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Is (i/x^2 d/dx) a Hermitian Operator?

  1. Sep 25, 2014 #1
    1. The problem statement, all variables and given/known data

    Hi, I'm doing a Quantum chemistry and one of my question is to determine if is hermitian or not? I am learning and new to this subject... Cant figure out how to do this question at all. Please helppp!!!

    ^Q= i/x^2 d/dx is hermitian or not?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 25, 2014 #2

    CompuChip

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    Hi arcoon, welcome to PF. Maybe you could expand your "Relevant equations" section with the definition / properties of a Hermitian operator?
     
  4. Sep 25, 2014 #3
    problem set 2 hermitian.jpg
    This is the whole question. This is probably very easy for some people. But not so easy for me at all.
     
  5. Sep 25, 2014 #4

    Orodruin

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    I believe that what Compu is gently hinting at is that you need to show some sort of effort more than simply stating the problem as per forum guidelines. For example, your relevant equations might contain information not explicitly given in the problem such as what the definition of a hermitian operator is. This you should be able to find in you course litterature. Under #3 you could list your thoughts on how to use this to check if it is hermitian or not.
     
  6. Sep 25, 2014 #5
    My math is not that advanced(Only have first year calculus). I been trying to find simple way to start this kind of problems but cant type the symbols yet. I know the simple logic but cant even type it here. All the information out there is so complex for me to understand at this point. That's why I could use help solving the problem. If i see the answer at least I will have an idea how to solve future problems.My instructor seems to know his stuff but not helpful at all.
     
  7. Sep 25, 2014 #6
    Such problem involves some integration to figure out whether it looks as a Hermitian operator or not. There's a useful thing that may help you: For a Hermitian operator we can show that < psi | H^ | phi > = <phi | H^ | psi >*, where H^ is a Hermitian operator sandwiched between the kets | psi > and | phi >, with | psi > and | phi > being the basis vectors of the space. Of course, for the sake of simplicity you can take both kets the same.
     
  8. Sep 25, 2014 #7

    Fredrik

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    Rewrite ##\int f(x)^*(\hat Q g(x))\mathrm{d}x## in the form ##\int (\hat Af(x))^*g(x)\mathrm{d}x## and determine if ##\hat A=\hat Q##.

    Normally, we would make you tell us that before we can help, but since you're new here, I'm willing to tell you that much. I will also tell you that the rewrite involves the product rule for derivatives, or equivalently, integration by parts.

    Edit: The alternative approach that I suggested was bad, so I have deleted it from this post. Then I noticed that I had forgot to include the complex conjugation in the equation at the top, so I corrected that too.

    If you hit the LaTeX preview button below, you will find a link to our LaTeX guide in the lower right corner.
     
    Last edited: Sep 25, 2014
  9. Sep 25, 2014 #8
    I am already there.cant do the math part. If i fail this quiz which is next week i will drop the course and never take it again. I though i was very interested in this subject but everybody else seems to think I have to know everything to start with.These are suppose to prepare us for quiz but I been trying to see few examples for last 4 hours. I still got nothing.
     
  10. Sep 25, 2014 #9

    Fredrik

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    I see. Unfortunately we can't just tell you the solution, because of the rules for the homework forum. You have to show us your work up to the point where you're stuck. I think it's OK to just copy and paste this integral sign ∫ and type the rest in plain text. (But be careful with parentheses). No need to type hats above the operators. You can denote the adjoint of an operator A by A*.
     
  11. Sep 25, 2014 #10
    I am so tired. Anyway you don't need to help me this way .... just torture. I though this site could help me get better but I should go with google or try something else.This is not just homework it is learning tool. Thanks anyway.
     
  12. Sep 25, 2014 #11

    Fredrik

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    I wouldn't blame the site. It's 100% your choice to not accept the excellent help that you could get here.
     
  13. Sep 25, 2014 #12

    Orodruin

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    So here is my take on things: We could just tell you yes or no. However, what would you learn from that? What would you do the next time you have to determine whether an operator is hermitian or not? Well, we can try making that experiment:
    "No."
    Now is the operator ##-\frac{d^2}{dx^2}## hermitian or not?
    This is why most people on this forum insist on helping people to solve problem themselves and straighten out question marks, not to solve the problems for people, which means that we must ask you about what your take on the problem so far is.

    In your particular case, you say that you have signed up for a quantum chemistry course after first year calculus. Now I do not know how things are done at your university, but where I work course descriptions generally include a listing of the prerequisite knowledge necessary to be able to follow a course (i.e., a list of knowledge that the course will assume that you already have) and for a quantum chemistry course I would at least expect students to know basic quantum mechanics. Did you check whether this is the case or not?

    The fact that you say that several people in your class already know "everything" may simply be that they have taken one or more courses on quantum mechanics and regardless of the angle you look at things, this forum cannot replace the knowledge gained from taking a QM course in one single thread. It may simply be that you have chosen to take a course that assumes some prerequisites that you do not yet have and that is nobody's fault (apart from perhaps the administration in letting you know what the prerequisites were before you signed up). Also your other thread is a basic QM question that would have been covered in any course on QM.

    Another option is of course that this type of subject just is not your cup of tea in the sense that you do not find it interesting - personally I have been there when I was trying to take a course in ancient greek some 15 years ago ...
     
