# Kronecker Delta and Dirac Delta

1. Jan 7, 2012

### coki2000

Hello PF,

When I was studying Quantum mechanics, I realized that this equality should be true,

$$<{\psi}_{n} \mid {\psi}_{m}>=\int {\psi}_{m}^*{\psi}_{n}dx={\delta }_{mn}$$

So ${\psi}_{m}^*{\psi}_{n}$ must be equal to dirac delta function so that we provide the kronecker delta as a solution of the integral.

Therefore, this equation must be true, mustn't it?

$$\int \delta (x-x')dx={\delta }_{mn}$$

Or, if it is wrong, what is the expression ${\psi}_{m}^*{\psi}_{n}$ equal to?

Thanks for your opinions and helps.

Last edited: Jan 7, 2012
2. Jan 7, 2012

### Pengwuino

No, it's equal to the Kronecker delta. The Dirac delta does not enter in at all here unless you know something about the actual functions $\Psi$

$\int \delta(x - x')dx = 1$ only if you integrate over x = x' within your integration. That's all you can say about that equation.

3. Jan 7, 2012

### coki2000

Sorry, I mean this ${\psi}_{m}^*{\psi}_{n}$ must be equal to dirac delta. I corrected it now.

4. Jan 7, 2012

### Pengwuino

Not quite.

$\int_{All Space} \psi^*_m\psi_n dx = \delta_{mn}$ where the integration is over the whole space that you're looking at. Nowhere in this does a Dirac delta come into play and it's only valid when you look at the integration over the whole space. Look at a specific, easy example like the simple harmonic oscillator with say, n = 0 and m = 1. The product $\psi^*_{m = 1} \psi_{n = 0}$ will clearly not be 0, but the integration will be.