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Intro to Quantum Homework Help! Please!

  1. Oct 4, 2005 #1
    I am having trouble with 2 problems about the harmonic oscilator.

    First of all the question tells me to use the wave function for when n=1

    So I have:
    Wavefunction = A[1]*r^1/2 * x * e^-((r^2)/2)

    where r= (mw/(hbar))^1/2
    I am wondering when I am multiplying this by the complex conjugate, is the complex conjugate going to have a e^+((r^2)/2). I dont think I am but for some reason, in a solution finding <x> which equals ](integral) psi* (times) x (times) psi] those exponential functions did not appear. So I was wondering how they cancelled out when you calculate <x>

    also it asks me to calculate <p>, which I know equals m*d<x>/dt. My question is, if my <x> does not depend on t, because the time dependent part of the equation was cancelled out when solving for <x>, how can I find d<x>/dt?

    Any help would be appreciated, I'm sure my questions or equations aren't too clear, so if you have any questions, I will try to clarify. Thank you.
     
    Last edited: Oct 4, 2005
  2. jcsd
  3. Oct 4, 2005 #2
    bump thanks in advance
     
  4. Oct 4, 2005 #3

    Tom Mattson

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    Just a little nitpick: You forgot the 'x'.

    No, [itex]e^{-r^2/2}[/itex] is a real-valued function. Real quantities are equal to their own complex conjugates.

    That's because those exponential functions (I assume you mean [itex]e^{-ikx}[/itex]) are complex-valued, and in general are not equal to their own complex conjugate.

    For a state function [itex]\psi(x)[/itex] and an operator [itex]\hat{O}[/itex] the expectation value [itex]\left<\hat{O}\right>[/itex] is given by:

    [tex]\left<\hat{O}\right>=\int_{-\infty}^{\infty}\psi^*\hat{O}\psi dx[/tex]

    Use your wavefunction and the fact that the momentum operator is [itex]\hat{p}=-i\hbar\frac{d}{dx}[/itex]. Try to use a symmetry argument when it comes time to integrate.
     
  5. Oct 4, 2005 #4
    Ok, so this is what I tought,

    but what I ment was that the real expoential value does not show up in the answer I was provided.


    The problem is I was told by a grad student that

    <x> = integral of a[1]^2*((hw)/(hbar))*x^3

    so somehow that e^(-(r^2)/2) that should have come up in both the wavefunction and the complex conjugate, dissapear when evaluating. Also I was told I could use <p>=m*d<x>/dt, and use <T> (kenetic energy) = <p^2>/2m

    My answer, and theirs, for <x> does not depend on t, so I am dumbfounded on how they got their solution and how they expect me to apply it.

    I came here thinking the same thing you told me. I dont understand how the intstructor and company were able to come up with that solution for <x> and also, how I can use <x>=m*d<x>/dt. My teacher is a real stickler to keep wth his method, do you know of any way they could be correct? Thank you for you help.
     
    Last edited: Oct 4, 2005
  6. Oct 4, 2005 #5

    Tom Mattson

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    What, the answer for <x>? Of course it doesn't appear there, because you're doing a definite integral. That doesn't mean that a real exponential does not equal its own complex conjugate.

    Have you tried to evaluate it? If so, then how far have you gotten?

    Why are you dumbfounded? Surely you know how to differentiate a constant function!

    I can't say if your teacher is correct without knowing his answer. What did he get for <x> and for <p>?
     
  7. Oct 4, 2005 #6

    ya, let me clarify, what he gave me was that
    <x> = Integral {Sumation: a[n]*((mw)/(hbar)*x^2n+1 dx}

    I got rid of the sumation becuase N=1 here, so I evaluated this to
    <x> = integral { a[1]*(mw/(hbar)*x^3 }

    so my answer was <x> = a[1]*((mw)/(hbar)*1/4*x^4

    so thats why I have gotten confused tho exponent, from the time independent wavefunction (e^(-r^2)/2) dissapears before i integrate according to him. The only reason for that, that I could come up with was that if somehwere in psi* I had e^(r^2/2).






    Well wouldnt that mean that then <p>=0? and thus kenetic energy=0 and I was told that KE and PE were going to both equal to 1/2 the total energy. So i dont think i can have a zero KE.


    He didn't really give me the the final answer, he just gave me that hint. And I asked him ok, well when I calculate <p>, can I use <p>=mdx/dt, and he said yes. But I got a hold of someone and said to use the operator function for <p>. So as you can see im a little confused :(.
     
