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.