Eigenvalues and eigenstates in quantum mechanics

[@syedrumman3920]

You have no idea how much you are helping to fix misconceptions for students! Thanks a lot for the content!!

[@ProfessorMdoesScience]

Thanks, comments like this motivate us to keep going :)

[@expandingflames5225]

I love how you have started from the very basics i.e. the group theory and are slowly building it up from first principles. Thank you so much for this brilliant series of videos.

[@ProfessorMdoesScience]

Thanks for watching and for the kind words! :)

[@nirmalbaishnab8054]

Please keep these awesome content coming. These are really pushing the limit what we can find over KZbin.

[@ProfessorMdoesScience]

Thanks for the comment, and glad you like it! :)

[@armalify]

Very nice when you always trying to find the connections between the mathematical steps and relating them to their physical meanings. This approach, from my point of view is the key success in teaching. Thank you.

[@TheWingEmpire]

I believe mathematics and physics should be approached together in a same point of view

[@doingrandomshit]

Absolutely fantastic video and great explanation. For me, this video has cleared up a lot of confusion and made me feel like I actually understand the mathematical framework. You are by far the best quantum mechanics channel on youtube. If I could subscribe twice, I would.

[@ProfessorMdoesScience]

Thanks for your support, and glad you like our videos!!

[@richardthomas3577]

just watched for review (whilst eating dinner) and have to say you are so very clear and precise. Excellent!

[@ProfessorMdoesScience]

Glad you liked it, and bon appetit ;)

[@zacharycormier9898]

I'm studying for my quantum mechanics midterm right now, thank you for this!

[@ProfessorMdoesScience]

Glad it helps!

[@richardd7614]

Thank you very much for taking time to help students like this. Very much appreciated.

[@ProfessorMdoesScience]

Glad you like the video! :)

[@abhasoodan7982]

I just came across this video and this helps so much!!!!! thank you so much, you've reduced the difficulty a lot!

[@ProfessorMdoesScience]

Glad you liked it!

[@davidroux7987]

Even though I teach this stuff myself, I still learn from your videos! Thank you for the refreshing, enjoyable presentation.

[@ProfessorMdoesScience]

Thanks for your kind words and glad you find it useful!

[@TheWingEmpire]

They don't teach a sixteen years old kid quantum mechanics in my place, a lot of help is your videos

[@ProfessorMdoesScience]

Thanks! And I just realized that you had found this video (I referred you to it in your comment in the video on "operators") :)

[@TheWingEmpire]

@@ProfessorMdoesScience yes

[@full_legit191]

Thanks you for this content

[@board_runner]

< edit append(top'end) > | *Edit:* | I completely forgot to tell _*the author*_, "_Professor M_", that this was an exceptionally good explanation that not only explained a rather intimidating expression, but gave me renewed energy in the field of 'teaching'. | I somehow forgot that it is not only "rep'N'prep.", but figuring out how to explain 'logic' as simple as this video has done it... Wow ...I'm looking forward to teaching, so much.. So thank you for that! |< end edit > < main message > Has anyone else experienced this??... This is now the second time I watch this video, last time was yesterday... same thing happened from about 0:15 The strangest thing, my left monitor started flickering. » _At first i thought i was out of RAM or something, since i had a tad more than 100 'browser'-tabs open(plus whatever), since my browser didn't contain my entire screen then It seemed strange that my entire screen was flickering_ « After some sleep and a nice clean-up, the same thing happened, same video(and *only* this one) and same behavior. Entire monitor, no matter window size, BUT only the one monitor and only this video..?!??? Have anyone experienced anything like this? *OR* maybe you have a theory to what might cause this strange phenomenon? - Is it my browser, GPU, encoding, RAM, ... i have no idea where to start..! HOLD ON... ...Just realized the problem.... Im running windows 10, *_DOUGH!!_* /dan # ps # but seriously, leave a comment/respo if u have any idea(wht,so evR) to what might cause this disturbance/interference. # /out

[@ProfessorMdoesScience]

I'm afraid I cannot help with the problem you've experienced with the video. But glad you enjoyed it! :)

[@nasszelle534]

what I need is a playlist or recommendation in which order someone should go trough all the videos.

