In order to remember whether has the plus or minus in the exponential, think of a video game where you collect experience points (xp) -- their value only goes up, therefore has the plus sign and its conjugate has the minus sign :D
@@zokalyx the difference is, nobody uses the left hand rule, and I know for a fact that some people define p as i hbar d/dx in coordinate space. I think knowing where the sign comes from is more important when you actually talk about QM with somebody, or read an article.
@@zoltankurti Usually, p^mu = i ħ d^mu (1) (upper indices!), so that when mu=1,2,3, we get, px = - i ħ d/dx py = - i ħ d/dy (2) pz = - i ħ d/dz because, d^mu = (d/dt, -d/dx, -d/dy, -d/dz) (3) d_mu = (d/dt, d/dx, d/dy, d/dz) and therefore the minus sign in eq. (2). To be honest, I’ve never seen eq. (1) with the opposite sign. Reference: en.wikipedia.org/wiki/Four-gradient#As_a_component_of_the_Schr%C3%B6dinger_relations_in_quantum_mechanics
Just stumbled onto your channel and I can confidently say don't understand most of it (ive only taken up to calc2 so far) but you're super funny and very entertaining :)
You don't know how much you have influenced and inspired me to dream of a career in Physics!The videos on how you study for your Finals or how to be good at Math,the books on classical mech,Quantum mech and Math methods you told in your videos and gave the pdf link in the description-I have downloaded all of them and am trying to get through them(btw I am in the 12th grade).I love them a lot!Very helpful,especially the Griffith's Electrodynamics,John R.Taylor's Classical Mechanics.I have started with Shankar's Principal of Quantum Mechanics and the book on Quarks and Leptons.Just like you, I am a FAN of Quantum theory and stuff and dream to be a Quantum Physicist someday(that's exaggeration,I know).Your psychology towards Studying Physics and Math have encouraged me to think the same way you do.I want to lead a life just like you.Just wanted to say a BIG THANK YOU to you for influencing my life and changing it from a sluggish,monotonous,shitty one to a life full of Potential energy and encouraging me to convert it to the kinetic one.Wish I could meet you some day after I have built my dream carrier! One thing I forgot to mention!Never Stop making this awesome useful videos!I don't want to lose contact with such a mind boggling Physicist in this HUGE world.Keep up this great job!Greetings!Can you make an educational series on Quantum Physics for noobs?That would be a great help!
Really appreciate the nice comment! Maybe in the future I can make those types of videos. I have too many unfinished series as it is right now, though.
My feelings exactly, these videos make me feel I'm actually involved with someone teaching a topic, not just watching a show for entertainment, which is good, but different.
Love your videos, youre such a good resource for someone like me who wants to go into phhsics. You present us with the real life of a physics student, and I myself take it as a good sign that i dont get bored or scared by your videos but rather i bevome more intrigued by the field. Keep doing what you do.
I follow another approach. As you probably know the momentum operator is defined as the generator of space translation. that means =exp(-i/hxp)exp(i/hxp)=exp(i/hxp)=exp(i/hxp)
There is a more straightforward proof by rewriting the quantum state in terms of position(or momentum) coordinates by inner products formula. And then rewrite the quantum state in the inner product as momentum(or position) specification, and finally compare to the Fourier transform formula to get the result. The proof is much simpler.
Cool video, very understandable. just one thing, What did < phi | X^ | psi> mean? Is it the value of the operator when the system goes from state psi to phi? Thanks
It's just what it says, the inner product of |phi> with X|psi>. What you're talking about sounds like if instead of X you had the time evolution operator U. However, you have to be careful when thinking about system "going" from psi to phi. What people actually mean when they say this is not that psi changes into phi, but that it develops non-zero amplitude along phi. I would suggest looking at a textbook. Shankar is very good, and will help you understand this stuff. Good luck!
you have no fucking idea how relevant this video was too me for my quantum mechanics final. I actually understood this, except there's one thing I don't understand. I have a very poor background in fourier theory. Could you possibly make a video explaining why psi(x) and phi(p) = those integrals and how you can derive them without Fourier transforms unless that's the only way???
