Delta function well: bound state – uncertainty principle

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.25.

We’ve seen that the bound state of a particle in a delta function potential wellhas the wave function

$\displaystyle \psi(x)=\frac{\sqrt{m\alpha}}{\hbar}e^{-m\alpha\left|x\right|/\hbar^{2}} \ \ \ \ \ (1)$

We can work out the mean values of position and momentum, and of their squares, in the usual way.

For ${\left\langle x\right\rangle }$, since the wave function is even and the function ${f(x)=x}$ is odd, we have

 $\displaystyle \left\langle x\right\rangle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}x\psi^{2}(x)dx\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (3)$

since the integrand is odd.

For the mean square position, we get

 $\displaystyle \left\langle x^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}x^{2}\psi^{2}(x)dx\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m\alpha}{\hbar^{2}}\int_{-\infty}^{\infty}x^{2}e^{-2m\alpha\left|x\right|/\hbar^{2}}dx\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2m\alpha}{\hbar^{2}}\int_{0}^{\infty}x^{2}e^{-2m\alpha x/\hbar^{2}}dx\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar^{4}}{2m^{2}\alpha^{2}} \ \ \ \ \ (7)$

where we used software to do the integral.

Now for the momentum. We get

$\displaystyle \left\langle p\right\rangle =\frac{\hbar}{i}\frac{m\alpha}{\hbar^{2}}\int_{-\infty}^{\infty}e^{-m\alpha\left|x\right|/\hbar^{2}}\frac{d}{dx}e^{-m\alpha\left|x\right|/\hbar^{2}}dx \ \ \ \ \ (8)$

We can split this integral into two parts joined at the origin.

$\displaystyle \frac{d}{dx}e^{-m\alpha\left|x\right|/\hbar^{2}}=\begin{cases} -\frac{m\alpha}{\hbar^{2}}e^{-m\alpha x/\hbar^{2}} & x>0\\ \frac{m\alpha}{\hbar^{2}}e^{m\alpha x/\hbar^{2}} & x<0 \end{cases} \ \ \ \ \ (9)$

The derivative is therefore an odd function. Since the original wave function is even, the product of the two is odd, so ${\left\langle p\right\rangle =0}$.

Calculating ${\left\langle p^{2}\right\rangle }$ is a bit trickier, since the derivative above is discontinuous at the origin. If we define the step function

$\displaystyle H(x)=\begin{cases} 1 & x>0\\ 0 & x<0 \end{cases} \ \ \ \ \ (10)$

we can write the derivative above as

$\displaystyle \frac{d}{dx}e^{-m\alpha\left|x\right|/\hbar^{2}}=-\frac{m\alpha}{\hbar^{2}}e^{-m\alpha\left|x\right|/\hbar^{2}}\left(2H(x)-1\right) \ \ \ \ \ (11)$

We’ve seen that the derivative of the step functioncan be taken as the delta function

$\displaystyle \frac{dH}{dx}=\delta(x) \ \ \ \ \ (12)$

so the second derivative of the wave function is

 $\displaystyle \frac{d^{2}}{dx^{2}}\psi(x)$ $\displaystyle =$ $\displaystyle \frac{\sqrt{m\alpha}}{\hbar}\frac{d^{2}}{dx^{2}}e^{-m\alpha\left|x\right|/\hbar^{2}}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\left(m\alpha\right)^{3/2}}{\hbar^{3}}\frac{d}{dx}\left[e^{-m\alpha\left|x\right|/\hbar^{2}}\left(2H(x)-1\right)\right]\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left(m\alpha\right)^{5/2}}{\hbar^{5}}e^{-m\alpha\left|x\right|/\hbar^{2}}\left(2H(x)-1\right)^{2}-2\frac{\left(m\alpha\right)^{3/2}}{\hbar^{3}}e^{-m\alpha\left|x\right|/\hbar^{2}}\delta(x) \ \ \ \ \ (15)$

Observing that ${\left(2H(x)-1\right)^{2}=1}$ everywhere, we get

$\displaystyle \frac{d^{2}}{dx^{2}}\psi(x)=e^{-m\alpha\left|x\right|/\hbar^{2}}\left[\frac{\left(m\alpha\right)^{5/2}}{\hbar^{5}}-2\frac{\left(m\alpha\right)^{3/2}}{\hbar^{3}}\delta(x)\right] \ \ \ \ \ (16)$

After all this we can now calculate ${\left\langle p^{2}\right\rangle }$:

