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Issue regarding the orthogonality of eigenvectors for Hermitian

  1. Feb 18, 2006 #1
    At the risk of arrousing the ire of the moderaters for posting the same topic in two forums, I again ask this question as no one in the quantum forum seems to be able to help. So...
    Regarding a proof of the orthogonality of eigenvectors corresponding to distinct eigenvalues of some Hermitian operator [itex]A[/itex]:
    Given [itex]A|\phi_1\rangle = a_1|\phi_1\rangle[/itex] and [itex]A|\phi_2\rangle = a_2|\phi_2\rangle[/itex] we have
    [tex]\begin{array}{lll} 0 & = & \langle\phi_1|A|\phi_2\rangle - \langle\phi_2|A|\phi_1\rangle^* \\ & = & a_1\langle\phi_2|\phi_1\rangle - a_2\langle\phi_1|\phi_2\rangle^* \\ & = & (a_1 - a_2)\langle\phi_2|\phi_1\rangle \\ \end{array}[/tex]
    Now, I understand that this shows the inner product of the eigenvectors to be 0 (given distinct eigenvalues) and therefore that the eigenvectors are orthogonal. But, if someone could please help, how do we arrive at line 2 of the equivalence from line 1. That is, replacing the operator with its corresponding eigenvalues. There appears to me to be some gymnastics with the positions of multiplicands. If someone could possibly detail this for me (and please assume I am an idiot, not such a stretch really) I would again be eternally grateful and would show just how grateful I am by possibly responding with a thank you. No, make that most definitely responding with a thank you.
     
  2. jcsd
  3. Feb 18, 2006 #2

    matt grime

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    Looks like a mistake in whatever source this is from, not a criminal one, just swap over the a_i, the result still holds (note, i loath bra and ket notation, finding it most unintuitive, so I may be missing something in my self enforced ignorance which makes it correct as stands).

    My proof would be

    (Ax,y)=(x,Ay) since A is hermitian, hence for distinct e-vectors x,y with e-values s and t respectively s(x,y)=t(x,y) which can only happen if (x,y)=0.

    This seems so much more understandable than the bra/ket notation.
     
  4. Feb 18, 2006 #3

    George Jones

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    Matt, I agree that it seems that the author of the (quite good) reference made a bit of a slip, and I, too, find bra ket notation to be unintuitive in situations like this. I translate to standard inner product notation, just as you did. Surprised?

    There are other situations, however, where I find bra ket notation to be quite useful.

    Regards,
    George
     
  5. Feb 18, 2006 #4

    matt grime

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    Moderately surprised, I must admit, since I generally find those used to Quantum Mechanics to be completely aghast that I find bra and ket stuff confusing. Part of the reason is that no one ever explained what it meant to me until after I was supposed to have used it as an undergrad. If only someone had said: we're just saying that l^2 is reflexive, then I'd have been so much happier, though what is wrong with the normal dual basis stuff is beyond me.

    Even now I'm still not very sure how to do bra and ket stuff and I'm so far divorced from QM that I don't feel the need to find out the definitions since any problem like this can usually be explained by linear algebra, and anything more QM based would be beyond me anyway.

    Actually, this case is a very good example: this result is just a statement about linear algebra (well, possibly bilinear algebra) and the notation just hides this fact.
     
    Last edited: Feb 18, 2006
  6. Feb 19, 2006 #5
    You are absolutely right Mr. Grime, which is why I thought perhaps this question was better suited under a linear algebra thread. Line 2 is correct, as near as I can tell anyways, I just do not understand the reason for this particular form. Oh well, I'll just have to give it some thought.
     
  7. Nov 29, 2006 #6
    Hello Mr. Grime.
    The question addressed in this thread is actually a problem I am working on in one of my classes. Unfortunately, though, I am confused by your notation. Would you mind possibly rewriting your proof in a different notation or verbally explaining what it is your roof is saying?
     