[@ProfessorMdoesScience]

You can find all our playlists here: studio.kzbin.info/door/ZqRWM99ixVKA-ydebf_tMQplaylists A good way to get started is in this order: 1. The postulates: kzbin.info/aero/PL8W2boV7eVfmMcKF-ljTvAJQ2z-vILSxb 2. The harmonic oscillator: kzbin.info/aero/PL8W2boV7eVfmdWs3CsaGfoITHURXvHOGm 3. Angular momentum: kzbin.info/aero/PL8W2boV7eVfmm5SZRjbhOKNziRXy6yIvI There are some extra dependencies, so you may want to add some videos as suggested in the "background" section of the description. Eventually we hope to setup a website with a clearer way to guide self-study, and also incude problems etc.

[@davidcarrero1414]

Soy latino y se que soy tercermundista pero usted me hace entender cuantica, que basado, like y nuevo sub.

[@ProfessorMdoesScience]

Gracias por el comentario, y suerte!

[@sagarpatel3043]

Thank you for helping us. Your videos means so much to me. I will suggest you teach us how to approach some challenging problems in quantum mechanics.

[@ProfessorMdoesScience]

Glad you find it useful! We are preparing additional material to complement our videos, including problems+solutions, so stay tuned for those!

[@masoncamera273]

This deserves more views

[@ProfessorMdoesScience]

Thanks for your support!

[@speedspeed121]

I am taking my second QM class next month. This really helps conceptualize a lot of topics that just seemed like they were math equations.

[@ProfessorMdoesScience]

Glad to help! What topics does your course cover? We have a growing list of videos covering typical QM course material.

[@speedspeed121]

@@ProfessorMdoesScience Thank you for the reply. Next quarter, QM2 covers "Angular momentum and spin; hydrogen atom and atomic spectra; perturbation theory; scattering theory." It seems like QM1 is pretty standard among most schools. You have most of the topics from QM1 in your list. My professor is a General Relativity-QFT researcher. I find that the more advanced the professors are, the less they talk about conceptual things. It's usually, "this is how we mathematically derive ____." That's fine, but it helps when someone paints us a picture. You seem to do that well. Thank you

[@ProfessorMdoesScience]

Glad you like our approach! The maths is essential to make quantum mechanics work, but we think it is also important to gain a good conceptual understanding :) We are currently working on videos on the hydrogen atom, and we already have a series on general angular momentum. Spin and perturbation theory will come later... We also have some more advanced topics that may come in handy in your QM3 course, such as second quantization! :)

[@tomgraupner171]

Nice one! Thanks for this content. I would like to see such a video for operators, which have "pseudo eigenvectors" like the operators "x" or "p".

[@ProfessorMdoesScience]

Thanks for the suggestion! We do plan to have a video discussing the subtleties associated with the "x" and "p" basis states...

[@tomgraupner171]

@@ProfessorMdoesScience Wonderful! thanks again for the great job you do!

[@nomanahmadkhan7791]

While discussing Q.M. Euclidean space often serves as an easy example to understand abstract ideas (although not all of them) and you also have used it in your videos beautifully. I was wondering whether there are some examples of operators, eigenvectors with eigenvalues too in Euclidean space? All that comes to mind are physical quantities with standard equations as we are used to write in C.M.

[@ProfessorMdoesScience]

Eigenvalues and eigenvectors of operators are also useful in Euclidean space. For example, you could use them to describe the transformations of rigid bodies (e.g. rotations and stretches).

[@canhdominh3255]

Thank you so much for your video. I wonder which book should I read to know about these stuffs in detail.