BeyondBelief basically momentum is related to the wavenumber which is found by finding the modes which make up a wavefunction. To find these modes we take a Fourier transform. To find why a Fourier transform tells you which modes make up a function look it up but that’s the point of it.
You've assumed the answer in the Fourier transform that you've boxed, and one can show that in one line! Take "Psi" to be a momentum eigenstate itself with momentum "p". Its wave functions in the position and momentum basis are, by definition, Psi(x) = and Psi(p') = delta(p' - p). Plug these into the first Fourier relation you've assumed: = 1/sqrt(2pi) * integral delta(p' - p) e^(ip'x) dp' = 1/sqrt(2pi) * e^(ipx) And there you go.
Yes, I realized my original comment was *also more complicated than it needed to be, so I deleted it. Sorry! The two resources I mentioned were Modern Quantum Mechanics by Sakurai and these notes by Littlejohn bohr.physics.berkeley.edu/classes/221/1819/221.html. I prefer the notes. I don't know if these sources are at the appropriate level for a first exposure. I've heard great things about Townsend's book on QM for the undergrad level, but have never used it. My friend swears it's as good as Sakurai but still accessible to a beginner. I mentioned the notes because they discuss the momentum operator in what is, imo, the most "fundamental" way: the generator of translations. Any treatment of questions like what is should start by asking what the momentum operator actually is to avoid circular reasoning like in this video. I assume there are many other ways of justifying things, and ultimately, because QM is mathematically consistent, you can take many points as the "starting" point. Happy learning!
I see what you're saying, but I don't think the argument is circular, but it is not a derivation. I'm not saying we're proving the wave functions in different representations are fourier transforms of eachother by assuming that they're fourier transforms of eachother. I'm saying that under the assumption that they are, we can extract what the inner product must be.
@@AndrewDotsonvideos Yeah I get that, maybe circular isn't the right word. I'm saying your argument is overly complicated and basically just amounts to saying is what it is. Here's an even simpler way to make the point, just multiply by the identity! |psi> = integral dp |p> take the inner product with |x> = integral dp Compare this last equation with the first Fourier relation you use, and you have the answer directly. Based on this argument, you can see you're actually just saying what is when you write down the Fourier relationship. This doesn't illuminate at all why* is what it is, though. Please keep the vids coming. I like them a lot even though I don't like your approach in this one.
Great video on a great topic! I think you got your logic a bit backwards. Don't get me wrong, all of what you said is completeley right. Basically, if I were teaching QM, I would introduce this topic differently. First, I would fimd the eigenfunctions of x and p in coordinate space. Than talk a bit about representing abstract vectors, and finally trying to change the basis to momentumspace. This way, calculating would just be an integral in coordinate space, and it should be already clear that it will be a basis transform coefficient. I guess that integral is the easy way you talked about, but I know tough guys don't do those... :D Clearly.
I see exactly what you're saying. The reference about these being transformation coefficients was mostly a wink and nod to those who really understand what those are. I think already assuming the wave functions are fourier conjugates to eachother might have been a pretty big assumption in the first place too, but it's also what people are probably already familiar with.
@@AndrewDotsonvideos |psi> = int dp |p> ---> = int dp , so if you assume that the wave functions are related by Fourier transform, you must have = 1/sqrt(2pi) * e^(ipx). This shows that the assumption is just re-stating what is in a more complicated way. You haven't shown anything. This entire video is circular and misleading.
Could you have done this when you finished your undergrad or this is sth that you learned in your graduate course? I'm currently taking QM I now and I'm starting to worry since this looks hard for me lol
A lot of undergrad courses in quantum introduce dirac notation pretty late on. Definitely something an undergrad can handle, but you have to be fluent in diracs bra-ket notation first. Don't worry, it wasn't my strongest point in undergrad!
It's basically saying, the state |p> has a well-defined momentum. Then, basically gives you a wave function for a free particle which has a well defined momentum (a plane wave)
@@GBY13 I believe you can. You can say, the state |x> has a well defined position, so gives you the wave function which has a well defined position in momentum space. So, taking , you can either think of it as a wave function in coordinate space with defined momentum, or of complex conjugated wave function in momentum space with defined position. It's interesting that those 2 things are mathematically the same. Also, you really can qualitatively see Heisenberg's uncertainty principle from this. Since gives you a plane wave, By considering this as a wave function with defined momentum, you get completely undefined position (a plane wave), and by considering this as a wave function with defined position, momentum is, again, completely undefined (aka in momentum space, wave function is a plane wave).