    Last edited: Nov 29, 2006
  8. Nov 29, 2006 #7
    Would the following be what you want?

    Given [itex]A|\phi_1\rangle = a_1|\phi_1\rangle[/itex] and [itex]A|\phi_2\rangle = a_2|\phi_2\rangle[/itex] we have
    [tex]\begin{array}{lll} 0 & = & \langle\phi_1|A|\phi_2\rangle - \langle\phi_2|A|\phi_1\rangle^* \\ & = & a_2\langle\phi_1|\phi_2\rangle - a_1\langle\phi_2|\phi_1\rangle^* \\ & = & (a_2 - a_1)\langle\phi_1|\phi_2\rangle \\ \end{array}[/tex]
     
  9. Nov 30, 2006 #8
    After asking my teacher about it, apparently that is exactly what I want. Thank you very much.
     
    Last edited: Nov 30, 2006
  10. Dec 3, 2006 #9

    HallsofIvy

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    Since this is, after all, the "mathematics" and not "quantum physic" section, here is a mathematician's view of it:

    If A is a linear transformation from one inner product space,U, to another, V, its "adjoint" is the linear transformation, A* from V to U such that <Au,v>U= <u,A*v>V where < , > represents the inner product (in the appropriate vector space).

    In particular a linear transfromation on a vector space, V (that is, from V to V) is said to be "self-adjoint" (Hermitian to quantum physicists) if and only if it is its own adjoint: <Au, v>= <u, Av> for all u, v in V.
    First, it is easy to show that all eigenvalues of A are real: Let [itex]\lambda[/itex] be an eigenvalue of A corresponding to unit eigenvector v. Then
    [tex]\lambda= \lambda<v, v>= <\lamba v, v>= <Av, v>= <v, Av>= <v, \lamba v>= Conj(
    \lambda )<v, v>= Conj(\lambda)[/tex]
    where [itex]Conj(\lambda )[/itex] is the complex conjugate. Since [itex]\lamba= Conj(\lambda)[/itex], [itex]\lambda[/itex] is real.

    Now, suppose u and v are eigenvectors corresponding to two distinct eigenvalues, [itex]\lamba_1[/itex] and [itex]\lambda_2[/itex], respectively.
    Then [itex]\lamba_1<u, v>= <\lamba_1 u, v>= <Au, v>= <u, Av>= <u, \lamba_2 v>= \lamba_2<u, v>[/itex] (the last equality because [itex]\lamba_2[/itex] is real. Since [itex]\lamba_1 <u, v>= \lamba_2 <u,v>[itex], [itex]\lamba_1<u,v>- \lamba_2<u,v>= (\lamba_1- \lamba_2)<u,v>= 0[/itex]. Since [itex]\lamba_1[/itex] and [itex]\lamba_2[/itex] are distinct, [itex]\lamba_1- \lamba_2[/itex] is not 0 and we must have <u, v>= 0. (That's essentially the same proof doodle gave, in Linear Algebra terminology rather that quantum physics.)
     
  11. Dec 3, 2006 #10
    The Quantum Computation notation is easier to follow. Thanks though.
     
    Last edited: Dec 4, 2006
  12. Dec 4, 2006 #11

    HallsofIvy

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    Hey, you said you're a MATH student!!!
     
  13. Dec 4, 2006 #12
    Math and physics are not mutually exclusive. I am beginning to find your posts offesive; please be considerate enough to not put someone down for having an interest in physics (which, if you think about it, is applied mathematics). Thank you.
     
    Last edited: Dec 4, 2006
  14. Dec 4, 2006 #13

    matt grime

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    Uh? What on earth is easier to grasp about <x|A and A|y> instead of x^tAy, i.e just multiplying a row vector into a matrix into a column vector? Since being taught inner products is year one of an undergrad course, and is incredibly elementary (look, coordinates!), and bra and ket notation is used on reflexive hilbert spaces, I don't see that the latter is at all easy, unless one has perverse notions of what constitutes understanding. The normal linear algebra notation has less baggage to it. The only reason for the bra and ket notation is to distinguish between what is physically meaningful to put in the left or right space in the inner product <?,?>.
     
