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Can expectation value of observables be imaginary?

  1. May 18, 2008 #1
    I am quite new to Quantum Mechanics and I am studying it from the book by Griffiths, as kind of a self-study..no instructor and all....

    For a gaussian wavefunction [tex]\Psi[/tex]=Aexp(-x[tex]^{2}[/tex]), I calculated
    <p[tex]^{2}[/tex]> and found it to be equal to ah[tex]^{2}[/tex]/(1-2aiht/m)

    (By h I mean h-bar..not so good at latex yet)

    Can it be correct? How can expectation value of p[tex]^{2}[/tex] be imaginary?
  2. jcsd
  3. May 18, 2008 #2


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    That's not what I got. How did you go about calculating it?
  4. May 18, 2008 #3
    What did you get? <p^2>= ah ?
    we know psi(x,0) = Aexp(-x^2)
    so i first calulated phi(k) integrating (1/[(2*pi)^(1/2)]) psi(x,0)*exp(-ikx) dx from -inf to inf
    then got psi(x,t) (1/[(2*pi)^(1/2)]) phi(k)*exp(i(kx-((h^2)*(k^2)*t)/2m) dk from -inf to inf

    till that part im correct, since the answer is given in the book and I verified.

    For the next part, I got <x>=<p>=0 and <x^2>=1/(4*w^2) where w=[a/(1+4a2h2t2/m2)] (the 2's are squares)
    It's fine till then

    Then got <p^2> by doing (-h^2)* integral psi(x,t)* (d^2/dx^2) psi(x,t) dx
    the problem is that i get imaginary constant (1-2aiht/m) each time i differentiate psi(x,t)
    and this doesn't get cancelled out later.
  5. May 18, 2008 #4
  6. May 18, 2008 #5


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    but you take the derivative twice, then you should not have any imaginary part left.
  7. May 18, 2008 #6
    (My previous post was a mistake ignore it..)

    Here's how I did it..
    Now we find [tex]\psi(x,t)[/tex] as
    where A=[tex](\stackrel{2a}{\overline{\pi}})^{\stackrel{1}{\overline{4}}}[/tex]
    (A should be in the numerator)

    So [tex]\psi(x,t)*=A\stackrel{e^{\stackrel{-ax^{2}}{\overline{1-\stackrel{2aiht}{\overline{m}}}}}}{\overline{\sqrt{1-\stackrel{2aiht}{\overline{m}}}}}
    where C is an imaginary factor which gives real factor when combined with the similar factor from [tex]\psi(x,t)[/tex]

    But the imaginary exponential factor comes out in the form of [tex](1+\stackrel{2aiht}{\overline{m}})^{2}[/tex] when we compute <p[tex]^{2}[/tex]> and doesn't get cancelled by some other real factor. Or am I making some calculation mistake?
  8. May 18, 2008 #7


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    I cant follow what you are doing, it is just a mess..

    you want to do <p^2> when you have the wave function in position representation. Then you do:

    [tex] <p^2> = \int_{-\infty}^{\infty} \psi (x)^* (-i\hbar\frac{d}{dx})^2\psi (x) dx [/tex]

    since [tex]\hat{p} = -i\hbar\frac{d}{dx} [/tex] in position representation.

    Where [tex] \psi (x) = \Psi (x,t=0) = A\exp(-x) [/tex]

    That is all you have to do man.

    You can take the wave function in momentum representation:

    [tex] \phi (k) = \dfrac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \psi (x) \exp{-ikx} dx [/tex]

    Then the expectation value of p^2 is:

    [tex] <p^2> = \int_{-\infty}^{\infty} \phi(p)^* \hat{p} ^2\phi (p) dp [/tex]

    where [tex]\hat{p} [/tex] is the momentum operator in MOMENTUM representation (i.e [tex] \hat{p} = p [/tex]), compare with the position operator in position representation: [tex] \hat{x} = x [/tex].

    compare also with:
    [tex] <x^2> = \int_{-\infty}^{\infty} \psi(x)^* \hat{x}^2\psi (x) dx [/tex]