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In order to remember whether has the plus or minus in the exponential, think of a video game where you collect experience points (xp) -- their value only goes up, therefore has the plus sign and its conjugate has the minus sign :D
I'm impressed
I think it's better not to memorise it this way, the sign is just a convention depending on wheter the momentum operator is written with +i or -i.
@@zoltankurti it is good to memorize even if it's a convention. like the right hand rule.
@@zokalyx the difference is, nobody uses the left hand rule, and I know for a fact that some people define p as i hbar d/dx in coordinate space. I think knowing where the sign comes from is more important when you actually talk about QM with somebody, or read an article.
@@zoltankurti Usually,
p^mu = i ħ d^mu (1)
(upper indices!), so that when mu=1,2,3, we get,
px = - i ħ d/dx
py = - i ħ d/dy (2)
pz = - i ħ d/dz
because,
d^mu = (d/dt, -d/dx, -d/dy, -d/dz) (3)
d_mu = (d/dt, d/dx, d/dy, d/dz)
and therefore the minus sign in eq. (2). To be honest, I’ve never seen eq. (1) with the opposite sign.
Reference: en.wikipedia.org/wiki/Four-gradient#As_a_component_of_the_Schr%C3%B6dinger_relations_in_quantum_mechanics
So glad I found this video 6 months after I needed it.
Just stumbled onto your channel and I can confidently say don't understand most of it (ive only taken up to calc2 so far) but you're super funny and very entertaining :)
Thanks a lot!
You don't know how much you have influenced and inspired me to dream of a career in Physics!The videos on how you study for your Finals or how to be good at Math,the books on classical mech,Quantum mech and Math methods you told in your videos and gave the pdf link in the description-I have downloaded all of them and am trying to get through them(btw I am in the 12th grade).I love them a lot!Very helpful,especially the Griffith's Electrodynamics,John R.Taylor's Classical Mechanics.I have started with Shankar's Principal of Quantum Mechanics and the book on Quarks and Leptons.Just like you, I am a FAN of Quantum theory and stuff and dream to be a Quantum Physicist someday(that's exaggeration,I know).Your psychology towards Studying Physics and Math have encouraged me to think the same way you do.I want to lead a life just like you.Just wanted to say a BIG THANK YOU to you for influencing my life and changing it from a sluggish,monotonous,shitty one to a life full of Potential energy and encouraging me to convert it to the kinetic one.Wish I could meet you some day after I have built my dream carrier!
One thing I forgot to mention!Never Stop making this awesome useful videos!I don't want to lose contact with such a mind boggling Physicist in this HUGE world.Keep up this great job!Greetings!Can you make an educational series on Quantum Physics for noobs?That would be a great help!
Really appreciate the nice comment! Maybe in the future I can make those types of videos. I have too many unfinished series as it is right now, though.
My feelings exactly, these videos make me feel I'm actually involved with someone teaching a topic, not just watching a show for entertainment, which is good, but different.
This was exactly what I needed when I needed it. Thank you.
Love your videos, youre such a good resource for someone like me who wants to go into phhsics. You present us with the real life of a physics student, and I myself take it as a good sign that i dont get bored or scared by your videos but rather i bevome more intrigued by the field.
Keep doing what you do.
u so noice ty
I just put this in a playlist so that I don't forget where to find this video when i'll inevitably forget where the expression comes from again
Kazoo kid is always a welcome sight, though a bit unexpected here! :'D
Great video!!
watching this before my graduate quantum final. THANK YOU
Thanks for the video. It was nice to know the significance isn't obvious. It wasn't to me anyway.
I will simply say: Thank you!
Very nice presentation! Thank you for that!
You got a new subscriber! =D
Omg I always saw those in my stats class but didn't really know what they were and now I know! :D
Kenn Beary in stats the brackets probably refer to the mean expectation value
I follow another approach. As you probably know the momentum operator is defined as the generator of space translation.
that means =exp(-i/hxp)exp(i/hxp)=exp(i/hxp)=exp(i/hxp)
wow. perfect explanation!
amazing video i like it so much thx from north korea
Nice explanation, thank you 😊😊
Just great
Great explanation!!
