How Does Quantum Mechanics Describe Free-Particle Probability Amplitudes?

[rar0308]

Problem10.1, Introductory QM,Liboff.

Homework Statement

If \psi (\mathbf{r},t) is a free-particle state and b(\mathbf{k},t) the momentum probability amplitude for this same state, show that
\iiint \psi^* \psi d \mathbf{r}=\iiint b^* b d \mathbf{k}

Homework Equations

\psi_\mathbf{k} (\mathbf{r},t) = Ae^{i(\mathbf{k} \cdot \mathbf{r} - \omega t)} (10.14)
\hbar \omega = E_k
\delta (\mathbf{r} - \mathbf{r'}) = \frac{1}{(2 \pi)^3} \iiint e^{i \mathbf{k} \cdot (\mathbf{r} - \mathbf{r'})} d \mathbf{k} (10.20)
d \mathbf{k} = dk_x dk_y dk_z
\psi (\mathbf{r},t)=\frac{1}{(2 \pi)^{3/2}}\iiint b(\mathbf{k},t)e^{i(\mathbf{k} \cdot \mathbf{r}-\omega t)} d \mathbf{k} (10.22)
b(\mathbf{k},t) = \frac{1}{(2 \pi)^{3/2}}\iiint\psi (\mathbf{k},t) e^{i(\mathbf{k} \cdot \mathbf{r} - \omega t)} d \mathbf{r} (10.23)
d \mathbf{r}=dxdydz

The Attempt at a Solution

1.I substituted eq 22 into left-hand side of problem's equation. Then I don't know how to go further. I think there will be some manipulation on the equation but I'm lacking some knowledge how to do it.

 

[dextercioby]

Your question is a simple case of Plancherel theorem. What is B equal to, if you're given its Fourier transformation ?

 

[rar0308]

b(\mathbf{k},t) = \frac{1}{(2 \pi)^{3/2}}\iiint\psi (\mathbf{k},t) e^{i(\mathbf{k} \cdot \mathbf{r} - \omega t)} d \mathbf{r} (10.23)
d \mathbf{r}=dxdydz

Is this a Fourier transform?

 

[dextercioby]

Yes. Then what is b equal to ? Can you perform that integration, once you know psi ?