  15. Dec 4, 2006 #14

    HallsofIvy

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    Would it have helped if I had put a :rolleyes: beside it? My major was in mathematical and applied mathematics. (Although I do feel I should point out that saying that "physics is applied mathematics" is putting physics down!)

    You say "posts", plural. I'm wondering if you mean other posts in which I have expressed the point that I dislike mathematics posts that use physics terminology. I will stand by that- If you post in a physics thread, use physics terminology. If you post in a mathematics thread, use mathematics terminology.
     
  16. Dec 4, 2006 #15
    Okay. . . I have completed the problem. Thank you for your help. Matt, I am glad that you understand the Matrix Theory notation, but obviously, I don't; I do however understand the Quantum Computation notation. Seeing as how the purpose of such threads is to make learning for those asking questions conducive (i.e., to make learning conducive for me), the notation that is best understood and should be used is the notation that I can understand. If you find something else easier, good for you, but I'm the one trying to learn this material and it needs to be explained so that I can understand. My understanding has nothing to do with your's and to put me down for not thinking the same way that you do is not only childish, but pointless. In no way does insulting me help me better understand a concept. Please stop.
    Also, understand that many students have learning disabilities. If I am dyslexic, then I will find Ivey's notation more confusing because it is a string of symbols that I cannot easy break up into understandable sections. It essentially "blows up in my face," hence why the quantum computation notation is easier for me to understand: fewer symbols to get screwed up in my head. If you are going to teach, be understanding of your students, not insulting towards them.
    Ivey, your posts should explain concepts to students. If a student understands something in terms of physics, then you should explain in terms of physics, because (and I reitterate this point) the purpose of these threads is to aid students, not stroke anyone's ego about how they think certain notations should be used. The notation that best aids a student is the notation the student understands. It is unfair to say that physics has no place in the world of math.
     
    Last edited: Dec 4, 2006
  17. Dec 5, 2006 #16

    matt grime

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    Since this wasn't your thread, or your question, what has your choice of notation got to do with anything? Further, who insulted you? Who said the notations of physics have no place in maths? (Never mind you assertion that we said physics has no place in maths!) Surely I was the first person to confess that I can't understand something, namely the notation of QM. It is *my* weakness of learning that makes me prefer simple linear algebra terms. I can't explain it in QM symbols because I have never learned what they mean. I am not putting you down; I genuinely cannot comprehend how QM notation is easier than simple linear algebra.

    The content of the question is truly trivial (and that isn't an insult) in linear algebra terms. You should take time to learn them, since it will greatly ease your studies of QM. Many apparently complicated statements in QM are trivial statements of linear algebra, and linear algebra is something we are all taught at a very early age. If you know the definition of an inner product, this result follows instantly from it. I cannot write that in QM symbols *because I do not know how*.

    The notation that should be used is the one that demonstrates why something is true in the most elementary fashion. If you simply put some symbols together without understanding the meaning of the symbols and why you can make the manipulations then you've not understood what's going on. The linear algebra ideas demonstrate this elementary result without overcomplicating it with unnecessary symbols.
     
