Daily Archives: Tue, 23 June 2015

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}.

Triangular wave function: probabilities

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.4.

The square modulus of the wave function which is the solution to the Schrödinger equation is interpreted as a probability density. As an example consider the wave function given by

\displaystyle  \Psi\left(x,0\right)=\begin{cases} A\frac{x}{a} & 0\le x\le a\\ A\frac{b-x}{b-a} & a\le x\le b\\ 0 & \mbox{otherwise} \end{cases} \ \ \ \ \ (1)

We can normalize {\Psi} by requiring

\displaystyle  \int_{0}^{b}\left|\Psi\right|^{2}dx=1 \ \ \ \ \ (2)

Plugging in the formula and doing the integral gives

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

where we’ve taken the positive real root for {A}. Note that {A} could also be multiplied by a phase factor {e^{i\delta}} for any real {\delta} without affecting normalization. This can be important in some applications where we need to add together wave functions.

Given this value for {A}, we can plot 1. Here, we’ve taken {a=1} and {b=3}:

Since {\Psi} has its maximum at {x=a}, that is where the particle is most likely to be found. The probability of the particle being found to the left of {x=a} is

\displaystyle  P_{x<a}=\frac{3}{b}\int_{0}^{a}\frac{x^{2}}{a^{2}}dx=\frac{a}{b} \ \ \ \ \ (6)

If {b=a}, then {\Psi} drops to zero at {x=a} so {P_{x<a}=1}. If {b=2a}, then {\Psi} is an isosceles triangle symmetric about {x=a} so {P_{x<a}=\frac{1}{2}}.

The expectation value of {x} is

\displaystyle  \left\langle x\right\rangle =\int x\left|\Psi\right|^{2}dx=\frac{a}{2}+\frac{b}{4} \ \ \ \ \ (7)

where we used Maple to simplify the integration. If {b=2a}, then {\left\langle x\right\rangle =a} as expected.