# Integral involving delta distribution

1. Jan 22, 2008

### J.D.

1. The problem statement, all variables and given/known data

Solving a problem about the variational method I came across one nasty integral. Here goes:

$$\bar{H} := \frac{ < \hat{0} | H | \hat{0} > }{< \hat{0} | \hat{0} >}$$

2. Relevant equations

$$H = -\frac{ \hbar^2 }{2m} \frac{\partial ^2}{\partial x^2} + \frac{1}{2} m \omega ^2 x^2$$

$$< x | \hat{0} > = \text{e}^{-\beta |x|}$$

$$< x' | \left( -i \hbar \frac{\partial}{\partial x} \right)^n | \alpha > = (-i \hbar)^n \frac{\partial^n}{\partial x'^n} < x' | \alpha >$$

$$\int f(x) \delta^{(n)}(x-a) dx = - \int \frac{\partial f}{\partial x} \delta ^{(n-1)} (x-a) dx$$

3. The attempt at a solution

Pretty straightforward I got

$$< \hat{0} | \hat{0} > = \frac{1}{\beta}$$

and so:

$$\bar{H} = \beta \int \int dx' dx'' < \hat{0} | x' > < x' | \left( -\frac{ \hbar^2 }{2m} \frac{\partial ^2}{\partial x^2} + \frac{1}{2} m \omega ^2 x^2 \right) | x'' > < x'' | \hat{0} >$$

$$\Leftrightarrow \bar{H} = \beta \int \int dx' dx'' \text{e}^{-\beta |x'|} < x' | \left( -\frac{ \hbar^2 }{2m} \frac{\partial ^2}{\partial x^2} + \frac{1}{2} m \omega ^2 x^2 \right) | x'' > \text{e}^{-\beta |x''|} = I_1 + I_2$$

Thus I ended up with two terms:

$$I_2 = 4 \beta \int_0^{\infty} \int_0^{\infty} dx' dx'' \text{e}^{-\beta x'} < x' |\frac{1}{2} m \omega ^2 x^2 \right) | x'' > \text{e}^{-\beta x''} = \frac{2 \beta \omega ^2}{m} \int_0^{\infty} \int_0^{\infty} dx' dx'' \text{e}^{-\beta x'} x''^2 \delta (x' - x'')$$

$$\Leftrightarrow I_2 = 2 \beta m \omega ^2 \int_0^{\infty} dx' \text{e}^{-2 \beta x'} x'^2 = 2 \beta m \omega ^2 \cdot \frac{2!}{(2\beta)^3} = \frac{m \omega^2}{2 \beta^2}$$

$$I_1 = 4 \beta \int_0^{\infty} \int_0^{\infty} dx' dx'' \text{e}^{-\beta x'} < x' |-\frac{ \hbar^2 }{2m} \frac{\partial ^2}{\partial x^2} | x'' > \text{e}^{-\beta x''} = \frac{ - 2 \beta \hbar^2 }{m} \int_0^{\infty} \int_0^{\infty} dx' dx'' \text{e}^{-\beta x'} \delta'' (x' - x'') \text{e}^{-\beta x''}$$

$$\Leftrightarrow I_1 = \frac{ - 2 \beta^3 \hbar^2 }{m} \int_0^{\infty} \int_0^{\infty} dx'' \text{e}^{-2 \beta x''} = \frac{ - 2 \beta^3 \hbar^2 }{m} \cdot \frac{1}{2 \beta} = \frac{- \beta^2 \hbar^2 }{m}$$

We finally arrive at

$$\bar{H} = I_1 + I_2 = \frac{- \beta^2 \hbar^2 }{m} + \frac{m \omega^2}{2 \beta^2}$$

Does this seem right to you guys?

2. Jan 22, 2008

### J.D.

Actually I am a little uncertain here if I should calculate either

$$\frac{\partial ^2 \delta ( x' - x'')}{\partial x''^2}$$

or

$$\frac{\partial ^2}{\partial x''^2} \left( \delta ( x' - x'') \text{e}^{-\beta x''}\right)$$

in $I_1$.