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Expectation Value Dependence

  1. Nov 24, 2004 #1
    I had thought that the expectation value would be the same....whether you did it in momentum space or position space. Could someone explain what is going on in this particular problem?

    [tex]
    \psi (x) = \sqrt{b} e^{-b |x| + i p_0 x / \hbar }
    [/tex]

    Taking the Fourier transform, I can get this function in momentum space.

    [tex]
    \phi (p) = \sqrt{\frac{2b}{\pi \hbar}} \thickspace \frac{\hbar^2 b}{\hbar^2b^2+(p-p_0)^2}
    [/tex]

    Now, if I compute [itex] <p^2> [/itex], I should get the same value no matter which space I choose to do it in. However, this does not seem to be the case. I computed these by hand...and by Mathematica....so I am fairly confident that they are correct (just confusing).

    [tex]
    \begin{align*}
    <p^2> &= \int \phi^* p^2 \phi \, dp = b^2 \hbar^2 + p_0^2\\
    <p^2> &= \int \psi^* \left(\frac{\hbar}{i}\right)^2 \frac{\partial^2 \psi}{\partial x^2} \, dx = -b^2 \hbar^2 + p_0^2
    \end{align*}
    [/itex]

    As you can see...and please carry out the computations if you doubt them...the expectation value seems to depend on the space in which I computed it. I thought that this was NOT supposed to happen. Any ideas?
     
  2. jcsd
  3. Nov 24, 2004 #2

    Galileo

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    I get:

    [tex]
    \langle p^2 \rangle = b\sqrt b(\hbar^2 b+p_0^2)[/tex]
    I'm having trouble getting Maple to evaluate the other one...
     
  4. Nov 24, 2004 #3

    dextercioby

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    Are u both sure with those integrals??I mean,the initial wave function is different for the intervals minus infinity to zero and o to plus infinity.This should be taken into consideration for all three integrals (the Fourier transform and the 2 EV-s).
    I would advise you to do those integrals by hand,or maybe with the help of an integrals book.
    One other thing:
    That's definitely wrong.Check out the units.The constant "b" has the dimension length to the minus one.
     
  5. Nov 24, 2004 #4
    Hi

    The wave function you consider has no derivative for x=0.
    Its second-derivative should be a Dirac function for x=0.

    I think your strange results are related to this.
    Probably you dropped the terms related to the singular behaviour at x=0.

    Also, be careful with Mathematica because such problems can easily be missed.
    By the way, I would be interrested to know how you did the integral in Mathematica.
    How did you let Mathematica know that the integral converges for x->Infinity?

    Michel
     
  6. Nov 25, 2004 #5
    Here we go with the details:

    [tex]
    \psi (x) = \sqrt{b} e^{-b |x| + i p_0 x / \hbar }
    [/tex]

    Then,

    [tex]
    \begin{align*}
    \phi (p) &= \frac{1}{\sqrt{2\pi \hbar}} \int_{-\infty}^{\infty} \psi(x) e^{-i p x /
    \hbar} \, dx \\
    &= \frac{1}{\sqrt{2\pi \hbar}} \left( \int_{-\infty}^{0} \sqrt{b} e^{bx + ip_0 x/\hbar}
    e^{-i p x / \hbar} \, dx + \int_{0}^{\infty} \sqrt{b} e^{-bx + ip_0 x/\hbar}
    e^{-i p x / \hbar} \, dx \right) \\ &= \sqrt{\frac{2 b}{\pi \hbar}} \thickspace
    \frac{\hbar^2 b}{\hbar^2b^2 + (p - p_0)^2}
    \end{align*}
    [/tex]

    Now,

    [tex]
    \begin{align*}
    <p^2> &= \int_{-\infty}^{\infty} \phi^* p^2 \phi \, dp\\&\\
    &= \frac{2b^3\hbar^3}{\pi} \int_{-\infty}^{\infty}
    \frac{p^2}{[b^2\hbar^2+(p-p_0)^2]^2} \, dp\\&\\
    &= \frac{2b^3\hbar^3}{\pi} \, \frac{\pi (b^2\hbar^2+p_0^2)}{2b^3\hbar^3}\\&\\
    &= b^2 \hbar^2 + p_0^2
    \end{align*}
    [/tex]

    If you want to do this in Mathematica, you don't have to do any special tricks.
    Just set up the problem and integrate. However, you must use the Assuming[]
    command. For example:

    Assuming[ b > 0 && pz > 0 && h > 0, Integrate[p^2 /(b^2 h^2 + (p -pz)^2)^2, {p, -Infinity, Infinity}]]

    In the code above, I am letting [itex]h[/itex] be [itex] \hbar [/itex] and
    [itex]pz[/itex] be [itex] p_0 [/itex].

