [-2,1]. Let me know if there’s anything else that is unclear.
-QuantumSense
@@quantumsensechannel Understood it now, thanks!
PS: Love your channel. This series has been super informative.
Can someone please explain to me, how did we get to: braket(psi) = intg(dx.psi(x).bracket(x))??? Because last I remember from the previous videos, it was: bracket(psi) = intg(dx.c(x),bracket(x)).
I'm getting really frustrated with this point; someone please help
that c(x) is actually psi(x). It was just to generalize
Just here to boost the algorithm
Hi everyone. I am a highschool school student and I have stuck since 12:45. Can anyone please explain it to me? Thanks a lot!
12:58 can someone explain how the chain rule is applied here?
chain rule: ( f(g(x)) )' = f'(g(x) * g'(x).
You can see f'(...) at the left and the right part is the derivative of the square root with respect to c, which is g'(x) in this case. He applied the chain rule twice, the second time being to find the derivative of √(k^2-c^2), but he just wrote the result, if you need the full steps there are derivative calculators on the internet for free that show the steps too.
I'll check that out, thanks! @@ChrisK-ef8np
Oh! Going for a PhD? Congratulations!
I just got to MSc..
Oh yeah, the math when we are physicists, not mathematicians, Jim! 🖖🤓🤝
I saw some of the comments and the details for these particular Hilbert spaces need more maths. Heh.
Going more for particle lab approach, it anyways has a constraint as you pointed out, maximum energy and that's the H=1
And of course, we have a multi-field model with deltas, and we use Fourier Transformer more than meets the eye, and we are going from one space to another, that's a way to see this. So, one of the most important things is that of putting baseline at least from 2nd harmonic. There is. You did the expansion and taking the derivatives for c1 and c2.
So partial differential system.
Then there's the Fast Fourier transformation, the delta. Everything is 0 except that one component. From, let's say a linear momentum measurable we get our eigenvalue, right?
Ah, you just said "homework" and there's the expected values and everything.
Now it's a question of getting some data to get the distribution.
And it's finally coming, there's what will get us going, the second H of the process of building out observable. Hilbert Hermite Hamilton.
Anyways, lab has more "doctor who's hitting the tardis console with a hammer to make her BEHAVE.
Ike if the observable doesn't work we just tweak the parameters and after getting the results, we reverse engineer a new expression for it, with the new parameters. Normally it's just a little tweak, but not always. I mean, there could be a huge mistake but the observable fits. Of course, it's not like it will work just for that feature of the q Framework 🤓🖖🤝
Love love love from pakistan
Nature often shows the physicist what the rules should be. Mathematicians then follow with a rigorous proof. God is the supreme mathematician.
=1
stop allowing ads, Google puts them with no time limits and so we cannot watch!
Amazing video! So from the 12 minute mark, you're using implicit function theorem - correct? Basically c_n = g(c_1, ..., c_(n-1)).
And F(c_1, ..., c_(n-1)) = f( c_1, ..., c_(n-1), g(c_1, ..., c_(n-1)) ). So ∂F/∂c_i = ∂f/∂c_i + ∂f/∂g . ∂g/∂c_i
Just trying to double-check my understanding..