Expectation value of the position for $\psi(x,0)=Ax^2(x^2-L^2)$ in an infinite square well [closed]

Consider a particle of mass $𝑚$ in an infinite square well of width $𝐿$. The wave function of the particle at $𝑡 = 0$ is $$\psi (x,0)=Ax^2(x^2-L^2), \quad 0\leq x \leq L$$

a.) What is $\psi(x,t)$ for $t \geq 0$?

b.) At some time $t >0$, what is the probability of measuring the particle to have energy $16\pi ^2\hbar ^2/(2mL^2)$? Does it depend on time?

c.) Calculate the expectation value of the position.

So for a the first thing I did was to find the normalization constant $A$ using the normalization condition $$\langle\psi (x,0)|\psi (x,0)\rangle = \int_{-\infty}^{\infty} \psi^*(x,0) \psi (x,0)dx =1$$ which after evaluating gives me $$A= \sqrt{\frac{105}{8L^7}}$$

then I find the expansion coefficients $$C_n = \langle E_n|\psi (x,0)\rangle = \int_{-\infty}^{\infty} \varphi_n^*(x)\psi (x,0)dx$$ for which i got $$C_n = \frac{3\sqrt{105}}{n^3\pi ^3}(-1)^n$$ now the time dependent wave function can be written down as $$\psi (x,t) = \sum_n \frac{3\sqrt{105}}{n^3\pi ^3}(-1)^n exp({-\frac{in^2\pi ^2\hbar t}{2mL^2}}) \sqrt{\frac{2}{L}}sin(\frac{n\pi x}{L})$$

I also did part b.) and got some really small probability that didnt depend on time. The part im confused about is part c.), I dont know what to do with the summations when you square the wave function in $$\langle x \rangle = \int_{-\infty}^{\infty} x |\psi (x,t)|^2 dx$$

closed as off-topic by Emilio Pisanty, Kyle Kanos, stafusa, Jon Custer, Daniel GriscomNov 7 '17 at 19:35

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• Is it clear you need $\langle x\rangle$ as a function of time or just at $t=0$? (For any $t$ this is a real mess and likely needs to be summed numerically.) – ZeroTheHero Nov 7 '17 at 4:41
• I'm not really sure, I've posted the question as it was given to me. I was also thinking that maybe the question might have meant to calculate the position expectation value when the particle is in the $|E_4 \rangle$ eigenstate since that is the eigenstate that it gives in part b. In that case $\langle x \rangle$ would just be $L/2$, but it doesn't read like that so i dont know – Elvis Nov 7 '17 at 4:45
• I’ve added my 2-cents as an answer. Your intuition is quite correct for $\langle x\rangle$ since the probability distribution at $t=0$ is symmetric about the midpoint. – ZeroTheHero Nov 7 '17 at 4:50
• I don't understand this question. The set-piece reads quite clearly (to me) as asking for the position expectation value in the initial state, i.e. the last integral in this post, which can be integrated directly and without recourse to infinite summations (and indeed from pure symmetry considerations and without doubt any integrals). – Emilio Pisanty Nov 7 '17 at 22:05

We can factor out the summation operators in the following way:

$$\int_{-\infty}^{\infty} x \left|\sum_n f_n(x,t)\right|^2dx = \int_{-\infty}^{\infty} x \left(\sum_n f^*_n(x,t)\right)\left(\sum_m f_m(x,t)\right)dx$$

$$= \int_{-\infty}^{\infty} x \sum_n \sum_m f^*_n(x,t) f_m(x,t)dx$$

Then, Fubini's theorem allows us to exchange summation and integration operators:

$$= \sum_n \sum_m \int_{-\infty}^{\infty} x f^*_n(x,t) f_m(x,t)dx$$

The calculation is not too difficult from here.

If you are looking for $\langle x\rangle$ at $t=0$: $$\langle x\rangle = \int_0^L dx x \psi(x,0)\psi^*(x,0) \tag{1}$$ by definition. No need to do any expansion in this case. Using (1) directly with $\psi(x,0)$ is obviously much simpler that expanding in eigenstates.

Otherwise you have to proceed as you have done. It’s not clear there is a shortcut since $\psi(x,t)\psi^*(x,t)$ is explicitly time dependent and has no obvious symmetries for $t\ne 0$. Integration of the type $$\int_0^L x\sin(k_n x)\sin(k_mx)$$ will be non-zero in general so you will have lots fo terms of the type $$\sum_{m,n} c_m c_n e^{-iE_nt/\hbar} e^{iE_mt/\hbar} \int_0^L dx\,x\sin(k_n x) \sin(k_m x)$$ to resum, and I don’t think this is doable analytically (but someone else might be able to do it.)