# Quantum mechanics inner products

1. Oct 29, 2011

### v_pino

1. The problem statement, all variables and given/known data
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 $$psi_k(x)=\frac{1}{\sqrt{2 \pi}} e^{ikx}$$.

2. Relevant equations

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

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

3. 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?

2. Oct 30, 2011

### 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.

3. Nov 1, 2011

### 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?

4. Nov 1, 2011

### vela

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

5. Nov 2, 2011

### 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?

6. Nov 2, 2011

### vela

Staff Emeritus
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.

7. Nov 2, 2011

### 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.

8. Nov 2, 2011

### vela

Staff Emeritus
Yes, that's correct.