# How can we prove this system is in a stationary state?

I'm having trouble using the given hint to solve the problem. The problem statement is as follows:

At instant $t=0$, the probability distribution of a particle under a potential $V(x)$ is $\rho(x,0)=|\phi_n(x)|^2$, where $\phi_n$ is the n-th stationary state of the system. Can we infer that the state of the particle is a stationary state?

Hint: use the fact that $|e^{ikx}|=1$

This is my attempt.

Since the potential is not time-dependent, the stationary states are of the form $\psi_i(x,t)=\phi_i(x)e^{-iE_it/\hslash}$, but let's consider the general case in which the state of the system is a linear combination of them so, $$\psi(x,t)=\sum_{1\le i\le k} C_i\phi_i(x)e^{-iE_it/\hslash}$$

Then

$$\rho(x,t)=\psi^{*}(x,t)\psi(x,t)=\sum_{1\le i\le k} |C_i|^2|\phi_i(x)|^2+\sum_{i\neq j} 2Re( C_iC_j^{*}\phi_i(x)\phi_j^{*}(x)e^{-i(E_i-E_j)t/h})$$

and $$\rho(x,0)=\psi^{*}(x,0)\psi(x,0)=\sum_{1\le i\le k}|C_i|^2|\phi_i(x)|^2+\sum_{i\neq j}2Re(\ C_iC_j^{*}\phi_i(x)\phi_j^{*}(x))$$

Since $\rho(x,0)=|\phi_n(x)|^2$, then we can infer that one solution is $C_i=1$ for $i=n$ and $C_i=0$ for $i\neq n$, and the state of the system is a stationary state, but are there any other solutions?

And, how can we use the given hint (if I haven't already used it in my proof) to reach the same conclusion?

• Think about what states can be used to make $\rho(x,0)=|\phi_n(x)|^2$. Obviously $\psi(x,0) = \phi_n(x)$ works, but can you think of any others (e.g. a state that might include something from the hint)? – Punk_Physicist Mar 30 '15 at 22:32
• Thanks for the quick response. I know what you mean, but if I multiply $\phi_n(x)$ by $e^{ikx}$, don't I get another eigenfunction for the same stationary state? (since the expected values of the energy, position, etc are the same) So, no matter what complex number of magnitud one I multiply $\phi_n(x)$, I will always get the same stationary state. – alfdc80 Mar 31 '15 at 0:45
• Multiplying your entire state by a constant (global) phase doesn't change the state, but $e^{ikx}$ is not a constant. As a way to see this is not the same physical state, you can think about what the average momentum $\langle p\rangle\propto \int\text{d}x\phi\frac{\text{d}\phi}{\text{d}x}$ looks like (i.e. $\frac{\text{d}}{\text{d}x}\phi\ne\frac{\text{d}}{\text{d}x}\phi e^{ikx}$). – Punk_Physicist Mar 31 '15 at 17:29
• You are totally right. Even though their expected values of energy, position and the probability distributions are all the same, the expected value of momentum is NOT the same. Therefore, they are indeed different stationary states. But, it is still a stationary state. So my question is (referring to my first question), can I find a combination of coefficients $C_1,C_2,...C_k$ different from the one I gave such that for those coefficients $\rho(x,0)=|\phi_n(x)|^2$ but the state of the system is not a stationary one (which is what we are asked)? – alfdc80 Mar 31 '15 at 22:30
• Ok, I think I'm getting confused by the terms used in the problem statement. Where it reads "is the $n-th$ stationary state", it should read "is the $n-th$ eigenfunction of the wave function of the n-th stationary state". In any case, I get the whole point of what we are told to do. – alfdc80 Mar 31 '15 at 23:32