[@ProfessorMdoesScience]

General eigenvalues and eigenvectors can be studied in any book on maths for science. For specific applications to quantum mechanics, we mostly use the books by Sakurai, Shankar, and Cohen-Tannoudji.

[@workerpowernow]

great video! for others looking for a text to strengthen their familiarity with the math employed, I'd highly recommend "Linear Algebra Done Right" by Sheldon Axler. I worked through that text and then watching this series maps nearly perfectly to the concepts and mode of presentation there

[@ProfessorMdoesScience]

Thanks, and great tip!

[@-thesmartboard8984]

This was really helpful thank you

[@ProfessorMdoesScience]

Glad it was helpful! :)

[@vaishnavichandilkar9064]

Best video to learn easily even u don't have much background

[@ProfessorMdoesScience]

Glad you find this useful! :)

[@williamberquist3923]

I'm struggling with the step at 10:35 where you move the right-hand side of the equation to the left and set it all to 0. I don't see how you were able to get the del_ij and remove some of the terms to get the simplified final equation.

[@ProfessorMdoesScience]

Thanks for watching! I essentially skipped a step, so thanks for pointing this out. What we start with in the right hand side is: lambda* = lambda*c_i, where you are correct that there is no delta_ij here. The step I skip is that we can multiply the full expression by the identity operator: lambda* = lambda*identity*. With this expression, the matrix elements of the identity operator are delta_ij, so the full matrix multiplication of the identity with the vector can be written as: lambda*identity* = lambda * sum_j delta_ij c_j. I hope this helps!

[@williamberquist3923]

@@ProfessorMdoesScience Thank you so much! I see it now. I really appreciate you taking the time to explain this to me!

[@williamberquist3923]

​@@ProfessorMdoesScience One last thing actually. How does the c_i in the lambda*c_i term become a c_j term after the identity operator is multiplied with the vector?

[@ProfessorMdoesScience]

The easiest way to see this is to realize that they are the same. Note that: sum_j delta_ij c_j = c_i In practice, the way to get to these expression is to use the matrix formulation of quantum mechanics, and we go over the details in this video: kzbin.info/www/bejne/rXran5VnocmMiqs I hope this helps!

[@williamberquist3923]

@@ProfessorMdoesScience thank you!

[@moinakdey9268]

This is an excellent lecture professor but I have a question out of the topic that is where can I get a course on Tensor Calculus online in detail. I have books on Tensor Calculus but I want a course. Thank you.

[@ProfessorMdoesScience]

I am not familiar with any such online course, so cannot make a good recommendation there. However, they must exist, so it should be possible to find one. In the meantime, we touch on the role of tensor products in quantum mechanics in this video: kzbin.info/www/bejne/oauWY2NsiJd1bLM

[@moinakdey9268]

@@ProfessorMdoesScience Thanks a lot sir. (-:

[@jupironnie1]

if possible, please update brief information on video clip to indicate Pauli Matrices as available rather than as coming soon.

[@ProfessorMdoesScience]

Thanks for spotting this, just updated!

[@jokerbatman6177]

I am new in quantum mechanics and I wanted to ask that in continuous basis, lets say for position basis in one dimension then does it means that each position on x axis in our space correspond to a ket in Hilbert space and if that is true than does it means that in small neighbourhood of "x" say dx that there dx numbers of eigen kets corresponding to that position "x" and so when we take then kronickar delta changes to dirac delta bcz we now require whole sum (of all the dx eigenstates) to be 1 ??

[@ProfessorMdoesScience]

For your example, then yes, we have a ket for every x in the range dx. Mathematically, orthonormality for a continuous basis is given by

[@jokerbatman6177]

@@ProfessorMdoesScience If it is true that any two states are orthogonal [even within dx] then if we take inner product of ket x with itself that is and now as you said that in continuous basis we take infinitesimal interval around x , so now if we take inner product at position x with itself and we are taking dx interval so the inner product will be (1 times no of states in width dx which are infinite) 1.infinity , and we want the total sum to be 1 (area under curve for dirac delta function is 1) so it will (1/dx).dx =1 Is this correct??