    It seems to me that you have mixed the two representations, you can't do that.
    Last edited: May 18, 2008
  9. May 18, 2008 #8


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    and if you want to include time dependence, you apply the time evolution operator, and your position wave function as a function of time, becomes:

    [tex] \Psi (x,t) = A\exp(-x)\exp{(-iEt/\hbar)} [/tex]

    Then the expectation value becomes:
    [tex] <p^2>(t) = \int_{-\infty}^{\infty} \Psi (x,t)^* (-i\hbar\frac{d}{dx})^2\Psi (x,t) dx =<p^2>\exp{(Et/\hbar)}[/tex]
  10. May 18, 2008 #9
    yeah thats what I have done..just that [tex]\psi(x,t)[/tex] is quite clumsy..
    So, whats is the answer u got?

    In the book its given as [tex]<p^{2}>=ah[/tex] but I can't find the mistake I'm making so I'm assuming that the author has calculated it for t=0.
  11. May 18, 2008 #10


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    is [tex] A = \frac{2a}{\sqrt{\sqrt{\pi}}} [/tex] ?

    Can you tell me what
    [tex] (-i\hbar\frac{d}{dx})^2\psi (x) = (-i\hbar\frac{d}{dx})^2A\exp(-x^2) [/tex]

    is? (I know, but do you?)
  12. May 18, 2008 #11
    No, I got A as [tex]
    A = \frac{\sqrt{\sqrt{2a}}}{\sqrt{\sqrt{\pi}}}

    I'll do [tex]-\hbar^{2}\frac{d^{2}}{dx^{2}}(Aexp(-kx^{2})[/tex]
    This would be [tex]
    -A\hbar^{2}(-2k)\frac{d}{dx}(xexp(-kx^2)) = 2kA\hbar^{2}(-2kx^{2}exp(-kx^{2})+exp(-kx^{2}

    But I guess my problem occurs only when we consider time dependance, in which case k has an imaginary part as well.(In fact I got k as [tex]\frac{a}{1+\frac{2aiht}{m}}[/tex])
    Last edited: May 18, 2008
  13. May 18, 2008 #12


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    I don't follow.
    You said:
    Where does the k come from?
  14. May 18, 2008 #13
    The value of psi(x,t) I obtained, which is correct, as per the book, is
    \Psi(x,t)=A\frac{e^{\frac{-ax^{2}}{1+\frac{2ai\hbar t}{m}}}}{\sqrt{1+\frac{2ai\hbar t}{m}}}

    So basically [tex]k = \frac{a(1-\frac{2ai\hbar t}{m})}{1+\frac{4a^{2}\hbar^{2}t^{2}}{m}}=w^{2}(1-\frac{2ai\hbar t}{m})[/tex]

    \Psi(x,t)=A\frac{e^{-w^{2}x^{2}(1-\frac{2ai\hbar t}{m})}}{\sqrt{1+\frac{2ai\hbar t}{m}}}
    \Psi(x,t)*=A\frac{e^{-w^{2}x^{2}(1+\frac{2ai\hbar t}{m})}}{\sqrt{1-\frac{2ai\hbar t}{m}}}
  15. May 18, 2008 #14
    So [tex]\Psi(x,t)*\hat{p}^{2}\Psi(x,t) = A\frac{e^{-w^{2}x^{2}(1+\frac{2ai\hbar t}{m})}}{\sqrt{1-\frac{2ai\hbar t}{m}}}
    (2\hbar^{2}A\frac{e^{-w^{2}x^{2}(1-\frac{2ai\hbar t}{m})}}{\sqrt{1+\frac{2ai\hbar t}{m}}}w^{2}(1-\frac{2ai\hbar t}{m})
    (1-2w^{2}(1-\frac{2ai\hbar t}{m})
    x^{2})) [/tex]
    [tex]=2\hbar^{2}A^{2}w^{4}(1-\frac{2ai\hbar t}{m})
    (1-2w^{2}(1-\frac{2ai\hbar t}{m})

    Now, if I denote [tex](1-\frac{2ai\hbar t}{m}) [/tex] by z ,
    [tex]\Psi(x,t)*\hat{p}^{2}\Psi(x,t) = \frac{2\hbar^{2}A^{2}w^{4}}{\sqrt{a}}z

    Now integration doesn't remove its dependence on z, which is imaginary. In which step am I making a mistake? I am not able to spot it...its very difficult to spot silly mistakes in such cluttered maths..
    And what is the value I should get? or what is the value you guys got?
  16. May 18, 2008 #15


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    You have wrong wave function...you are mixing the two representations!

