How to Show Inner Products in Quantum Mechanics Using Fourier Transforms?

[v_pino]

Homework Statement

Given (\psi_1, \psi_2)=\int dx \psi_1^*(x) \psi_2(x), show (\psi_1, \psi_2)=\int dk \phi_1^*(k) \phi_2(k), where \phi_{1,2}(k)= \int dx \psi_k^*(x) \psi_{1,2}(x) and psi_k(x)=\frac{1}{\sqrt{2 \pi}} e^{ikx}.

Homework Equations

\psi (x)= \int dk \phi(k) \psi_k(x)

\psi(x)=\int dk \phi(k) \psi_k(x)

The Attempt at a Solution

(\psi_1 , \psi_2)= \int dx \left \{ \int dk \phi_1^*(k) \psi_k^*(x) \right \}\left \{ \int dk \phi_2(k) \psi_k(x) \right \}

=\int dx \left \{ \int dk \phi_1^*(k) \frac{1}{\sqrt {2 \pi}}e^{-ikx} \int dk \phi_2(k) \frac{1}{\sqrt {2 \pi}}e^{ikx} \right \}

= \frac{1}{2 \pi}\int dx \left \{ \int dk \phi_1^*(k) \int dk \phi_2(k) \right \}

Is this correct so far? How do I proceed from here? It looks like a Fourier Transform with the 1/2pi. And I have two integrals within another one for the dx. Can I separate them some how?

 

[diazona]

When you substitute in the integrals for \psi_1^*(x) and \psi_2(x), the k in one integral isn't the same variable as the k in the other integral. If you use the same variable for both, you're almost certainly going to confuse yourself. In this case, you can't cancel out the e^{-ikx} factor from one integral with e^{ikx} from the other.

 

[v_pino]

=\int dx \left \{ \int dk_1 \phi_1^*(k_1) \frac{1}{\sqrt {2 \pi}}e^{-ik_1x} \int dk_2 \phi_2(k_2) \frac{1}{\sqrt {2 \pi}}e^{ik_2x} \right \}

I have revised my equation above. But now that the exponents don't cancel, how should I proceed with this?

 

[vela]

You want to use the fact that \delta(k_1-k_2) = \frac{1}{2\pi}\int_{-\infty}^\infty e^{i(k_1-k_2)x}\,dx

 

[v_pino]

(\psi_1, \psi_2)=\delta(k_2-k_1) \int dk_1 \phi_1^*(k_1) \int dk_2 \phi_2(k_2)

Is the delta function equal to 1 in this case? And has the equation given in the problem (\psi_1, \psi_2)=\int dk \phi_1^*(k) \phi_2(k) combined k_1 and k_2?

 

[vela]

The delta function depends on k₁ and k₂. You can't just pull it out front like that.

By using the delta function, you can perform one of the integrations. I suggest you read about the delta function to see how that works.

 

[v_pino]

I obtained (\psi_1, \psi_2)=\int dk_2 \phi_2(k_2) \int dk_1 \phi_1^*(k_1) \delta(k_2-k_1) = \int dk_2 \phi_2(k_2) \phi_1^* (k_2) by using \psi(x)= \int dx' \psi(x') \delta(x-x')

Is it correct? Thanks for the help once again.

 

[vela]

Yes, that's correct.