I have a problem of a particle in a box of width $a$. I am trying to find the probability of finding the particle on the left side of the box at time $t$. The wavefunction at $t=0$ is given by

$$|\psi(0)\rangle = \frac{1}{\sqrt{2}} |\phi_1\rangle + \frac{1}{2}|\phi_2\rangle + \frac{1}{2} |\phi_3\rangle,$$

where $\phi_n$ is the wavefunction at order $n$.

My Attempt: I have calculated and normalized the time-independent wavefunctions $\phi_n$ and used the time evolution operator to write down the time-dependent wavefunctions

$$|\psi(t)\rangle=\frac{e^{i E_{1} t / \hbar}}{\sqrt{2}}\left|\phi_{1}\right\rangle+\frac{e^{-i E_{2} t / \hbar}}{2}\left|\phi_{2}\right\rangle+\frac{e^{-i E_{3} t / \hbar}}{2}\left|\phi_{3}\right\rangle,$$

with the famous result

$$ |\phi_n \rangle = \sqrt{\frac{2}{a}} \sin\left( n \pi \frac{x}{a} \right).$$

To find the probability I have tried to compute

$$P(0 < x < a/2) = \left| \int_0^{a/2} \psi^*(t) \psi(t) dx \right|^2 \\ = \frac{\left(20 \sqrt{2} \cos \left( \frac{(E_2-E_1)}{\hbar} t \right) + 12 \cos \left(\frac{(E_3-E_2)}{\hbar} t \right)+15 \pi \right)^2}{900 \pi ^2} $$

To check that my answer makes sense, first I verified that


But now when I look at the left-side and right-side probabilities they do not add up to one

$$P(0<x<a/2) = 0.86, P(a/2<x<a) = 0.005$$

What am I doing wrong?

  • $\begingroup$ Apart from this your question contain too much abuse of notations like $$\phi_n(x)=\langle x|n\rangle =\sqrt{\frac{2}{L}}\cdots $$ $\endgroup$ Mar 22, 2021 at 15:15

1 Answer 1


The probability density is given by $$\rho(x,t)=|\Psi(x,t)|^2=\Psi^*(x,t)\Psi(x,t)$$ and so $$P(a\leq x\leq b)=\int_a^b\Psi^*(x,t)\Psi(x,t) dx$$

As far I can see you have taken a mod square which is wrong, in the $4$th equation from below.


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