Wave functions in quantum mechanics

[@armalify]

It is clearly obvious how much effort was expended in clarifying such important concepts in Q.M. Every single second worth listening. May I suggest to you to write a book ! Why not if it is possible

[@jagatramkanwar3119]

Professor M does quantum mechanics extremely well and simple as well.

[@ProfessorMdoesScience]

Thanks for this! :)

[@paichethan]

Fantastic. You are explaining exactly what I needed. I was wasting my time on schdr equation. I couldn't even get a opportunity to learn Matrix formulation. Now I think you made my life easier. Thank you very much.

[@ProfessorMdoesScience]

We now have our own take on the Schrödinger equation, needed to describe the time evolution of quantum states: kzbin.info/www/bejne/eXzTqWyeoLZmfq8

[@kka107]

I don’t even have a physics major but I do love physics and I also happen to have a strong background in linear algebra. So you can see how much your videos benefit me. my deep appreciation to you and to the work you put behind this. We are living in a time where people can learn anything by just watching great videos like yours on KZbin. 🙏

[@ProfessorMdoesScience]

Thanks for your kind words, glad to hear you like our videos!

[@anindyarastogi3741]

The highlight of the lecture in my opinion is at 2:29 with the definition of what is a wave function : the expansion coefficient of the state \psi (that represents the physical state of a QM system) in the basis formed by the eigenkets of the position operator . A big shout out to both Professor M's for these very clear and crisp videos.

[@ProfessorMdoesScience]

Glad you like them! :)

[@workerpowernow]

wow, this is definitely the clearest explanation i've encountered for why the change of basis for the representation of an state vector from position to momentum space is encoded in the fourier tranform

[@ProfessorMdoesScience]

Glad you find it useful! :)

[@quantum4everyone]

Just a note that there is a simpler way to compute . Just write

[@ProfessorMdoesScience]

Thanks for the suggestion, looks like a nice alternative approach!

[@bishalthapa2179]

you expansion are very easy to understand

[@學習中的哈密瓜]

Thanks for this! Great explanation

[@ProfessorMdoesScience]

Glad you like it!

[@LifeForAiur]

This is great at just what I needed. I only have two questions. Why is | psi > equal to the integral over space of < x | psi >? Next, does | x > span all of space?

[@ProfessorMdoesScience]

Glad you find it useful! |psi> is the abstract state, and when we write it over the continuous |x> basis, the expansion coefficients are given by , hence the form of the integral. We discuss this in more detail in the video on representations: kzbin.info/www/bejne/qKGVnGWlh5qNh5o And yes, for all purposes |x> is a basis spanning the full space. The real mathematical answer is somewhat more subtle because |x> is not normalizable, so it is not a vector in the physical state space which requires normalizable states. If you want to learn more about this latter subtlety, I suggest you read about "rigged Hilbert spaces". I hope this helps!

[@LifeForAiur]

@@ProfessorMdoesScience Thanks again! I'll check out rigged Hilbert spaces. First time I've ever heard of it.

[@manujsharma1432]

Thank you for such a nice explanation 👍

[@ProfessorMdoesScience]

Thanks for your continued support! :)

[@nasszelle534]

How do I know that an operator (say Position op.) not only maps to a subspace? In contradiction to that psi can be reconstructed in equation @2:15

[@ProfessorMdoesScience]

If I understand your question correctly, then it is possible that in the expansion of |psi> in the {|x>} basis there are some basis states that do not contribute. But this would be automatically taken care of because the expansion coefficients would then vanish. Another way of understanding this is that |psi> may be orthogonal to some basis states, so that =0. Those basis states then don't contribute to the expansion of |psi>. I hope this helps!

[@jozsefkele7858]

Since watching these videos I have added about 20 pages to my own set of LaTeX formatted quantum notes. Even including these videos in the references section lol.

[@ProfessorMdoesScience]

Great to hear! :) What are these notes for? Self-study or are you using them for teaching?

