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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:

    http://en.wikipedia.org/wiki/Hamiltonian_(quantum_mechanics [Broken])

    [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 by a moderator: May 3, 2017
  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 by a moderator: May 3, 2017
  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?
  22. May 18, 2008 #21


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    "Because in the case of free particle, the value of you get from solving the time-independent SE is not normalizable. "

    [tex] \hat{H}\psi (x) = E\psi(x) \Rightarrow -\hbar^2\frac{1}{2m}\frac{d^2\psi(x)}{dx^2} = E\psi(x) \Rightarrow [/tex]

    [tex] \psi (x) = A e^{i\sqrt{\frac{2mE}{\hbar^2}}x} [/tex]

    A can not be found, since this is not a wave function that CAN be normalized (not all wave functions can be normalized). However, the Gaussian shaped WF of yours can be normalized.

    So the plane wave is a special case.
  23. May 19, 2008 #22
    Yes that is ok.. But we know only the initial wavefunction psi(x,0). We know that that is gaussian. So If we are to find out how it evolves with time(we want to find psi(x,t) we should multiply it by exp(-iEt/hbar) . Now that gives the wavefunction for a particular value of energy E.

    The actual WF should be a supersposition of all such states with all possible values of E. So, here we should integrate with respect to k (that is, dk). Since E depends on k, we cannot take the E term outside.

    Besides, <p^2>(t) = <p^2>exp(-iEt/hbar) doesn't make sense because E itself depends on p.
    That is why I think we should 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}}}
  24. May 19, 2008 #23


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    1. By definition, the expectation value of any observable is real. Let
    P(x,t) = W(x,t)* W(x,t), which is the probability density to find the particle at x at time t, given that W(x,t) is the wave function. Clearly P is real. Let <O> denote the expectation value of an observable O. As O is self-adjoint, and/or Hermitean, and has real eigenvalues, then surely

    <O> = ∫dxP(x,t) O(x,t)

    is real. So any computation that yields an imaginary component to <O> is in error.

    2. The rumors that plane waves, exp(ikx-iEt) cannot be normalized are without foundation -- just takes a flip or two of the wrist, first done by Hermann Weyl during the first part of the last century in a rather monumental effort. Then, Dirac made it easy with his delta function, ultimately made fully rigorous with Laurent Schwartz's Theory of Distributions. (If you have any doubts about this matter, I highly recommend the superb book, Fourier Analysis and Generalized Functions by M.J. Lighthill, Cambridge U. Press. The controversy on this issue was solved a century ago. So, not to worry.)

    3. Since there seems to be a bit of a hangup with Gaussian integrals, I suggest an easier problem; one that should illuminate the entire process of determining expectation values. That is, let

    W(x,t) = a exp(i(p1)x-iE1t) + b exp(i(p2)x-E2t)

    and find the expectation value of p and p^^2. (If you are still confused about the normalization of continuous spectra, then use a big box to contain the system.)

    4. Then, armed with the confidence of new knowledge, do the Gaussian thing in the momentum representation.

    Reilly Atkinson
  25. May 19, 2008 #24


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    It might be worth saying precisely what you mean -- naively, "[itex]\psi[/itex] is normalized" would mean that the inner product [itex]\langle \psi, \psi \rangle[/itex] evaluates to 1, but the inner product cannot be (continuously) extended to a pair of Schwartz distributions.
  26. May 20, 2008 #25
    In this case, [tex]
    \frac{d^{2}W}{dx^{2}} [/tex]

    [tex]= \frac{d}{dx}(ai p_{1}e^{i(p_{1}x-E_{1}t)}+bi p_{2}e^{i(p_{2}x-E_{2}t)}) [/tex]

    = -(ap_{1}^{2}e^{i(p_{1}x-E_{1}t)}+bp_{2}^{2}e^{i(p_{2}x-E_{2}t)})

    So, [tex]
    W*\hat{p}^{2}W = -\hbar^{2}[ae^{ip_{1}x+iE_{1}t} + be^{ip_{2}x+iE_{2}t}

    = a^{2}p_{1}^{2}\hbar^{2}e^{2ip_{1}x}+b^{2}p_{2}^{2}\hbar^{2}e^{2ip_{2}x}+abp_{2}^{2}\hbar^{2}e^{i(p_{1}+p_{2})x}e^{i(E_{1}-E_{2})t}+abp_{1}^{2}\hbar^{2}e^{i(p_{1}+p_{2})x}e^{i(E_{2}-E_{1})t}

    Are my steps correct till now?
    Last edited: May 20, 2008
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