    [tex] \Psi (x,t) = A\exp(-x^2)\exp{(-iEt/\hbar)} [/tex] This is not what are calculating with, you use: [tex] \Psi(x,t)=A\frac{e^{-w^{2}x^{2}(1-\frac{2ai\hbar t}{m})}}{\sqrt{1+\frac{2ai\hbar t}{m}}} [/tex]

    Can you explain why you use that?

    [tex] <p^2>(t) = \int_{-\infty}^{\infty} \Psi (x,t)^* (-i\hbar\frac{d}{dx})^2\Psi (x,t) dx = <p^2>exp{(Et/\hbar)}[/tex]

    Then why did you do:


    Your wave function was: Aexp(-x^2), that was why I asked you to do:

    [tex] (-i\hbar\frac{d}{dx})^2\psi (x) = (-i\hbar\frac{d}{dx})^2A\exp(-x^2) [/tex]
    Last edited: May 18, 2008
  17. May 18, 2008 #16
    I use that because that is the wavefunction I have. [tex]
    \Psi (x,t) = A\exp(-x^2)\exp{(-iEt/\hbar)}
    is not the wavefunction. This part has been verified by the book as well. I have to use the superposition of states with all energies. How can I use only a single energy E?
  18. May 18, 2008 #17
    The question states that:
    "A free particle has the initial wave function
    [tex]\Psi(x,0)=Ae^{-ax^{2}}[/tex] "

    So this is the INITIAL wavefunction.
    To find [tex]\Psi(x,t)[/tex] we need to integrate it over all energies right?

    \Psi(x,t)=\frac{1}{\sqrt{2\pi}\int_{-\infty}^{\infty}\phi(k)e^{i(kx-\frac{\hbar k^{2}}{2m}t)}

    Isn't this how we should represent [tex]\Psi(x,t) [/tex]?
    Last edited: May 18, 2008
  19. May 18, 2008 #18


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    does the energy depend on x?

    Time evolution of a state is:


    [tex] \Psi (x,t) = \Psi (x, t=0) e^{-iEt/\hbar} [/tex]

    isn't [tex] E = \frac{p^2}{2m} = \frac{\hbar^2k^2}{2m} [/tex] In your case?

    And since the energy don't depend on x:
    [tex] <p^2>(t) = \int_{-\infty}^{\infty} \Psi (x,t)^* (-i\hbar\frac{d}{dx})^2\Psi (x,t) dx \exp{(Et/\hbar)}[/tex]

    [tex] <p^2> = \int_{-\infty}^{\infty} \psi (x)^* (-i\hbar\frac{d}{dx})^2\psi (x) dx [/tex]

    [tex] \psi (x) = \Psi (x,t=0) = A\exp(-x^2) [/tex]
    Last edited: May 18, 2008
  20. May 18, 2008 #19
    This is what I did not get... Dont you have to take superpostition over all states/ all energies?
    Because in the case of free particle, the value of [tex]\psi(x)[/tex] you get from solving the time-independent SE is not normalizable. So in this case should not we use [tex]

    \Psi(x,t)=\frac{1}{\sqrt{2\pi}\int_{-\infty}^{\infty}\phi(k)e^{i(kx-\frac{\hbar k^{2}}{2m}t)}

    [/tex] and then apply <p^2> on that [tex]\Psi(x,t)[/tex] we obtain?
    Last edited: May 18, 2008
  21. May 18, 2008 #20


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    The time evolution of a state is obtained by applying the time evolution operator, always. It is a fundemantal cornerstone in non relativistic quantum mechanics.

    The energy is [tex] E = \frac{p^2}{2m} = \frac{\hbar^2k^2}{2m} [/tex] right?
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