# [Q]Some confusing about Dirac Delta Function

1. Oct 21, 2008

### good_phy

Hi.

Recently day, I tried to solve quantum mechanics problem in liboff fourth version to prepare

But what make me be confused a lot is Dirac Delta Function.

One of my confusing on Dirac Delta is what i wrote below.

-One of the formula describing Dira Delta Function is $$\int_{\infty}^{\infty}e^{-2{\pi} (k_{2}-k_{1})}dt = \delta (k_{2}-k_{1})$$

If, we are replacing $\infty$ some finite constance, it means integration range
changed to some finite range, Do we still get $\delta (k_{2}-k_{1})$

This is important problem because a lot of quantum problem have finite range, so orthogonality is damaged if delta function can not be derived from finite integral.

please assist to me. Thank you.

Last edited: Oct 22, 2008
2. Oct 22, 2008

### clem

Your formula is wrong. The exponent should be -2it pi(k_2-k_1).
The limits must be infinite to get the delta function.
In box normalization, as an alternative to delta function normalization,
the limits are +/- L and then the limit L-->infinity is taken.
If a problem really does have a finite range, then the efs will not be simple exponentials.

3. Oct 22, 2008

### good_phy

I see, Thank you So thank to your assistant. I can make conclusion that

Hermition operator produces orthogonal eigen function but, orthogonality only persists in

case of that domain of eigen function is infinity,

In other word, orthogonality is break up with physical restriction.(Real physics, there is

almost no infinity demain)

Please anybody correct my careful conclusion.

4. Oct 22, 2008

### Avodyne

No, not correct. In a box, the eigenfunctions are orthogonal, but not all momenta are allowed. The simplest case, mathematically, is periodic boundary conditions. In a box of length L, the eigenfunctions are exp[2 pi i n x/L], n=...,-1,0,1,..., and these are orthogonal.

Last edited: Oct 22, 2008
5. Oct 22, 2008

### Ben Niehoff

The delta function can always be expressed as a linear combination of eigenfunctions for whatever potential you have. As noted, when the box is finite, the eigenfunctions themselves are different.