[@enricolucarelli816]

So, if I got this right, a wave function makes mostly sense if we are dealing with a Hilbert vector space of infinite dimension, because in a vector space of finite dimension, say dimension n, the wave function would be a function phi(x) where x can only take the values 1 to n, have I got this right?

[@ProfessorMdoesScience]

You are correct, the "position representation" has an infinite (and in fact continuous basis). An example of a finite-dimensional state space is that spanned by spin, for example it is 2-dimensional for spin-1/2 particles. We are working on a series on spin-1/2, so hopefully we'll soon have concrete examples of this!

[@enricolucarelli816]

@@ProfessorMdoesScience All right! Thank you! For me this has been an important clarification. I kept struggling trying to understand the difference between wave function and state vector.👍

[@AAKASHAAKASH-uv6ru]

Thanks for all the explanations! I have a problem at 12:55 and hoping that you can clarify. While solving for differential equation, we had the relation that N=e^C, where C was any arbitrary constant in complex plane, as we are solving an differential equation with complex coefficient on right hand side at 10:28. So, N should be a complex number. In the video, we had |N|^2=1/2pihbar, so how can we say N=1/sqrt(2pihbar) from |N|=1/sqrt(2pibhar)? Also, can you refer me to the video where we derived or gave an explanation to why momentum/position operators are hermitian and why their commutator value is i*hbar? I am unable to find them in your videos. Again, thank you very much!!

[@ProfessorMdoesScience]

Good insight! In general we cannot go from |N|=1/sqrt(2pibhar) to N=1/sqrt(2pihbar), as there is an extra phase factor arising from the fact that N can be a complex number. However, in quantum mechanics an overall phase factor does not change the physics. Therefore, we are free to choose any phase factor we want, and the convention is to choose the phase factor such that we end up with a real number. This is exactly what we've done here. For a discussion about overall phase factors in quantum mechanics, I refer you to our video: kzbin.info/www/bejne/pmLdmGCZZtOprbM I hope this helps!

[@FasAntick]

Great videos, although i'm confused the integration at 5:54. Are all the terms on the RHS of the scalar product under the integral? Any chance you could explain the steps in this integral? I know the integral of the dirac delta over all space equals 1, but how does this lead to the next line?

[@ProfessorMdoesScience]

Glad you like them! Yes, all terms are under the integral, so the integral reads as follows: integral dx integral dx' psi*(x) phi(x') delta(x-x') = integral dx psi*(x) integral dx' phi(x') delta(x-x') To get to the next line, we need the properties of the delta function, which you are correct to say that it integrates to 1, but we are actually using a slightly more general definition of it, which says the following: integral f(x) delta(x-x0) dx = f(x0) This means that, for any function f(x), the integral of f(x) multiplied by the delta function delta(x-x0) gives the value of f at x0. Going back to our integral, our "f" function is phi(x'), and when we then integrate over the x' variable, and as we have delta(x-x'), this means that we get phi(x). Specifically: integral dx' phi(x') delta(x-x') = phi(x) Recovering the full expression, we get: integral dx integral dx' psi*(x) phi(x') delta(x-x') = integral dx psi*(x) phi(x) I hope this helps!

[@coldmine9480]

Thank you so much. My quantum mechanics class starts in january so I need a channel badly to make me understand the Quantum mechanics. 😊😊😊

[@ProfessorMdoesScience]

Happy to help :)

[@coldmine9480]

@@ProfessorMdoesScience If wave function is a complex valued function, then it shows real and imaginary values that's are stored in it. Then why still we can't see the wave functions and why it collapse on observation if they consist of real parts?

[@ProfessorMdoesScience]

In general I would not say that because a quantity is a real number or a real function that means that its something we can observe. For example, if you are familiar with electromagnetic theory, you will know about the magnetic vector potential, which is a very useful real mathematical object but it is not measurable, what we can measure is the magnetic field which is calculated as the curl of the magnetic vector potential. The same happens with the wave function: it is a mathematical function that appears in the theory, but which we cannot direcly measure, irrespective of whether it is real or imaginary. However, just like we can build the observable magnetic field from the unobservable magnetic vector potential, we can build observable quantities from the unobservable wave function. For example, the absolute value squared gives the probability density of finding a particle at a given position.

