# Infinite square well – combination of two lowest states

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.5.

As an example of an explicit case of a particle in the infinite square well, suppose we have a particle that starts off in a combination of the two lowest states:

$\displaystyle \Psi(x,0)=A\left[\psi_{1}(x)+\psi_{2}(x)\right] \ \ \ \ \ (1)$

To normalize, we find ${A}$ by using the orthonormal property of the stationary states, so:

 $\displaystyle \int|\Psi(x,0)|^{2}dx$ $\displaystyle =$ $\displaystyle A^{2}\int(\psi_{1}^*+\psi_{2}^*)(\psi_{1}+\psi_{2})dx\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A^{2}\int(|\psi_{1}|^{2}+|\psi_{2}|^{2})dx\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2A^{2}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (5)$

So ${A=1/\sqrt{2}}$.

Using ${\omega\equiv\pi^{2}\hbar/2ma^{2}}$, we have for the full wave function:

$\displaystyle \Psi(x,t)=\frac{\sqrt{2}}{2}\psi_{1}(x)e^{-i\omega t}+\frac{\sqrt{2}}{2}\psi_{2}(x)e^{-4i\omega t} \ \ \ \ \ (6)$

and

 $\displaystyle 2|\Psi(x,t)|^{2}$ $\displaystyle =$ $\displaystyle (\psi_{1}^*(x)e^{i\omega t}+\psi_{2}^*(x)e^{4i\omega t})(\psi_{1}(x)e^{-i\omega t}+\psi_{2}(x)e^{-4i\omega t})\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \psi_{1}^{2}+\psi_{2}^{2}+2\psi_{1}\psi_{2}\cos(3\omega t)\ \ \ \ \ (8)$ $\displaystyle |\Psi(x,t)|^{2}$ $\displaystyle =$ $\displaystyle \frac{1}{2}(\psi_{1}^{2}+\psi_{2}^{2}+2\psi_{1}\psi_{2}\cos(3\omega t)) \ \ \ \ \ (9)$

The stationary states are real functions, so we can drop the * notation on the complex conjugates. Note that ${\int|\Psi(x,t)|^{2}dx=1}$ for all times, since the cosine term integrates to zero due to orthogonality.

The average position is:

 $\displaystyle \left\langle x\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\int_{0}^{a}(x\psi_{1}^{2}(x)+x\psi_{2}^{2}(x)+2x\psi_{1}\psi_{2}\cos(3\omega t))dx\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{a}{2}-\frac{16a}{9\pi^{2}}\cos(3\omega t) \ \ \ \ \ (11)$

The particle’s mean position oscillates about the midpoint of the well with an amplitude of ${16a/9\pi^{2}\approx0.18a}$.

The mean momentum can be found the quick way by taking the derivative of ${\left\langle x\right\rangle }$.

$\displaystyle \left\langle p\right\rangle =m\frac{d\left\langle x\right\rangle }{dt}=\frac{8\hbar}{3a}\sin(3\omega t) \ \ \ \ \ (12)$

where we have used the definition of ${\omega}$. Doing it the long way using integration does give the same answer, as can be checked using Maple (or by hand).

The two possible energies are ${E_{1}}$ and ${E_{2}}$ and since the wave function consists of equal contributions from the corresponding stationary states, they occur with equal probability. Thus

$\displaystyle \left\langle H\right\rangle =\frac{1}{2}(E_{1}+E_{2})=\frac{5\pi^{2}\hbar^{2}}{4ma^{2}} \ \ \ \ \ (13)$

Again, this can be obtained the long way through integration.

## 16 thoughts on “Infinite square well – combination of two lowest states”

1. growescience

If you mean equation 7, I believe it’s correct. It leads to equation 9 which agrees with the solution in your reference (the equation below your equation 5).

1. Cindy

you describe equation 9 by this
” since the cosine term integrates to zero due to orthogonality”
but i don’t get it @@ cosine function has this property!!??
i thought that it’s equal 1 just because the normalization of wavefunction

1. gwrowe Post author

The functions ${\psi_{1}}$ and ${\psi_{2}}$ in the cosine term ${2\psi_{1}\psi_{2}\cos(3\omega t)}$ are orthogonal, not the cosine.

2. cindy

Why should we find Cn using eq.37 in example2 but we can just put on the time dependent term in (6) in this problem??

And I tried to find Cn using eq37 and found it to be zero@@???

3. cindy

Well the equation 37 I mean is that in this book noted as 37
Using “fourier’s trick” to get the coefficient at arbitrary given time at first
And the example I mean is also the book’s example2 in chapter2

4. David

I am confused on where the sqrt(2)/2 comes from in eqn 6. Is that Cn? If so how is it calculated. Thanks for your help and your solutions are a great companion to the book.

1. gwrowe Post author

It’s just ${A}$ as calculated in equations 2 through 5.

1. gwrowe Post author

I’ve plugged in the stationary state wave function

 $\displaystyle \psi_{n}(x)$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2}{a}}\sin\frac{n\pi x}{a} \ \ \ \ \ (1)$

I did the integrals using Maple, but if you want to do them by hand they are fairly standard integrals – you’d need to use a trig identity or two and integration by parts (or you could just look them up in tables).

1. gwrowe Post author

Both stationary states ${\psi_{1}}$ and ${\psi_{2}}$ have the same coefficient ${A}$ in equation 1, so they have equal contributions to the overall wave function.