There is a more straightforward proof by rewriting the quantum state in terms of position(or momentum) coordinates by inner products formula. And then rewrite the quantum state in the inner product as momentum(or position) specification, and finally compare to the Fourier transform formula to get the result. The proof is much simpler.
Thanks..it really helps!
Thank you for the memes.
Dope. This will help me.
Wow! Thank You soo much..
Cool video, very understandable. just one thing, What did < phi | X^ | psi> mean? Is it the value of the operator when the system goes from state psi to phi?
Thanks
It's just what it says, the inner product of |phi> with X|psi>. What you're talking about sounds like if instead of X you had the time evolution operator U. However, you have to be careful when thinking about system "going" from psi to phi. What people actually mean when they say this is not that psi changes into phi, but that it develops non-zero amplitude along phi. I would suggest looking at a textbook. Shankar is very good, and will help you understand this stuff. Good luck!
you threatened those states with a hermitian dagger at the start lmao
Thank you for this since Sakurai won't tell me nicely how to do this
Amazing
you have no fucking idea how relevant this video was too me for my quantum mechanics final. I actually understood this, except there's one thing I don't understand. I have a very poor background in fourier theory. Could you possibly make a video explaining why psi(x) and phi(p) = those integrals and how you can derive them without Fourier transforms unless that's the only way???
It comes directly from Fourier transforms with momentum and position related because of the debroglie wavelength
BeyondBelief basically momentum is related to the wavenumber which is found by finding the modes which make up a wavefunction. To find these modes we take a Fourier transform. To find why a Fourier transform tells you which modes make up a function look it up but that’s the point of it.
I dont know any of this yet but for starters....square matrix? Completeness? Coord vs Work space?
I could use a little convergence
You've assumed the answer in the Fourier transform that you've boxed, and one can show that in one line! Take "Psi" to be a momentum eigenstate itself with momentum "p". Its wave functions in the position and momentum basis are, by definition, Psi(x) = and Psi(p') = delta(p' - p). Plug these into the first Fourier relation you've assumed:
= 1/sqrt(2pi) * integral delta(p' - p) e^(ip'x) dp' = 1/sqrt(2pi) * e^(ipx)
And there you go.
Hey! :) Thanks again haha. You mentioned a book in your last comment, but I was unable to find your old comment, could you tell me? ^^Thanks!
Yes, I realized my original comment was *also more complicated than it needed to be, so I deleted it. Sorry! The two resources I mentioned were Modern Quantum Mechanics by Sakurai and these notes by Littlejohn bohr.physics.berkeley.edu/classes/221/1819/221.html. I prefer the notes. I don't know if these sources are at the appropriate level for a first exposure. I've heard great things about Townsend's book on QM for the undergrad level, but have never used it. My friend swears it's as good as Sakurai but still accessible to a beginner.
I mentioned the notes because they discuss the momentum operator in what is, imo, the most "fundamental" way: the generator of translations. Any treatment of questions like what is should start by asking what the momentum operator actually is to avoid circular reasoning like in this video. I assume there are many other ways of justifying things, and ultimately, because QM is mathematically consistent, you can take many points as the "starting" point.
Happy learning!
I see what you're saying, but I don't think the argument is circular, but it is not a derivation. I'm not saying we're proving the wave functions in different representations are fourier transforms of eachother by assuming that they're fourier transforms of eachother. I'm saying that under the assumption that they are, we can extract what the inner product must be.
@@AndrewDotsonvideos Yeah I get that, maybe circular isn't the right word. I'm saying your argument is overly complicated and basically just amounts to saying is what it is.
Here's an even simpler way to make the point, just multiply by the identity!
|psi> = integral dp |p>
take the inner product with |x>
= integral dp
Compare this last equation with the first Fourier relation you use, and you have the answer directly.
Based on this argument, you can see you're actually just saying what is when you write down the Fourier relationship. This doesn't illuminate at all why* is what it is, though.
Please keep the vids coming. I like them a lot even though I don't like your approach in this one.
this is great
That dagger!