  14. Sep 25, 2014 #13
    prerequisites for this course is only Practical Spectroscopy or phys chem and first year calculus(I only have spectroscopy which has nothing to do with greek letters and math) . However I passed these courses overall Grade 'A'. Quantum chemistry has all bunch of symbols which we havent been explained properly. I don't even know what complex conjugate and fancy greek letters yet . we are 2 weeks into this course and I am smart but nobody to show ME SOME EXAMPLES YET including our instructor. He writes simple question skips the math and writes down the answer. However he demands us to do homework which I clearly cant do. I can appreciate the concept of this site. If I already know stuff and need some guidance this works. But having say that this should be simple problem for you guys. If you can walk me through your answer I may have chance to be successful and be good at it.
     
  15. Sep 25, 2014 #14

    vanhees71

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    Then it's important to read a good textbook. Which one are you recommended to use?
     
  16. Sep 25, 2014 #15

    Fredrik

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    Complex numbers are often covered in high school, so it's understandable that the book assumes that you know them. If you don't, then you should drop everything else and study them right away. The Greek letters are just a notational convention. You can use f and g instead of ##\psi## and ##\phi##. (I often do).

    We can help you work through this example, but we will only do it in accordance with the rules of this forum. It's a textbook-style problem, so it will be treated as homework even if it's not. I have told you where to start, so the ball is in your court. You will have to tell us where you're stuck.

    If you have general questions about e.g. wavefunctions, inner products or complex numbers, you can ask them in the appropriate forum (quantum physics, linear algebra, general math), where you can get complete answers.
     
  17. Sep 25, 2014 #16

    Orodruin

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    If you have passed all the prerequisite courses and feel that that is not enough, this is definitely a thing that makes me angry with your university and you should be too. You should definitely point this out to whoever is coordinating the courses and setting the prerequisites. I know this is not helping you at the moment, but I just want to underline that this is definitely the fault of the coordinators. In the normal case, if you have passed all the prerequisites you should not have too much difficulty to pass as long as you spend a reasonable amount of time. With that being said, I second essentially everything Fredrik said.
     
  18. Sep 25, 2014 #17

    Fredrik

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    You know what, I don't see the harm in me showing you how to do the simplest problem of this kind, the one with the momentum operator instead of the ##\hat Q## in your problem. I think you will need to see this, because the standard argument involves some non-rigorous "physicist's math" that you can't be expected to guess unless you've seen it in class.

    The standard argument goes like this:
    \begin{align}
    &\int f(x)^*\left(-i\frac{d}{dx}g(x)\right)\mathrm{d}x =-i\int f(x)^*\left(\frac{d}{dx}g(x)\right)\mathrm{d}x =-i\int \left(\frac{d}{dx}\big(f(x)^*g(x)\big)-\left(\frac{d}{dx}f(x)^*\right)g(x)\right)\mathrm{d}x\\
    & =i\int \left(\frac{d}{dx}f(x)^*\right)g(x)\mathrm{d}x =\int \left(-i\frac{d}{dx}f(x)\right)^* g(x).
    \end{align}
    The operator acting on the f in the final expression is by definition the adjoint of the operator acting on g in the first expression. Since these are the same, the conclusion is that the operator acting on g in the first expression is self-adjoint.

    You're probably wondering what happened to the other term that showed up at the end of the first line. This argument isn't meant to be rigorous, but the idea is that for all wavefunctions f, we have ##\int|f(x)|^2\mathrm{d}x<\infty##, and this implies that f is either super weird or such that ##|f(x)|\to 0## as ##x\to\pm\infty##. If we ignore the super weird wavefunctions, it seems reasonable to do this:
    $$\int \frac{d}{dx}(f(x)^*g(x))\mathrm{d}x =\left[f(x)^*g(x)\right]_{-\infty}^{+\infty} =0-0=0.$$ This argument is so ugly that it hurts a little to type it, but this is how it's done in physics books.
     
  19. Sep 25, 2014 #18
    thank you so much Fredrick. That does help me a lot to understand some basic concepts. I tried to book time with my prof. He said he could see me in 8 days from today? Easier to talk to president then my prof.
     
  20. Sep 25, 2014 #19
    Ok this where i am at
    ∫f*Qgdx = ∫gQ*f*dx
    ∫f*(i/x^2 d/dx) = ∫g(-i/x^2 d/dx)
    from here i am trying to use integration by parts
    ∫u dv = uv - ∫v du
    for the left hand side
    u= f* i/x^2 dv= d/dx
    du= can you please help me with this step not really sure how to handle star(math help center people didn't know how to do this)
    v=x ? is this correct?
    should I do same thing with the right hand side?
     
  21. Sep 26, 2014 #20

    Fredrik

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    I don't have a lot of time today. I'll be home about 8-10 hours from now. Right now I only have time to tell you this:

    You wrote ∫f*(i/x^2 d/dx), but that expression is missing a g on the right (and the integration "dx"). Also, you shouldn't write ∫f*(i/x^2 d/dx)g dx = ∫g(-i/x^2 d/dx)f dx, because we still don't know if this holds. Your task is to find out if it does.

    Integration by parts is an application of the product rule:
    $$\int_a^b u(x)v'(x)\mathrm{d}x =\int_a^b ((uv)'(x)-u'(x)v(x))\mathrm{d}x =[u(x)v(x)]_a^b-\int_a^b u'(x)v(x)\mathrm{d}x.$$
    You don't need to know a lot about complex conjugation for this problem (but you really need to study it as soon as possible). You just need to know that for all ##z,w\in\mathbb C##, we have ##(zw)^*=z^*w^*##. The complex conjugate of a function f is defined as the function f* such that f*(x)=f(x)* for all x in the domain of f.
     
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