    Last edited: Dec 5, 2006
  18. Dec 5, 2006 #17
    I was unaware that I have to "own" a thread to ask a question. MY QUESTION was if the original response to the original question could be expressed in different terms. If you cannot do that, don't respond. MY QUESTION should be answered so that I can understand what's going on: I repeat: what has that go to do with your understanding? You don't understand QM notation; I do. You understand linear algebra; I don't. Seeing as how it is MY QUESTION, any answer needs to be in terms I understand. If all you know is linear algebra and I need it explained in terms of QM, you should not bother to try and answer. You are still being insulting; STOP IT. Whether you are trying to be obnoxious or not, you are succeeding. STOP IT. If you want to pick on someone, go make fun of a four year old, BUT STOP INSULTING MY UNDERSTANDING OF HOW THINGS WORK. Just because I understand a different explaination does not make the explanation wrong. If you are considering responding to this post, STOP IT. My question has been addressed; it was addressed a while ago, and ironically, it was not addressed by anyone on this thread. SO STOP IT. There is no point in continually harrassing me. NOT EVERY STUDENT IS CAPABLE OF UNDERSTANDING EVERY ABSTRACT IDEA. If you cannot understand that and how that affects the functioning of the minds of certain students, DO NOT TEACH. I reitterate: if YOU explain something, it should be so THE STUDENT can understand. If the student does not understand your explanation, either find a different way to explain the concept, or find someone else to explain it so the student can learn, BUT DO NOT INSULT, HARRASS, PUT DOWN, OR OTHERWISE ANTAGONIZE THE STUDENT. DOING SO ONLY MANAGES TO MAKE YOU LOOK LIKE AN IMMATURE PERSON. So STOP IT. Ivy, it is evident that after I expressed that I felt somewhat offended by your posts (and I mean the posts on this thread, not any other) that you expressed your intentions of not being offensive or insulting. I do appreciate that; thank you.:smile:
     
  19. Dec 6, 2006 #18

    matt grime

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    But I didn't answer you at all, did I?

    You have your opinion, I have mine, and I have not denigrated you for not understanding linear algebra. You won't learn linear algebra, I won't learn the niceties QM,. It's no skin off my nose if you don't want to learn linear algebra. I would suggest that the advice 'do it the linear algebra way' is probably the best pedagogical advice there is on this question. And spoon feeding people the answer is not the way, nor is demanding that we conform to your rules. If you want a thread on how to translate between the two notations, start one. You will benefit greatly from it. I will keep writing the statement in the most natural language for it, and which demonstrates the answer most elegantly, simply, and intuitively. If you don't want to learn that, well, again, it is nothing to do with me. But if you did it with a matrix you'd see it was straight forward, and you'd be able to do it in the basis free linear algebra way, thence the QM way since it is just saying the innerproduct of x and Ay is the same as the inner product of A*x and y (* for complext transpose, or adjoint).

    I have exhorted you to go away and learn it because many results in QM are elementary statements of linear algebra if you just wrote them out like that. Even George at the start of the thread, who is very much a physics/applied notation person points out that in a situation such as this QM notation is hiding an elementary fact.

    I'm sorry you feel insulted; you have not been.
     
    Last edited: Dec 6, 2006
  20. Dec 6, 2006 #19

    HallsofIvy

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    Math Student, while I am nervous about responding to a post when you specifically told me not to (do you really consider it polite to address a person and refuse to allow him to respond?) I would like to point out respectfully that this was posted in the "Abstract Algebra" thread. Why should I then assume that the person asking the question would not know linear algebra? The person who originally posted the question did not ever tell me that he did not understand linear algebra nor did he say he did not understand my explanation.

    Frankly you seem to have a very low opinion of the intelligence and knowledge of physicists and physics students since you assert that they are not capable of understanding certain things. The physicists and physics students I have known were quite capable of understanding both linear algebra and linear algebra notation and were anxious to learn different ways of looking at problems.

    You were the one who said "the physics notation is easier to follow". You did not say that YOU found it easier to follow or that Inquire4more, the original questioner, found it easier to follow, only the flat statement that it WAS easier to follow. I disagreed with that and then you started a diatribe.
     
  21. Dec 6, 2006 #20
    Both of you shut up. I am not going to be polite to two people who do nothing but make false assumptions about what it is I have said, and then tell me I'm wrong for what it is they think I've said. As I said AWHILE AGO, no thanks to either of you and your offensive resonses, I finished the problem and had it explained to me in such a way that I can understand it. Don't even bother responding to this because I won't read whatever false statements you have to make. If either of you are teachers, find a new job because teaching just is not for you.
     
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