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

    Reply
  2. George

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

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

    Reply
  3. 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.

    Reply
    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).

      Reply
          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 }.

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