    I suspect you are correct, but I would like some help figuring out that exact
    location of the mistake. I would like to think that the x=0 point wouldn't
    matter since it is set of measure zero...but perhaps not.

    Now, we do it in position space. Sorry, I don't have the in-between
    steps...basically you have to break the integral up into two separate integrals.
    I will provide Mathematica code to get the answer.

    [tex]
    \begin{align*}
    <p^2> &= \int_{-\infty}^{\infty} \psi^* \left(\frac{h}{i}\right)^2
    \frac{\partial^2 \psi}{\partial x^2} \, dx\\
    &= \int_{-\infty}^{0} \sqrt{b} e^{bx - ip_0 x/\hbar}\left(\frac{h}{i}\right)^2
    \frac{\partial^2}{\partial x^2}(\sqrt{b} e^{bx + ip_0 x/\hbar}) \, dx+
    \int_{0}^{\infty} \sqrt{b} e^{-bx - ip_0 x/\hbar}\left(\frac{h}{i}\right)^2
    \frac{\partial^2}{\partial x^2}(\sqrt{b} e^{-bx + ip_0 x/\hbar}) \, dx\\
    &=-\frac{1}{2}(b \hbar + i p_0)^2 + -\frac{1}{2}(b\hbar-i p_0)^2\\
    &=-b^2\hbar^2 + p_0^2
    \end{align*}
    [/itex]

    Sample mathematica code...I could have defined functions, but I took the lazy way out. This code should work in Mathematica 5.

    ---begin code----

    psineg = Sqrt Exp[b x + I pz x /h];
    psinegcon = Sqrt Exp[b x - I pz x /h];
    psipos = Sqrt Exp[-b x + I pz x /h];
    psiposcon = Sqrt Exp[-b x - I pz x /h];

    Assuming[h > 0 && pz > 0 && b > 0, Integrate[psinegcon (h/I)^2 D[psineg, {x, 2}], {x, -Infinity, 0}]]

    Assuming[h > 0 && pz > 0 && b > 0, Integrate[psiposcon (h/I)^2 D[psipos, {x, 2}], {x, 0, Infinity}]]

    ---end code---

    Then just add the two results...or have Mathematica do it.

    Anyway, what we see is that the expectation value is different depending on the space in which I computed it. I also believe that this could be related to the fact that [itex] \psi [/itex] is not differentiable at [itex] x=0 [/itex]. However, I also feel that the math is correct...so where did I go wrong? As far as [itex] \psi [/itex] not being differentiable at zero, it seems like this might not matter since I break the integral into two separate integrals....when I take the derivative over these secondary ranges, [itex] \psi [/itex] is differentiable---and the point x=0 is no longer an issue.
     
  7. Nov 25, 2004 #6
    Hello

    I think you can go through the same calculations, but you should not forget the contribution at x=0.
    You simply need to take the following rules into account (I think they are ok, but check it):

    [tex]\frac{d|x|}{dx} = 2H(x)-1[/tex]​
    [tex]\frac{dH(x)}{dx} = \delta(x)[/tex]​
    Where H(x) is the Heaviside function, (H(x)=0 for x<0 and 1 for x>0), and [tex]\delta(x)[/tex] is the Dirac distribution.

    Thanks for your indications on Mathematica, I will try it at home this evening.
    My version of Mathematica is more than 10 years old.
    It does not know the "Assuming" function.
    There is maybe a replacement, but I am not sure.

    I hope you will get the correct results.
    However, I am wondering now what is the physical meaning of all that?
    Clearly the singularity at x=0 is associated with high frequency components in the momentum spectrum.

    Your question triggered a vacuum in my recollections about quantum mechanics.
    I can imagine that operators in different representations (x or p) are equal if their averages are always equal too, of course.
    But why is it that p² in the different representations are obtained just by squaring in these two representations.
    I think this is true for any function of operator.
     
  8. Nov 25, 2004 #7
    It does matter if [tex]\psi[/tex] is not differentiable at 0. The only way it wouldnt matter is if as the integrand was evaluated tending to 0 from the right and tending to 0 from the left, the limit was the same. If this isn't the case, then you must find another way to evaluate the improper integral.

    masud.
     
  9. Dec 2, 2004 #8
    It doesn't matter what you are calculating the expectation value of, you should always get the same result in any basis. Your choice of space is just a choice of basis which you integrate out.
     
  10. Dec 6, 2004 #9
    A very good example to understand better the use of the unbounded operators and the basis change (or the fourier transforms if you prefer).

    You have not the same results because lim_p p^2.phi(p) is not zero thus p^2.phi(p) is not square integrable. p^2|phi> does not belong to the hilbert space.
    Thus you cannot apply simply the basis transformation to p^2.phi(p) (you need to use distribution theory,i.e. enlarge the hilbert space, to recover the result).

    Seratend.
     
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