[@coldmine9480]

@@ProfessorMdoesScience thank you so much 😊

[@SergeyPopach]

What about energy and time? We know that time and energy can be also related by Fourier transform. So, as the momentum is related to k value, p=hk, energy related to angular frequency, E=hw. Can we relate the energy representation to time representation in similar manner or that's not the case simply because time quantity is not an operator?

[@ProfessorMdoesScience]

Your last sentence is the key point: time is not an operator, it is a parameter, so we cannot do as you ask. However, one can still study the relationship between energy and time, for example we explore it in this video: kzbin.info/www/bejne/l6avi2WNhLCcp6c I hope this helps!

[@Bloodwork-ig6dg]

Delta(x-x‘) isn‘t the Dirac delta function, right?

[@ProfessorMdoesScience]

Yes it is, why would you say it isn't?

[@Iconic_Alex27]

Maybe he thought this way, because for x=x', it will make δ(0) which is infinity, but for usual defination of scalar field, = finite.

[@armalify]

Why exactly did you define the translation operator using the equation that appears at 07:51 ? Thank you

[@ProfessorMdoesScience]

I have taken a pragmatic approach to the definition of the translation operator both here and in the original video: kzbin.info/www/bejne/b2ibnoCdfL6mZ68 What I mean by "pragmatic" is that I have written out this definition without justification, and then simply shown that it does what we want: translating states. The reason why I have taken this approach rather than providing a more fundamental justification is that the maths associated with the more fundamental justification is rather more subtle than the level of this series of videos. In short, translations (and rotations) form a mathematical structure called a "group", and all elements of this group (i.e. our translations) can be generated by joining together infinitesimal elements of the group (in our case infinitesimal translations). In group theory, these infinitesimal elements are called the "generators" of the group, and in that context one finds that the generator of translations is the momentum operator. This is ultimately why the translation operator is essentially the exponential of the momentum operator. For rotations, a similar argument shows that angular momentum is the generator of rotations, and the rotation operator is similarly the exponential of the angular momentum operator. I do have plans for a series on group theory and its uses in quantum mechanics where I will go into these ideas in more detail, but I hope this helps in the meantime!

[@I5vvdoedibrvel]

Hello. Thank you for nice video lectures. At 8:25, I think that

[@ProfessorMdoesScience]

I think the expression on the video is correct. You can find the precise derivation in this video: kzbin.info/www/bejne/b2ibnoCdfL6mZ68 I hope this helps!

[@avi2005]

Thanks for the interesting videos. What is the recommended order to watch all you videos?

[@ProfessorMdoesScience]

Glad you like them! The best way to watch them is to follow the various playlists. And for those, a good order to start from the basics is: 1. The postulates of quantum mechanics: kzbin.info/aero/PL8W2boV7eVfmMcKF-ljTvAJQ2z-vILSxb 2. The quantum harmonic oscillator: kzbin.info/aero/PL8W2boV7eVfmdWs3CsaGfoITHURXvHOGm 3. Angular momentum: kzbin.info/aero/PL8W2boV7eVfmm5SZRjbhOKNziRXy6yIvI 4. Hydrogen atom (note we are still building this one): kzbin.info/aero/PL8W2boV7eVfnJbLf-p3-_7d51tskA0-Sa For more advanced topics, we have these two: 1. Identical particles: kzbin.info/aero/PL8W2boV7eVfnJ6X1ifa_JuOZ-Nq1BjaWf 2. Second quantization: kzbin.info/aero/PL8W2boV7eVfnSqy1fs3CCNALSvnDDd-tb Following the above is a core course on quantum mechanics. The other playlists we have go deeper into various topics, for example time dependence or the density operator. We are working on a website to facilitate watching the material, but in the meantime I hope this helps!

