Wave functions in quantum mechanics 

Thanks for this! :)

We now have our own take on the Schrödinger equation, needed to describe the time evolution of quantum states: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-CKpx9hkQ3HM.html

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

Glad you find it useful! :)

Glad you like it!

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

Glad you like them! :)

Glad you like it!

Thanks for your continued support! :)

Glad you like it! :)

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

Happy to help :)

@@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?

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.

@@ProfessorMdoesScience thank you so much 😊

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!

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: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-rp2k2oR5ZQ8.html 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!

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

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: ru-vid.com/group/PL8W2boV7eVfmMcKF-ljTvAJQ2z-vILSxb 2. The quantum harmonic oscillator: ru-vid.com/group/PL8W2boV7eVfmdWs3CsaGfoITHURXvHOGm 3. Angular momentum: ru-vid.com/group/PL8W2boV7eVfmm5SZRjbhOKNziRXy6yIvI 4. Hydrogen atom (note we are still building this one): ru-vid.com/group/PL8W2boV7eVfnJbLf-p3-_7d51tskA0-Sa For more advanced topics, we have these two: 1. Identical particles: ru-vid.com/group/PL8W2boV7eVfnJ6X1ifa_JuOZ-Nq1BjaWf 2. Second quantization: ru-vid.com/group/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!

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: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-auLZ2WOKiqE.html I hope this helps!

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!

I have taken a pragmatic approach to the definition of the translation operator both here and in the original video: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-978mMgGYs1M.html 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!

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!

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: ru-vid.com/group/PL8W2boV7eVfmMcKF-ljTvAJQ2z-vILSxb 2. Quantum harmonic oscillator, which provides a simple example of using these postulates: ru-vid.com/group/PL8W2boV7eVfmdWs3CsaGfoITHURXvHOGm 3. Angular momentum, which allows us to move from 1D to 3D: ru-vid.com/group/PL8W2boV7eVfmm5SZRjbhOKNziRXy6yIvI 4. Hydrogen atom, which is a realistic example of using quantum mechanics (not that this playlist is still under construction): ru-vid.com/group/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!

@@ProfessorMdoesScience Thank you very much!

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!

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

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: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-p1zg-c1nvwQ.html I hope this helps!

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!

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

No doubt is ever trivial! :)

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

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: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-EyGJ3JE9CgE.html And then use these to get the corresponding eigenfunctions in spherical coordinates here: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-Gk2XNmIHVwo.html Hope this helps!

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

I think the expression on the video is correct. You can find the precise derivation in this video: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-978mMgGYs1M.html I hope this helps!

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!

@@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.👍

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!

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

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.

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

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

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

@@rimon9697 Indeed :)

@@ProfessorMdoesScience Thanks :D

I hope you still found it interesting!

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

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.

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...