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Expectation value of x <x>

  1. Jan 8, 2014 #1
    1. The problem statement, all variables and given/known data

    This is a question I had in my Quantum Mechanics class but my problem is with the calculus which is why i am posting it here. The question is to find the expectation value of x given the wave function equals Ax^3 where 0 ≤ x ≤ a, 0 otherwise. The solution given in class is 7/8 and I have NO idea how to get that. I am sure my integration is correct but not sure where to go from there.

    2. Relevant equations

    <x>=<ψ| x ψ>

    3. The attempt at a solution
    ∫ψ*xψ dx
    ∫x(-Ax^3)(Ax^3)dx
    -A^2∫x^7 dx
    (-A^2 x^8)/8 from a→0

    I know I should know this but i have been out of school a couple of semesters and not doing math. If you tell me which technique to look up that will be fine. Thanks so much!
     
  2. jcsd
  3. Jan 8, 2014 #2

    Dick

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    It's not the math that's giving you a hard time. It's the concept. Your first job should be to normalize the wave function. You need to find a value of A such that ∫ψ*ψ dx=1 where you integrate x from 0 to a. Does that sound familiar?
     
  4. Jan 9, 2014 #3

    Ray Vickson

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    Your "answer" of 7/8 is incorrect; it must depend on 'a' as well. Are you sure it is not supposed to be ##(7/8)a##?
     
  5. Jan 9, 2014 #4
    ok yes Dick I was wondering about this. Here is my point of confusion...I guess this needs to be moved to the quantum section then...sorry! I have been re-reading the wave function section in my modern physics book but i am not getting it still (my quantum book is still en route). When I normalize I get x^6=1 then x=1...is this the A value? I got confused because at first the wave function was presented as |ψ>=x^3. But then he told us to find the expectation value for <x> and ψ=Ax^3. Then if <x>=<ψ| x ψ> why is <ψ| = -Ax^3 because there is no i present? I had ∫x(Ax^3)(Ax^3) for the initial setup, but was told this is wrong. I don't understand why I am so confused. And still not sure how to arrive at 7/8 and i think this is my math problem. When we integrate from 0→a is everything = to 1...this is the only way I can get 7/8 and it seems wrong to me.

    I really appreciate you taking the time to help me.
     
  6. Jan 9, 2014 #5
    Thanks Ray. The professor said the answer is 7/8 but maybe "a" was somehow implied? I am sure he thinks we should at least be doing well with this.
     
  7. Jan 9, 2014 #6

    DrClaude

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    You did something wrong here. Why is there no A in this equation? Also, you are doing the integration wrong: it is a definite integral, i.e., with bounds on x.
     
  8. Jan 9, 2014 #7

    Ray Vickson

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    If ##x## is real, ##\psi=A x^3## may be complex because ##A## may be a complex number. We have
    [tex] \langle\psi|O|\psi\rangle = \int_{x=0}^{a} A^* x^3 \, O(x) \, A x^3 \, dx
    = |A|^2 \int_{x=0}^a O(x) x^3 \, dx [/tex]
    for any (operator) function ##O(x)##. Here, ##A^*## is the complex conjugate of ##A##: ##A^* = A## if ##A## is real, ##A^* = -A## if ##A## is pure imaginary, and ##A^* = u-iv## if ##A = u+iv## with real ##u,v##. In all cases, ##|A|^2\equiv A^* A = u^2+v^2## and ##|A| = \sqrt{u^2+v^2} > 0##.

    Putting ##O = 1## we need to have ## \langle\psi|1|\psi\rangle = \langle\psi|\psi\rangle = 1,## so that determines ##|A|## (but not ##A## itself). Then put ##O(x) = x## to finish the problem. The answer does contain the value of 'a', so if your instructor tells you otherwise then either he/she has made an error or has made a change of units. You should go back to square one and do both the integrals very carefully. Doing it yourself is the only real way to learn.
     