[@zeusilver]

If I have a specific system and I don't projet the state on a base but I write only |ψ> , can the vector be written with specific numbers? In a Hilbert space, how can I distinguish a vector ψ1 from a vector ψ2 as I can't specify their components in a base?

[@ProfessorMdoesScience]

For practical calculations you will normally need to write the state in some particular basis. But you can still make general statements about quantum mechanics with general states, with several examples in our series on the postulates of quantum mechanics. I hope this helps!

[@stevenlin6106]

Excellent!

[@ProfessorMdoesScience]

Glad you like it!

[@rimon9697]

could you tell me why you have written < x | x prime > = delta(x-x prime), I have been scratching my head over that and just could not find it anywhere? Any help would be appreciated

[@ProfessorMdoesScience]

A basis of eigenstates is orthonormal. For a discrete basis {|u_n>} the orthonormality condition reads: =delta_nm. The position basis {|x>} is a continuous basis, and to adapt the orthonormality condition to the continuous case we get

[@rimon9697]

@@ProfessorMdoesScience Ahh... this is like extending the kronecker delta to continuous basis, if I am right?

[@ProfessorMdoesScience]

@@rimon9697 Indeed :)

[@rimon9697]

@@ProfessorMdoesScience Thanks :D

[@Teachwelles]

What is the recommended viewing sequence of the 95 videos. Is it suggested by viewing them from oldest to newest? Each video I have viewed refers to other videos. I find these very helpful in my learning quest!!!

[@ProfessorMdoesScience]

Glad you like the videos! A good way to view them is to follow the playlists, and for those the order could be: 1. The postulates (basic mathematical tools of QM): kzbin.info/aero/PL8W2boV7eVfmMcKF-ljTvAJQ2z-vILSxb 2. The quantum harmonic oscillator (a good illustration of the use of the postulates): kzbin.info/aero/PL8W2boV7eVfmdWs3CsaGfoITHURXvHOGm 3. Angular momentum (from 1D to 3D): kzbin.info/aero/PL8W2boV7eVfmm5SZRjbhOKNziRXy6yIvI 4. Central potentials (a good illustration of 3D QM): kzbin.info/aero/PL8W2boV7eVfkqnDmcAJTKwCQTsFQk1Air 5. Hydrogen atom (the culmination of single-particle QM): kzbin.info/aero/PL8W2boV7eVfnJbLf-p3-_7d51tskA0-Sa These are a good starting point that are roughly equivalent to an introductory course on quantum mechanics. Beyond that, other topics of interest are quantum mechanics of many particles, density matrices, etc. I hope this helps!

[@cunningham.s_law]

are there now other ways to write a translation operator? would they result in different results?

[@ProfessorMdoesScience]

Not sure I fully understand your question, can you clarify?

[@sayanjitb]

Dear sir, as it is evident from the expression that the wave function \psi(x) is the expansion coefficient of \ket{\psi} = \sum {\psi(x) |x>}. It is a coordinate in the coordinate space representation. So how is it possible for a coefficient to fully encode the feature of the full state vector as we see in problems? Am I missing something? TIA

[@ProfessorMdoesScience]

Note that the wave function is a function of x, which is a continuous variable that goes from minus infinity to plus infinity. So what we have is an infinite number of coefficients psi(x), one for every x. I hope this helps!

[@sayanjitb]

@@ProfessorMdoesScience Ahh right, thank you so much for answering my trivial doubt! 🙂

[@ProfessorMdoesScience]

No doubt is ever trivial! :)

[@shiqiangec1304]

Great video ! What happens with the error dependent on epsilon^2 ? My calculus basis are not solid, and now a part of my brain thinks that the Fourier relation between x and p is an approximation, which I'm sure is not the case.

[@ProfessorMdoesScience]

Good question! We *are* taking the limit epsilon --> 0, so that term vanishes and the relation becomes exact indeed. It is the same as to what happens when you define a derivative or an integral in mathematics via infinitesimals, and eventually taking all limits to zero. I hope this helps!

[@shiqiangec1304]

@@ProfessorMdoesScience Thank you very much, I am in peace now :)

[@Upgradezz]

What bra should act on a ket to give us a wavefunction in terms of r, theta, phi??