 $\displaystyle \left\langle p^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \left(\frac{\hbar}{i}\right)^{2}\int_{-\infty}^{\infty}\psi(x)\frac{d^{2}}{dx^{2}}\psi(x)dx\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hbar^{2}\frac{\sqrt{m\alpha}}{\hbar}\int_{-\infty}^{\infty}e^{-2m\alpha\left|x\right|/\hbar^{2}}\left[\frac{\left(m\alpha\right)^{5/2}}{\hbar^{5}}-2\frac{\left(m\alpha\right)^{3/2}}{\hbar^{3}}\delta(x)\right]dx\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\left(m\alpha\right)^{3}}{\hbar^{4}}\int_{-\infty}^{\infty}e^{-2m\alpha\left|x\right|/\hbar^{2}}dx+2\left(\frac{m\alpha}{\hbar}\right)^{2}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -2\frac{\left(m\alpha\right)^{3}}{\hbar^{4}}\int_{0}^{\infty}e^{-2m\alpha x/\hbar^{2}}dx+2\left(\frac{m\alpha}{\hbar}\right)^{2}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{m\alpha}{\hbar}\right)^{2} \ \ \ \ \ (21)$

Going from the second to the third line, we used the property of the delta function

$\displaystyle \int_{-\infty}^{\infty}f(x)\delta(x)dx=f(0) \ \ \ \ \ (22)$

The uncertainty principlefor the bound state of the delta function potential is therefore

 $\displaystyle \sigma_{x}\sigma_{p}$ $\displaystyle =$ $\displaystyle \sqrt{\left\langle x^{2}\right\rangle \left\langle p^{2}\right\rangle }\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{\sqrt{2}} \ \ \ \ \ (24)$

10 thoughts on “Delta function well: bound state – uncertainty principle”

1. Bob

Hi,

When you took the derivative of : [exp(-ma|x|/h^2) (2H(x)-1)]

how did you get a (2H(x)-1)^2 term? When I try to do it, only (2H(x)-1) term ever appears.

Thanks!
Bob

2. growescience

You need to apply the product rule:

$displaystyle frac{d}{dx}left[e^{-malphaleft|xright|/hbar^{2}}left(2H(x)-1right)right]=frac{d}{dx}left(e^{-malphaleft|xright|/hbar^{2}}right)left(2H(x)-1right)+e^{-malphaleft|xright|/hbar^{2}}frac{d}{dx}left(2H(x)-1right)$

Since:

$displaystyle frac{d}{dx}e^{-malphaleft|xright|/hbar^{2}}=-frac{malpha}{hbar^{2}}e^{-malphaleft|xright|/hbar^{2}}left(2H(x)-1right)$
the first term gets a factor of ${left(2H(x)-1right)^{2}}$.

3. George

I believe that in your mean squared average you are missing an x^2, like so:

$\langle x^{2}\rangle=\int_{-\infty}^{\infty}x^{2}\psi^{2}\left(x\right)dx$

4. Pero

I avoided the integration for p^2 by calculating the expected value of V, hence the expected value of T (E is given). V and T are both E/2. Then getting p^2 from that.

1. growescience

I don’t think that argument works here, since the average potential is negative (it’s a delta-function well) so you can’t set it equal to kinetic energy (which is always positive). You get

 $displaystyle leftlangle Vrightrangle$ $displaystyle =$ $displaystyle intpsi^{2}Vdx$ $displaystyle$ $displaystyle =$ $displaystyle -frac{malpha^{2}}{hbar^{2}}$

Even if you try to set the absolute value of this to ${leftlangle Trightrangle =leftlangle p^{2}rightrangle /2m}$, there’s still that factor of 2 which messes things up and gives ${leftlangle p^{2}rightrangle =2left(malpha/hbarright)^{2}}$ which is twice the correct answer. I think that ${leftlangle Trightrangle =leftlangle Vrightrangle }$ is true only for certain potentials (like the harmonic oscillator, where ${Vge0}$ everywhere).

1. Pero

We have $E = -\frac{m\alpha^2}{2 \hbar^2}$ and now $=2E$, so $= E - = -E$ and $= -2mE$ as required.

1. Pero

Sorry, wordpress is chewing up what I type. In words: E is the energy of the system, which came out when the equation was solved in the first place. The expected value of V can be easily derived, as you did above. The expected value of T must be the difference E-V. This gives the expected value of p^2 without any integration.

2. growescience

Ah, I see. Yes, that will work. What confused me was your saying that ${leftlangle Vrightrangle =leftlangle Trightrangle =frac{E}{2}}$. Actually, ${leftlangle Vrightrangle =-2leftlangle Trightrangle =2E}$, but other than that, your argument gives a clever way of finding ${leftlangle p^{2}rightrangle }$.