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<p> Operator on Probability Density in X-Space

  1. Oct 29, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider a particle whose wave function is:

    2\alpha^{3/2}xe^{-\alpha x} & \text{if} & x> 0\\
    0 & \text{if} & x\leq 0

    Calculate <p> using the [tex]\hat{p}[/tex] operator on probability density in x space.

    2. Relevant equations

    [tex]\hat{p}=-i \hbar \frac{\partial}{\partial x}[/tex]

    3. The attempt at a solution
    So I'm pretty sure I'm missing something, I figured since you take the derivative then do the integral, they will "cancel" each other, then you can just evaluate -ih|Psi|^2 from 0 to infinity, in which case you get 0. Here is my work:

    [tex]<p> =\int_{0}^{\infty}\hat{p}|\Psi(x)|^{2}dx=\int_{0}^{\infty}-i \hbar 4\alpha^{3} \frac{\partial}{\partial x} x^{2} e^{-2 \alpha x}dx=-i \hbar 4\alpha^{3}\int_{0}^{\infty}\frac{\partial}{\partial x} x^{2} e^{-2 \alpha x}dx [/tex]

    Then when you take the derivative, then integrate it over that range:

    [tex]=-i \hbar 4\alpha^{3} x^{2} e^{-2 \alpha x}|_{0}^{\infty}=0[/tex]

    And when you evaluate this it goes to 0, did I do something wrong? TBH, I did miss this lecture, and unfortunately my professor forgot to record it like he usually does, so I had to take notes from his PowerPoint, as well as a friend. If it matters, I have already found <p>, but only after I did the Fourier Transform from the position space. It turned out to be:

    [tex]<p> =\frac{-\alpha \hbar}{3\pi}[/tex]

    If I did it correctly. So can anyone give me insight on what I did wrong? Any help is appreciated.
  2. jcsd
  3. Oct 29, 2013 #2

    Simon Bridge

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    $$\langle \hat{p}\rangle = \int_0^\infty \psi^\star (x)\hat{p}\psi (x) dx$$ ... spot the difference.
  4. Oct 29, 2013 #3
    Oh, I see my problem, thank you, so I need to apply the p operator to psi before multiplying by the conjugate?
  5. Oct 29, 2013 #4

    Simon Bridge

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    That's the one - the relation you used only works for operators like position.
    In general, the operator goes in between the wavefunctions.

    In Dirac notation: <Q>=<ψ|Q|ψ>
  6. Oct 29, 2013 #5


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    Yes, and the complete derivation is as follows
    [tex]\langle A \rangle =\langle \psi|\mathbf{A}| \psi \rangle= \int_{\mathbb{R}} \mathrm{d} x \langle \psi|x \rangle \langle x|\mathbf{A}|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} x \; \psi^*(x) \hat{A} \psi(x).[/tex]
    Note that there is a subtle difference in my use of symbols. While [itex]\mathbf{A}[/itex] is the operator representing the observable [itex]A[/itex] in abstract (basis independent) Hilbert space, [itex]\hat{A}[/itex] denotes the representation of the operator in position representation. Your equation for [itex]\hat{p}[/itex] is of course correct.
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