# Delta function well: statistics

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.5.

The delta function well gives rise to a wave function that decays exponentially either side of the delta function:

$\displaystyle \Psi\left(x,t\right)=Ae^{-\lambda\left|x\right|}e^{-i\omega t} \ \ \ \ \ (1)$

We can normalize ${\Psi}$ in the usual way:

 $\displaystyle \int_{-\infty}^{\infty}\left|\Psi\right|^{2}dx$ $\displaystyle =$ $\displaystyle \left|A\right|^{2}\left[\int_{-\infty}^{0}e^{2\lambda x}dx+\int_{0}^{\infty}e^{-2\lambda x}dx\right]\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\left|A\right|^{2}\int_{0}^{\infty}e^{-2\lambda x}dx\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left|A\right|^{2}}{\lambda}\ \ \ \ \ (4)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \sqrt{\lambda} \ \ \ \ \ (5)$

By symmetry, ${\left\langle x\right\rangle =0}$ and

$\displaystyle \left\langle x^{2}\right\rangle =2\lambda\int_{0}^{\infty}x^{2}e^{-2\lambda x}dx=\frac{1}{2\lambda^{2}} \ \ \ \ \ (6)$

Therefore

$\displaystyle \sigma=\sqrt{\left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}}=\frac{1}{\sqrt{2}\lambda} \ \ \ \ \ (7)$

A plot of ${\left|\Psi\right|^{2}}$ is shown, with vertical yellow lines indicating ${x=\pm\frac{1}{\sqrt{2}\lambda}}$, for the case ${\lambda=2}$:

The probability that the particle lies outside ${x=\pm\frac{1}{\sqrt{2}\lambda}}$ is

$\displaystyle P_{\left|x\right|>\sigma}=2\lambda\int_{1/\sqrt{2}\lambda}^{\infty}e^{-2\lambda x}dx=\frac{1}{e^{\sqrt{2}}}=0.2431 \ \ \ \ \ (8)$

In this case, the probability of ${x}$ being greater than one standard deviation is a constant, independent of ${\lambda}$.

## 14 thoughts on “Delta function well: statistics”

1. chris

Why does the e^(-iwt) term disappear?
Is it because we square it, or does it have to do with the nature of the exponential…?

1. gwrowe Post author

When we calculate ${\left|\Psi\right|^{2}}$, we multiply ${\Psi}$ by ${\Psi^*}$ so ${e^{-i\omega t}}$ gets multiplied by ${e^{+i\omega t}}$ giving 1.

1. gwrowe Post author

Because of the absolute value ${\left|x\right|}$ in equation 1. For ${x<0}$, ${e^{-\lambda\left|x\right|}=e^{\lambda x}}$.

1. knowsNothing

So what is the purpose of the absolute value, if {x<0}, {e^{-\lambda\left|x\right|}=e^{\lambda x}}. Whether x is negative or positive, shouldn't the product of {-\lambda\left|x\right|} remain negative?

1. gwrowe Post author

It’s hard to read your Latex code (to get it to show properly, see here). Yes, if ${x<0}$, then ${e^{-\lambda\left|x\right|}=e^{\lambda x}}$ but we also want the wave function to decay exponentially for ${x>0}$, and in that case ${e^{-\lambda\left|x\right|}=e^{-\lambda x}\ne e^{\lambda x}}$, so we need the absolute value.

2. Sigex

When you do the integral, I get – 1/2L, but by you getting sqrt(L) you seem to have lost the minus sign some how. How did you lose the minus sign that comes from the expoential on e in the integral?

1. gwrowe Post author
 $\displaystyle \int_{0}^{\infty}e^{-2\lambda x}dx$ $\displaystyle =$ $\displaystyle -\frac{1}{2\lambda}\left.e^{-2\lambda x}\right|_{0}^{\infty}$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2\lambda}\left(0-1\right)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2\lambda}$

Note that since ${e^{-2\lambda x}}$ is always positive, the integral cannot be negative.

3. Cindy

hello
i have some questions about eq6
i used gamma function here and thought i was stupid after looking your calculation
did u do integration by parts in eq6 ??
the equation is simple but i have no idea to the answer only by “look”

and how to know
“By symmetry, =0

i wonder how to just have a look and get the results
(after i do the calculation and thought i was stupid again……

1. gwrowe Post author

You could use the gamma function to do equation 6 after transforming the variable to ${u=2\lambda x}$:

 $\displaystyle 2\lambda\int_{0}^{\infty}x^{2}e^{-2\lambda x}dx$ $\displaystyle =$ $\displaystyle \frac{2\lambda}{\left(2\lambda\right)^{3}}\int_{0}^{\infty}u^{2}e^{-u}du$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{4\lambda^{2}}\Gamma\left(3\right)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2!}{4\lambda^{2}}$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2\lambda^{2}}$

Or you could integrate by parts twice to get rid of the ${x^{2}}$.

As for ${\left\langle x\right\rangle }$, it’s the integral of an odd function (${x}$) multiplied by an even function ${\left(e^{-2\lambda\left|x\right|}\right)}$ over an interval symmetric about the origin, so the answer is zero.

4. Anonymous

I tried to verify = 0 by using eq 6 (replacing x^2 by x). Is this correct? (I am not getting zero, maybe I missed out something.)

1. gwrowe Post author

You can see that ${\left\langle x\right\rangle =0}$ by noting that since ${\Psi}$ is an even function, then ${x\Psi}$ is an odd function and the integral from ${-\infty}$ to ${+\infty}$ of any odd function is zero. If you want to modify equation 6, you need to replace ${x^{2}}$ by ${x}$ and also extend the lower limit on the integral to ${-\infty}$ but it’s better to use the symmetry argument.