# Transition Probabilities Between Two States

I'm currently working on the following problem for my Quantum Mechanics course, I could use some help with one of the integrals involved.

A particle of mass $m$ and charge $q$ is confined in a one-dimensional infinite square well of width $a$ in the $x$-direction. The particle starts at its ground state $(n=1)$. Under electromagnetic wave radiation with frequency $\omega$ and an electric field strength $E\hat x$, the particle has a probability to transit to higher energy levels.

What is the transition probability to the $n=2$ state? Express your answer in terms of frequency and time. Is the transition to $n=3$ allowed? Can you give a general transition rule?

The following is the work I have done. I will point out my question when the time comes. If you see a mistake in my work prior to my question, I'd appreciate some help with that as well.

So, $$\vec E=E\hat x \Rightarrow H' = -qE_0 x$$ $$\Rightarrow H'_{ba}=q\langle\psi_n|x|\psi_m\rangle E_0 = -\rho E_0$$

For $\rho=q\langle\psi_n|x|\psi_m\rangle$. The transition probability between states $n$ and $m$ is: $$c^{(1)}_b \approx \frac{-i}{\hbar} \int_0^t H'_{ba}e^{i\omega_0 t'}dt'= \frac{i}{\hbar} \rho E_0 \int_0^t e^{i\omega_0 t'}dt'= \frac{q}{\hbar \omega_0}\rho E_0(e^{i\omega_0 t} -1)$$

The following part is where I could use some help. I'm not sure how to evaluate the integral for $\rho$.

$$\langle\psi_n|x|\psi_m\rangle = \int_0^a \bigg ( \sqrt{\frac{2}{a}} \sin(\frac{n\pi x}{a})\bigg) x \bigg ( \sqrt{\frac{2}{a}} \sin(\frac{m\pi x}{a})\bigg) \\ = \frac{2}{a} \int_0^a x\sin(\frac{n\pi x}{a})\sin(\frac{m\pi x}{a})dx$$

I know that this integral is zero for $m$ is odd, which corresponds to even transition jumps in the physics. I'm just not sure how to calculate the general case for when $m$ is even (odd transition jumps). For the $n=2$ and $n=3$ states, it is easy enough to plug these values into WolframAlpha and get the correct probabilities, but I'd like to show the general equation before moving onto those parts.

Much appreciated.

It's easiest to do this kind of integral via tabular integration because one factor is a polynomial and the other is easy to integrate. $$\frac2a\int_0^ax\sin\left(\frac{n\pi x}a\right)\sin\left(\frac{m\pi x}a\right)dx=\frac1a\int_0^ax\left[\cos\left(\frac{(n-m)\pi x}a\right)-\cos\left(\frac{(n+m)\pi x}a\right)\right]dx$$ Now for our table: $$\begin{array}{ccc}\frac xa&&\cos\left(\frac{(n-m)\pi x}a\right)-\cos\left(\frac{(n+m)\pi x}a\right)\\ &\searrow+&\\ \frac1a&&\frac a{(n-m)\pi}\sin\left(\frac{(n-m)\pi x}a\right)-\frac a{(n+m)\pi}\sin\left(\frac{(n+m)\pi x}a\right)\\ &\searrow-&\\ 0&&-\frac{a^2}{(n-m)^2\pi^2}\cos\left(\frac{(n-m)\pi x}a\right)+\frac{a^2}{(n+m)^2\pi^2}\cos\left(\frac{(n+m)\pi x}a\right)\end{array}$$ So you start with the polynomial in the left column and the other factor on the right. You differentiate the polynomial until it's zero and integrate the other factor as many times. Then you multiply pairs diagonally downwards with alternating signs as indicated by the arrows. After adding up results we get \begin{align}\frac2a\int_0^ax\sin\left(\frac{n\pi x}a\right)\sin\left(\frac{m\pi x}a\right)dx&=\left[\frac x{(n-m)\pi}\sin\left(\frac{(n-m)\pi x}a\right)-\frac x{(n+m)\pi}\sin\left(\frac{(n+m)\pi x}a\right)\right.\\ &\left.+\frac{a}{(n-m)^2\pi^2}\cos\left(\frac{(n-m)\pi x}a\right)-\frac{a}{(n+m)^2\pi^2}\cos\left(\frac{(n+m)\pi x}a\right)\right]_0^a\\ &=\left[\frac{a}{(n-m)^2\pi^2}-\frac{a}{(n+m)^2\pi^2}\right]\left[(-1)^{n+m}-1\right]\\ &=\frac{4nma\left[(-1)^{n+m}-1\right]}{(n^2-m^2)^2\pi^2}\end{align} If you try doing that integral by parts without a way to efficiently lay out computations like tabular integration, it gets very messy.