[@ProfessorMdoesScience]

@@jokerbatman6177 Yes, indeed normalisation is still 1, and this is enabled as you say by the integral of the Dirac delta function being 1.

[@jokerbatman6177]

@@ProfessorMdoesScience Thank you very much professor.

[@jokerbatman6177]

@@ProfessorMdoesScience I am still skeptical about the idea, that the states are orthogonal even in dx interval bcz here we are multiplying dx which is very small number , as in calculus we say dx=vdt here we say that velocity is not changing in that very small time interval dt and is almost constant so can't we apply the same analogy here, as here we are writing ket as integral of (|x>)dx and as we know that multiplication is just a way of saying that this number is repeating itself this many times so here also dx is also small number multiplied in a sense that it says, this ket with this coefficient is repeating itself this many times and then adding up together these small small intervals to give a ket |a> ,just like we do any other integration so shouldn't be the kets in dx intervals should be parallel rather than perpendicular??

[@seabeepirate]

I’ve just found your channel and this math was over my head but I think only by a little bit so I’m going to watch your other videos and see if I can fill in the blanks. For my abstract understanding, would it be fair to say that the eigenstates are the clean resonation of wave functions from all points of a matrix? And further, the degenerate eigenstates are multiples of the simplest arrangement which also fit the framework of zero point crossings for the simplest arrangement?

[@ProfessorMdoesScience]

Glad you like the videos! I am not sure I completely follow your statements; could you please clarify what you mean by "clean resonation" and "all points of a matrix"?

[@nastyavicodin6229]

Best content!

[@ProfessorMdoesScience]

Thanks! :)

[@jackbeda521]

The alternative notation presented at 2:34 seems to me to be a little bit dangerous if the eigenvalues of an operator do not uniquely specify an eigenket. For example: suppose A |φ> = λ |φ>, and Α|ψ> = λ|ψ> (but with |φ> != |ψ> (even up to a constant, say)), it would then be ambiguous notation to write: Α |λ> = λ |λ> as it is unclear which ket is being referred to. How is this resolved? Is the state I describe above, with degenerate eigenvalues, just very rare?

[@ProfessorMdoesScience]

You are absolutely correct that with degenerate eigenvalues this can be ambiguous. In that case, one needs additional labels to distinguish the corresponding eigenstates. An example of this is provided in this video: kzbin.info/www/bejne/f5mtp4tqfZyroaM I hope this helps!

[@progra_kun4331]

me encantan tus videos amigo deberias traducirlos al español asi mas gente los aprovecha

[@ProfessorMdoesScience]

Gracias por la sugerencia!

[@user-hy7jl1mq7m]

how do i write the states for |jm> if j=1/2? which video should i looketo solve this?

[@ProfessorMdoesScience]

We have an overview of general angular momentum here: kzbin.info/aero/PL8W2boV7eVfmm5SZRjbhOKNziRXy6yIvI However, we also hope to publish videos specific to spin-1/2 in the near future, so stay tuned! :)

[@shakerali800]

Great work.

[@ProfessorMdoesScience]

Thanks! :)

[@eleonoraredaelli5420]

do you publish the pdf of your lessons somewhere? they would be very useful!

[@ProfessorMdoesScience]

We don't at the moment, but working on a platform to share more content to go with the videos, so stay tuned! :)

[@deekshasinha2999]

Why is the solution to the equation @16:30 c1= i ?

