Is the Kronecker Delta Related to the Dirac Delta in Quantum Mechanics?

[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.

 

[Pengwuino]

So <{\psi}_{n} \mid {\psi}_{m}> must be equal to dirac delta function

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.

 

[coki2000]

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.

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

 

[Pengwuino]

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

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.