    Last edited: Oct 4, 2005
  8. Oct 4, 2005 #7

    Tom Mattson

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    That's wrong, <x> is not a function of x. As I said before you're doing a definite integral over x. x cannot be in the answer.

    It sounds like whoever was helping you is extremely confused himself. Look, you've got an integral of the form:

    [tex]
    \int_{-\infty}^{\infty}x^3e^{-ax^2}dx
    [/tex]

    Since the integrand is odd, the integral--and hence <x>--is obviously zero.

    Yes, it would. You would get the exact same answer if you did it the way I recommended.

    The KE isn't zero, because:

    [tex]\left<\hat{T}\right>\neq\frac{\left<\hat{p}\right>^2}{2m}[/tex]

    Rather, we have:

    [tex]\left<\hat{T}\right>=\frac{\left<\hat{p}^2\right>}{2m}[/tex]

    which is completely different.

    You'll get the same answer no matter which method you use.
     
    Last edited: Oct 4, 2005
  9. Oct 4, 2005 #8
    So wait wouldn't I have: integral A[1]^2*(mw/(hbar)*x^3*e^-r^2. And you're saying that over -infinity to infinity that equals zero becuase the inegrand is odd? i dont see how thats odd or that becuase its odd, it would equal zero. but thats the least of my worries.



    ok so then does <p^2>=m d<x^2>/dt equal zero? becuase 0^2=0?

    how do I compute T then?'



    also I have that to find <v>

    where <v>=1/2*m*w^2*integral{psi*(times)x^2(times)psi}

    so now if I know that integral{psi*(times)x(times)psi}=0

    wouldn't <v>=0? or am i missing something?
     
    Last edited: Oct 4, 2005
  10. Oct 4, 2005 #9

    Tom Mattson

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    You're right, that should be x3 in there. There was a typo in my integral, but it's fixed now.

    The integrand is clearly odd, because if you substitute [itex]-x[/itex] in for [itex]x[/itex], you pick up an overall negative sign on the integrand. And it is a well-known fact from calculus that definite integrals over odd integrands vanish. You need to review your fundamentals.

    No, because [itex]\left<\hat{p}^2\right>\neq0[/itex].

    Put the expression for [itex]\hat{T}[/itex] into the following equation:

    [tex]\left<\hat{O}\right>=\int_{-\infty}^{\infty}\psi^*\hat{O}\psi dx[/tex]

    You will find that it is not zero.

    This looks very strange. What is v, and why are you putting x^2 in the integrand?

    This doesn't make any sense at all.

    It is true that:

    [tex]\int_{-\infty}^{\infty}\psi^*x\psi dx=0[/tex]

    But this does not in any way imply that:

    [tex]\int_{-\infty}^{\infty}\psi^*x^2\psi dx=0[/tex]

    and in fact the last integral is wrong.
     
    Last edited: Oct 4, 2005
  11. Oct 5, 2005 #10

    Tom Mattson

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    Never mind this bit, I get it now. I was thinking that v stood for velocity (which is why I thought it strange, this being QM and all). But clearly v stands for potential energy. I was thrown off because PE is usually represented by a capital V.

    The rest of my comments are unchanged though.
     
  12. Oct 5, 2005 #11
    ok, so I undestand now why <x>=0, and why <p>=0

    So lets see if I have <x^2> straightened out.

    <x^2>= integral {psi*(times)x^2(times)psi}

    = integral {a[1]((mw)/(hbar))*x^4*e^(-(mw)/(hbar))x^2


    and that wont equal zero becuase its a even function. I would probibaly solve this through integration by parts? It would take a few steps though...no?

    so once I have <x^2> I could try <p^2>=md<x^2>/dt

    OR

    <p^2> = Integral {psi* (times) -i^2 (times) hbar^2 (times) d^2psi/dx^2}


    after calculating that

    To find <T> I use:

    <T> = <p^2>/2m

    and then for average potential I use

    <V>= 1/2*m*w^2*[integral {psi* (times) x^2 (times) psi}


    Sry for the strange way I write the equations of the interweb, I gotta get me one of those equation editors :)
     
    Last edited: Oct 5, 2005
  13. Oct 5, 2005 #12

    Tom Mattson

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    Yes.

    No.

    Go to an integral table and look up the following class of integrals:

    [tex]\int_0^{\infty}x^ne^{-ax^2}dx[/tex]

    You can use that when calculating both the PE and the KE.

    The equation editor is built right into this website, and you can use it. If you click on any of my equations you will see a pop-up that has the source code. Follow this link tutorials and more samples.
     
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