[@ProfessorMdoesScience]

Eigenvectors only provide a "direction". This means that vectors that have the same direction but different lengths and phases are equally valid eigenvectors. For example, if the vector (1,1) is an eigenvector of a particular operator, then so is (2,2), or (i,i), etc., and in fact we can have an infinite number of solutions. What we typically do is to work with normalized eigenvectors because it is convenient, but this is just a choice. So in this case, we find that the eigenvector components must obey the equation -c2=ic1. This is one equation for two unkowns, which means that the equation cannot fully determine both c1 and c2. Instead, for a given c1, then c2 is given by the equation, but c1 could be anything. The fact that c1 could be anything is a reflection of the statement above that eigenvectors are only fixed up to their length or phase. So all we have to do is to make a convenient choice for c1, and then fix c2. In the video I say that "c1=i" is "a" solution, which means that I make this choice. Choosing c1=i then forces c2=1. So the total solution is (c1,c2)=(i,1). I could have made a different choice for c1, for example c1=1. Then we would have c2=-i, and (c1,c2)=(1,-i). Both (i,1) and (1,-i) are eigenvectors of the operator, because they are the same up to a phase, in this case a phase of pi/2, e^{i*pi/2}=i as i*(1,-i)=(i,-i^2)=(i,1). I hope this helps!

[@deekshasinha2999]

Omg thank you so much!!! I have been stuck on this for hours and I couldn’t think straight!! Thanks so much for the quick reply!!

[@ProfessorMdoesScience]

Really pleased that this helped!

[@ragibshahriar187]

Want more of this

[@ProfessorMdoesScience]

Thanks for watching! :) You can explore all our other videos in our channel, and happy to hear suggestions for future videos!

[@TheWingEmpire]

Can you please explain why quantum mechanical works are based on complex Hilbert space instead of real? Imaginary I had no physical meaning in nature right?

[@ProfessorMdoesScience]

You could approach this question from various angles (and we would need a lot of space to cover them properly). However, it boils down to the fact that a state in a real vector space is insufficient to explain many of the experimental observations about quantum mechanics, for example interference. For your second question, you are correct that physical properties are real, and indeed when you look at those in quantum mechanics they are always associated with real numbers, even if the states live in a complex vector space. For example, physical properties can only take values that are given by the eigenvalues of a Hermitian operator, and those are always real. You can find more details about this here: kzbin.info/www/bejne/jnrKdYicrtd_d7s I hope this helps!

[@TheWingEmpire]

@@ProfessorMdoesScience I give you my honest opinion, I have never ever seen anyone reply to my queries with so much care, thank you very much

[@quantum4everyone]

I would not say all physical observables are real. In ac circuits we often discuss things like complex admittance and so forth. Even driven-dissipative harmonic oscillators have complex observables. It is also true that quantum observables need not always be hermitian. Normal operators are sufficient and can be complex. In quantum computers, it is common to create circuits to measure complex-valued object such as green’s functions. But, I think the biggest issue is the standard theory for measurement of quantum mechanics doesn’t really apply to how we actually do measurement in quantum mechanics. There is a huge disconnect. This is why you won’t find quantum textbooks actually apply the theory of measurement to any real experiment. When von Neumann presented his theory in 1952, Schroedinger told him it was a beautiful experiment, but he did not think it applied to a single experiment. I think Schroedinger’s conclusion is still true today.

[@KaranveerSingh-xn4tv]

Hey. A question. We know that eigenvectors of an operator are orthogonal to each other. And also that any linear combination of eigenvectors is also an eigenvector of the operator. But inner product of linear combination state and one of the initial (original) eigenvectors will not be zero. Right? They are not going to be orthogonal. Can you tell me what is that I'm missing? I reckon it has something to do with gram-schmidt procedure.

[@ProfessorMdoesScience]

The second statement is not correct in general. Consider two different eigenvectors of an operator A, say A|psi_1>=lambda_1|psi_1> and A|psi_2>=lambda_2|psi_2>. If we now create a linear combination, say |psi>=|psi_1>+|psi_2>, then for this to also be an eigenvector of the operator A we would need to find a scalar lambda such that A|psi>=lambda|psi>. But if lambda_1 and lambda_2 are different, this is not possible. Therefore, it is not true that any linear combination of two eigenvectors is also an eigenvector of the operator. This statement is only true if the two eigenvectors correspond to the same eigenvalue. I hope this helps!