  9. Jan 10, 2014 #8
    Hi Dr. Claude...ok yes I am really not sure what I was writing there but what I meant was (x^7)/7=1. I really apologise...I was extremely sleep deprived when writing that post. And you are correct I was doing the integral from infinity to negative infinity and was thinking it would diverge and I wasn't understanding what I'm doing wrong... but my prof told me i need to normalize with the same bounds as the expectation value.
     
  10. Jan 10, 2014 #9
    Hello Ray,

    Thank you so much for taking the time to help me. I have been doing it myself but after the point that I wrote above is where i felt things were going awry. I spoke with my professor since my last post and he explained that we can find the value of "a" as 7/8 with the information given. He also told me to set the expectation value = to 1. I need to find a way to post my work. Give me a moment to post my handwritten work.
     
  11. Jan 10, 2014 #10

    Ray Vickson

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    I hope that is not what your professor told you, because it is wrong. We have
    [tex] \langle \psi | \psi \rangle=1 \Longrightarrow |A|^2 = \frac{7}{a^7} [/tex]
    so
    [tex] \langle x\rangle \equiv \langle \psi|x| \psi \rangle = \frac{7}{8} a [/tex]
    If you do set the expected value to 1 then you get ##a = 8/7.##
     
  12. Jan 11, 2014 #11
    Right. That is what I got when I set it equal to 1, but I still think there is something wrong in my method/reasoning, because I do not get the same solution u get for the normalization....I get ##a^7/7## if I put that in place of 1 and solve the equation without the A...I get 8/7. When I do it with the A I get A^2a=8/7. Maybe I am handling the A wrong and am not understanding what it is. I was under the impression that A is the amplitude set by the normalization, but I am treating it as variable and it seems I should not be doing that. I am apparently the only one in my class who cant get this right and he even wrote in my notebook beside my work to set the expectation value to one and solve to get 7/8 so this is really just messing up my understanding. I will post my work...I just need some batteries for my camera. Hopefully I will understand this soon. My textbook even arrived and I still feel unclear and it is really bothering me because this seems basic. I am sure when I post my work you will be able to spot immediately where I am going wrong and give me guidance.
     
  13. Jan 11, 2014 #12

    Ray Vickson

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    You have a wave function ##\psi(x) = A x^3 I_{[0,a] }(x),## where ##I_S(x)## is the indicator function of a set ##S##; that is,
    [tex] I_S(x) = \begin{cases} 1 & \text{ if } x \in S\\
    0 & \text{ if } x \not\in S
    \end{cases} [/tex]
    Here, ##x## is the variable, while ##A, a## are parameters; that is, they are constants throughout the problem. Part part of the problem is to figure out how they are related to each other. In other words, if somebody tells you the value of ##a## you can figure out what must be the value of ##A## by requiring normalization of the wave function on the interval ##0 \leq x \leq a##. (In fact, we cannot determine the "phase" of ##A## without more information, but at least we can get ##|A|##, and in this problem that is all we need.) Just remember: ##A## and ##a## are not variables.
     
  14. Jan 12, 2014 #13
    :biggrin: I think I have finally seen the light haha.
     
  15. Jan 12, 2014 #14
    Okay Ray so I think I was panicking too much about it so I decided to look at it again after sleep and recreation. I worked the problem again keeping what you told me in mind especially about how to handle A. Also I looked at what my professor wrote in my book and although he said "set the expectation value to 1" what he actually wrote in my book was the equation for normalization:
    http://imagizer.imageshack.us/v2/320x240q90/208/m2c0.jpg [Broken]
    So this is what I wrote this morning and like you, I'm convinced that the answer has to be in terms of ##a## although my prof made it clear to me that the answer is only 7/8 and for <x^2> the answer would be 5/6.
    http://imagizer.imageshack.us/v2/800x600q90/194/31zk.png [Broken]
     
    Last edited by a moderator: May 6, 2017
  16. Jan 12, 2014 #15
    It seems my problem was really with how I was handling "A" and I did not know how to apply the concept of normalization. What do you think?
     
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