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)


\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. 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.

  1. 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?

  2. Cindy

    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.

    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.


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