[@KaranveerSingh-xn4tv]

@@ProfessorMdoesScience Alright! But in the case where linear combination of eigenvectors is an eigenvector of the operator, wouldn't < psi | psi1> be non zero?

[@ProfessorMdoesScience]

If you have a degenerate eigenvalue, then you are correct that any linear combination within the degenerate subspace gives a valid eigenvector, and you can build non-orthogonal eigenvectors. In principle you could work with those, but the maths becomes more tedious. Therefore, we conventionlly choose linear combinations that make them orthogonal within the subspace too, and indeed the approach we use to construct orthogonal vectors is Gram-Schmidt orthonormalization. We describe some of these ideas in our video on Hermitian operators: kzbin.info/www/bejne/jnrKdYicrtd_d7s

[@KaranveerSingh-xn4tv]

@@ProfessorMdoesScience Thank you, Professor. I really appreciate the help.

[@jupironnie1]

Is there a basis to state that each measurement is an Eigenvalue ? (other than just stating it is a postulate). maybe there is history behind this idea ....Broglie ideas on matter waves ? or sheer guesswork to "fit the curve".? or from Planks black body oscillation ? cheers ps: I spent months trying to figure out Special Relativity till I read that a better name is Invariant Theory and Laue coined the term Special relativity in 1909. Things got better after that as i did not focus on term Relativity anymore.

[@ProfessorMdoesScience]

Measurements do appear as a postulate in the standard formulation of quantum mechanics, and we cover those in detail here: kzbin.info/www/bejne/q2K1ZJ6IjM1km80 kzbin.info/www/bejne/pZWvqIiOgL5jgNU Matter waves as suggested by de Broglie were later developed into the full formalism of quantum mechanics, where the state of a matter particle is described by a wave function. I hope this helps!

[@jupironnie1]

@@ProfessorMdoesScience Many thx. Debroglie extended the ideas of eigen shapes of a string vibration to an electron moving in a fixed orbit (as per Bohr model). This implies that the matter of an electron is "smeared/stretched" along the orbit. However, i think this idea was rejected. I am unable to picture an electron as both a matter with a solid particle and as a smeared wave. Somewhere in his thoughts, he incorporated Rayleigh's ideas of heat radiation in box. Rayleigh basis was to count the number of different waves shapes which can be enclosed in a box ....a la Fourier series. What does Quantum Mechanics in year 2022 picture ? . (I do need a physical picture rather than a axiom maths formulation). Please do correct me if my above narrative is wrong ? Cheers ps: The next question is if Born had a basis to indicate amplitude as probability. He just moved from deterministic basis to a probabilistic measure. The ideas of Boltzman to use probability in gas theory seems to make sense but using similar shape of energy function/amplitude of physical waves and electrons in double slit experiment in QM as basis of probability postulate is, for me, at the moment difficult to accept. I think Born claimed Einstein used this idea but Einstein did not agree.....I am not clear what exactly Einstein did.

[@kaushikgupta1410]

Hey, Really a big thanks from my side. Can u please suggest some books for practices?

[@ProfessorMdoesScience]

We hope to do a few videos on quantum mechanics books we like. In the meantime, these are a few titles: "Quantum Mechanics" by Merzbacher, "Quantum Mechanics" by Cohen-Tannoudji, "Modern Quantum Mechanics" by Sakurai, and "Principles of quantum mechanics" by Shankar.

[@kaushikgupta1410]

@@ProfessorMdoesScience ok thx , i will definitely look at these books.

[@tremendous6571]

Could you please send the pdf of the book you have mentioned in the discription

[@ProfessorMdoesScience]

We don't have the a pdf copy of the book. It is a pretty standard maths for physics textbook, so you should be able to find it in most libraries. Otherwise, most other maths for physics textbooks should also cover the same topic.

