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Expectation value of momentum

  1. Nov 2, 2011 #1
    Ψ(x)=(2/a)^(1/2) [csin(nπx/a)]

    The Expectation value of momentum <P>=∫Ψ*(x)[-ih d/dx ] Ψ(x) dx = 0
    the average momentum is zero.It means the particle is moving equally in the +x and -x.

    And
    if Ψ(x)=(2/a)^(1/2) {csin(πx/a)+dsin(2πx/a)}

    I calculate the average momentum is also zero,too,
    Is this means the same?
    My teacher didn't tell me the answer,and he say it is not zero "intuitively".
    But i think it is zero,isn't?

    Because Ψ(x)=(2/a)^(1/2) [csin(πx/a)]+ (2/a)^(1/2) [dsin(2πx/a)]
    according to the Ψ(x)=(2/a)^(1/2) [csin(nπx/a)] ---> <p>=0
    so <p>=0+0=0



    thanks.
     
    Last edited: Nov 2, 2011
  2. jcsd
  3. Nov 2, 2011 #2
    That's not quite what it means. It means that upon measurement, the particle has an equal likelihood of being found moving in either the +x or -x direction (I assume we're talking about a particle in a box).

    For any* wavefunction at all (even something weird like [itex]\psi(x) = 3x^x[/itex]), the expected momentum will be 0 so long as [itex]\psi(0) = \psi(a) = 0[/itex].

    Here's why:

    [tex]\langle p_x \rangle = \int_{0}^{a} \psi^{*}(x) \left[ -i \hbar \frac{d\psi}{dx} \right]\, dx [/tex]

    [tex]\langle p_x \rangle = -i \hbar \int_{0}^{a} \psi(x) \frac{d\psi}{dx}\, dx [/tex]

    [tex]\langle p_x \rangle = \frac{-i \hbar}{2}\left[ \psi(a)^2 - \psi(0)^2 \right] [/tex]

    [tex]\langle p_x \rangle = 0[/tex]

    *The above math does not verify if this is true for complex wavefunctions, but my guess is "probably". If someone wants to prove/disprove that, feel free.
     
    Last edited: Nov 2, 2011
  4. Nov 2, 2011 #3
    so if the Ψ(x)=(2/a)^(1/2) {csin(πx/a)+dsin(2πx/a)}
    do the <P>=0 ??
     
  5. Nov 2, 2011 #4
    Yep -- for a particle in a box. (Although I am assuming you meant to put an "n" in the argument of each sine function).
     
  6. Nov 2, 2011 #5
    thanks ^^
     
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