[@ProfessorMdoesScience]

An easy way to do this is to use x,y,z and then apply the usual conversion from Cartesian to spherical coordinates. We have an example of this for orbital angular momentum operators here: kzbin.info/www/bejne/e6qqe2aAep52nac And then use these to get the corresponding eigenfunctions in spherical coordinates here: kzbin.info/www/bejne/fZyViYGjfq2JrdE Hope this helps!

[@enricolucarelli816]

Great videos! Is there some way to go the right sequence? Looking at the date of the videos, this one seems to be the oldest one, but then you refer to other videos in it, so I’m not sure if going by time stamp is correct.

[@ProfessorMdoesScience]

Thanks! The best way to view the videos is by following the order in the various playlists. And between those, a good way to get started is the following: 1. Postulates of QM, which gives you the basic mathematical tools: kzbin.info/aero/PL8W2boV7eVfmMcKF-ljTvAJQ2z-vILSxb 2. Quantum harmonic oscillator, which provides a simple example of using these postulates: kzbin.info/aero/PL8W2boV7eVfmdWs3CsaGfoITHURXvHOGm 3. Angular momentum, which allows us to move from 1D to 3D: kzbin.info/aero/PL8W2boV7eVfmm5SZRjbhOKNziRXy6yIvI 4. Hydrogen atom, which is a realistic example of using quantum mechanics (not that this playlist is still under construction): kzbin.info/aero/PL8W2boV7eVfnJbLf-p3-_7d51tskA0-Sa I would say these provide a good basis. Other playlists provide additional topics, some of which are more advances (e.g. second quantization). I hope this helps!

[@enricolucarelli816]

@@ProfessorMdoesScience Thank you very much!

[@quentintsang]

Please teach us group theory.

[@ProfessorMdoesScience]

Thanks for the suggestion! We do hope to extend our videos to also discuss "mathematical methods for physics", including group theory.

[@band5680]

Hello, so if the continuum of coefficients \psi(x) is the wave function. What is the \ket{\psi}? Does it have some meaning in wave mechanics? (apart from the definition as a vector in a complex vector space) It doesn't contain any more information than the coefficients, as each coefficient is uniquely linked to a basis vector, right? So is it identical to the wave function? ...but it isn't, is it? Haha, an answer would be much appreciated. :) Thank you for the helpful videos! Max

[@ProfessorMdoesScience]

You are absolutely correct, they are different ways to represent the same quantity: the state of the system. However, a particular form may be more or less useful in a particular calculation, so that is how we typically decide which form to use. I hope this helps!

[@band5680]

@@ProfessorMdoesScience Thank you so much for the quick response!

[@quantum4everyone]

It is important in mathematics to differentiate a vector from the coordinates of the vector. The state vector psi is the abstract vector, which lives regardless of its coordinates. The wavefunction, is the coordinates of that vector in a p[articular basis. So, for example, one can find energy eigenfunctions as abstract vectors (such as raising operators acting on the ground state) which is the abstract vector and then multiplying by a position space bra, constructing the wavefunction. Though not done here, this allows you to compute eigenvalues without wavefunctions using the so-called Schroedinger factorization method.

[@rahulshastri9278]

Never seen this derivation of transformation function between x and p basis. Usually it is directly introduced by Fourier coefficient.

[@ProfessorMdoesScience]

I hope you still found it interesting!

[@garvitmakkar]

glad to learn

[@ProfessorMdoesScience]

Glad you like it! :)

[@rodrigoappendino]

OOOH. I get it now. I thought the wave function was the ket psi, and not that bra-ket product.

[@ProfessorMdoesScience]

Thanks for watching, and glad you like it! :)

[@kratzeni]

Typically I like your videos but this one is very hard to follow.

[@ProfessorMdoesScience]

Sorry to hear this, any particular part that is tricky to follow? Should also add this was one of our first videos, so likelihood is that we hadn't quite got to grips with the whole format yet...