Lol!!
thank u
i need more completeness in my life
the knife tho😂😂😂😂
Why am I watching this? I just took my quantum final I need no more of this lol
Now explain it like you were talking to a 5 year old pls
I thought that's what I was doing
Yes i also need the basics but ilike his teaching style he made the tough guy talk.
@@AndrewDotsonvideos :D
Great video on a great topic! I think you got your logic a bit backwards. Don't get me wrong, all of what you said is completeley right. Basically, if I were teaching QM, I would introduce this topic differently. First, I would fimd the eigenfunctions of x and p in coordinate space. Than talk a bit about representing abstract vectors, and finally trying to change the basis to momentumspace. This way, calculating would just be an integral in coordinate space, and it should be already clear that it will be a basis transform coefficient.
I guess that integral is the easy way you talked about, but I know tough guys don't do those... :D Clearly.
I see exactly what you're saying. The reference about these being transformation coefficients was mostly a wink and nod to those who really understand what those are. I think already assuming the wave functions are fourier conjugates to eachother might have been a pretty big assumption in the first place too, but it's also what people are probably already familiar with.
@@AndrewDotsonvideos |psi> = int dp |p> ---> = int dp , so if you assume that the wave functions are related by Fourier transform, you must have = 1/sqrt(2pi) * e^(ipx). This shows that the assumption is just re-stating what is in a more complicated way. You haven't shown anything. This entire video is circular and misleading.
WHERE WAS THIS 2 MONTHS AGO OMG
I don’t get why there’s a position space and momentum space. Wouldn’t they both just use cartesian basis
Could you have done this when you finished your undergrad or this is sth that you learned in your graduate course? I'm currently taking QM I now and I'm starting to worry since this looks hard for me lol
A lot of undergrad courses in quantum introduce dirac notation pretty late on. Definitely something an undergrad can handle, but you have to be fluent in diracs bra-ket notation first. Don't worry, it wasn't my strongest point in undergrad!
Notification Squaaaaaadd
? Is the p and x in the exponent of the final answer operators; is px the same as xp.
I understood is a Fourier transform coefficient. But anyone please tell me what the physical meaning of is?
It's basically saying, the state |p> has a well-defined momentum. Then, basically gives you a wave function for a free particle which has a well defined momentum (a plane wave)
@@mihajlosreckovic8404 Thanks for your reply. Can we also interpret the other way around, i.e.
@@GBY13 I believe you can. You can say, the state |x> has a well defined position, so gives you the wave function which has a well defined position in momentum space. So, taking , you can either think of it as a wave function in coordinate space with defined momentum, or of complex conjugated wave function in momentum space with defined position. It's interesting that those 2 things are mathematically the same.
Also, you really can qualitatively see Heisenberg's uncertainty principle from this. Since gives you a plane wave, By considering this as a wave function with defined momentum, you get completely undefined position (a plane wave), and by considering this as a wave function with defined position, momentum is, again, completely undefined (aka in momentum space, wave function is a plane wave).
For psi(x) is the exponent coefficient (hate this word) 1/h-bar or i/h-bar ?
Also, is there angular momentum space, boson/lepton number space etc?
Why did you apply delta function twice (1st on |x'> and 2nd time on x' ) when you wrote |x'>x' delta(x'-x)
here x' is eigenvalue
Why do you have to insert the one two times with x and x‘ i don’t really get that
I swear it would mean everything for me if you read. It's so interesting.
Tip: when writing outer products, |psi x psi| is easier than |psi>
Zio pera
Only a Jr in highschool, I don't understand anything but it's still interesting. 😂
Its the fourier transform bro
Ahh yes I know some of these words
what branches of physics and math did you study to obtain this knowledge?
Linear algebra, quantum mechanics, and fourier analysis (only to understand the integral relationships of the wave functions)
First!
Lol i thought this was a flammable maths vid 🤔
god damn it
LOL @ 5:30
Tfw I'm a cal legend but can barely multiply a matrix
Hi
I will never understand this series of vids, im only in high school
u will one day
Everybody who understands this was once a highschooler.
What black magic is this??? (I’m a Computer Science major)
Hit like if you confused this for a papa flammy video
? Is the p and x in the exponent of the final answer operators; is px the same as xp.