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Given a particle's wave function, what is the general method of finding the probability distribution of momentum (i.e., the probability of finding that particle with a particular value of momentum)?

For example, given $$ \psi(x) = a e ^ { ikx } + b e ^ { -ikx }, $$ what is the probability of measuring $ p_x = \hbar k $ ?

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    $\begingroup$ You should ask instead what is the probability to find momentum $p_x = \hbar k$ within a small interval $\hbar \, dk$. $\endgroup$
    – Cham
    Commented Jan 25, 2018 at 17:44
  • $\begingroup$ @Someone - The wave function is the sum of two momentum eigenstates, so the momentum can have only two values. (See Mike's answer) $\endgroup$
    – user155325
    Commented Jan 25, 2018 at 19:10
  • $\begingroup$ Then your wave function cannot be normalized, unless there are some boundary conditions specified (particle in a box ?). The wave function should then cancel at the boundary : $\psi(0) = \psi(L) = 0$. $\endgroup$
    – Cham
    Commented Jan 25, 2018 at 19:19
  • $\begingroup$ Well, I never quite said that the "momentum can have only two values"; I said that the answer is zero everywhere except those two values. The Dirac distribution is funny (like distributions in general). All we really know is that when you integrate over some region that includes its nonzero point, you get a nonzero result. So @Someone is right; I avoided commenting precisely because explaining the subtleties of the Dirac $\delta$ is outside the scope of my answer, and more in the scope of reading up on it for homework. $\endgroup$
    – Mike
    Commented Jan 25, 2018 at 20:08

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First of all, I assume that you know that the probability (density) distribution is given by the squared amplitude of the wavefunction. In your example, you are given the wavefunction in the position basis, so it gives you the position probability density. If you want the momentum probability density, you have to change basis to get $\psi(p)$.

The standard way to change basis is to use a "resolution of the identity". The manipulation is easiest in bra-ket notation. You start out with \begin{equation} \psi(x) = \langle x | \psi \rangle = a\, e^{i k x} + b\, e^{-i k x}. \end{equation} But you want $\psi(p) = \langle p | \psi \rangle$. You can find it with this manipulation: \begin{align} \langle p | \psi \rangle &= \int_{-\infty}^{\infty} \langle p | x \rangle \langle x| \psi \rangle\, d x \\ &= \int_{-\infty}^{\infty} \frac{1} {\sqrt{2\pi \hbar}} e^{-i p x / \hbar} \langle x| \psi \rangle\, d x \\ &= \int_{-\infty}^{\infty} \frac{1} {\sqrt{2\pi \hbar}} e^{-i p x / \hbar} \left( a\, e^{i k x} + b\, e^{-i k x} \right)\, d x. \end{align} Here, the resolution of the identity I used was $\int_{-\infty}^\infty |x\rangle \langle x |\, dx$. For a one-dimensional system, that's just the identity operator, so you should be able to just plonk it into any old expression you want.

I'll leave it as a homework exercise to do the integral and take the squared magnitude of the result. But here are a couple hints. First, I happen to know that your solution $\psi(x)$ is a sum of two momentum eigenstates, so you should almost always get zero probability of measuring any given momentum — except for those two particular momentum eigenvalues. That is, your answer will be zero everywhere except for two values related to $k$ and $-k$. Second, you might want to read up on the Dirac $\delta$ function — specifically how it can be expressed as an integral.

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