[@tremendous6571]

Thanks

[@alexanderheller2039]

legend

[@TheWingEmpire]

You just stole my comment. I agree with you 100%

[@alexaserna8330]

How to calculate the (1/sqrt(2)) in the last exercise) 16:50 !!! I can´t help

[@ProfessorMdoesScience]

That step is the normalization step. The magnitude (length) of a state |psi> is given by sqrt(). The state we have initially is: |psi> = column(i 1), where I hope the notation is clear. Remember that the bra is of magnitude 1, so as we start with a magnitude equal to sqrt(2), then to make it of magnitude 1 we need to divide by sqrt(2). Thus, we end up with the normalized state: column(i 1)/sqrt(2). I hope this helps!

[@alexaserna8330]

Thanks! I forgot a minus yesterday, I was very slept hahaha. When You forgot the minus, the length to the vector is equal to 0 😭hahaha Ang I want to say that I have saw all your videos and they have been very clear and beautiful ❤️ nobody teach me like these before So thanks 😊 When I have the opportunity to teach quantum mechanics in my country, I will remember your videos and I will try to be very clearly like you, just I will teach in Spanish so it'll be little bit different 😅 I will support you when I can! Thanks again

[@ProfessorMdoesScience]

Thanks for watching and for your kind words! We actually also speak Spanish :)

[@alexaserna8330]

@@ProfessorMdoesScience Amazing! Do you have a Spanish channel? I can share with my friends that not speak English!

[@ProfessorMdoesScience]

@@alexaserna8330 We don't have a Spanish channel, have you tried whether subtitles work?

[@amaljeevk3950]

❤

[@michaelschnell5633]

At kzbin.info/www/bejne/pmLdmGCZZtOprbM you describe how to find the n eigenvalues of an n*n matrix on C² and with that prove that the all do exist (of course they might be duplicate), and with this of course appropriate eigenvectors . Here: kzbin.info/www/bejne/rmHanGxuqrKsr9U we learn that the raising operator does not have eigenstates. Seemingly this is possible because the operator is an "infinite" matrix. Is there more that can be said about this seeming contradiction ? (I asked the same in the comments of the other video, as well.) Thanks for listening ! -Michael

[@ProfessorMdoesScience]

Really nice question, and I repeat here the answer we provided in the mirror question for completness: "This is a very interesting and insightful question, and a full answer would require much more than a comment. However, to get you started I would suggest looking at the first answer in this post on StackExchange: physics.stackexchange.com/questions/445144/eigenstates-of-the-creation-operator I hope this helps!"

[@narfwhals7843]

@@ProfessorMdoesScience From the link and a bit of further research I gather that the raising operator has _left_ eigenvectors.

[@ProfessorMdoesScience]

@@narfwhals7843 You are correct about these statements, which directly follow from the fact that the raising and lowering operators are each other's adjoints

[@budjy1]

Great help but the greek letter Psi is pronounced Si, like the word sigh. The p is silent. Thanks

[@ProfessorMdoesScience]

Thanks for that!

[@StergiosTsiormpatzis]

Hm... not really in Greek. Ψ(ψ) is pronounced as "πσι" in Greek, not "σάι" or "πσάι" as people in quantum used to pronounce it. It took me actually a few lectures in the university to realize to what the professors refer to when they speak about Greek letters (this is not only about Ψ but most of the letters, including Π(π), Ξ(ξ), Φ(φ) and Χ(χ).

[@StergiosTsiormpatzis]

This is quite correct Greek pronunciation kzbin.info/www/bejne/hWLIhoKbmtSIerc

[@babyoda1973]

Parth g sent me

[@ProfessorMdoesScience]

Welcome, and I hope you like our videos!

[@yjmsniper]

I HATE EIGENVECTORS - WHY MAKE THINGS HARD