Are Energy Eigenstate Coefficients Real-valued?

[LarryS]

TL;DR

Energy Eigenstates Coefficients Real-valued?

Given any system with discreet energy eigenstates, φ_n(x)e^-iEnt . The φ_n are functions only of position. But are they also almost always real-valued?Thanks in advance.

 

[Isaac0427]

Summary:: Energy Eigenstates Coefficients Real-valued?

Given any system with discreet energy eigenstates, φ_n(x)e^-iEnt . The φ_n are functions only of position. But are they also almost always real-valued?Thanks in advance.

I am assuming, given your question, by coefficients you mean the solutions to the TISE (which I would call the eigenfunctions, not coefficients). If that is the case, if you are talking about a one dimensional bound state where $\lim_{x\rightarrow \pm \infty}\phi_n(x)=0$ (which is almost always the case for bound states), they can always be found to be real. I'll be more specific though, and give you a sketch of the proof:

First, it must be noted that if $\phi_n(x)$ solves the TISE with energy $E_n$, then $\tilde{\phi}_n(x)=e^{is}\phi_n(x)$ (where s is a real constant) also solves the TISE with the same energy. However, ant time you take any expectation value or calculate any probability, the constant phase (i.e. the $e^{is}$) makes no difference. Therefore, they represent the same physical state. So the energy eigenstate CAN be imaginary, but as you will see we can always chose it to be real while keeping full generality.

Next part is this: you can prove that for bound states in one dimension where $\lim_{x\rightarrow \pm \infty}\phi_n(x)=0$, any two solutions of the TISE are physically the same solution (that is, if two functions $\phi_n(x)$ and $\Phi_n(x)$ both solve the TISE with the same energy $E_n$, then it is necessarily true that $\Phi_n(x)=e^{is}\phi_n(x)$ where s is a real constant). This is why we always stop at one solution for each energy-- there can never be two physically different bound state solutions in one dimension (assuming $\lim_{x\rightarrow \pm \infty}\phi_n(x)=0$).

Finally, the last thing to prove is this: if $\phi_n(x)$ solves the TISE with an energy $E_n$, then the function $\phi_{R,n}(x)=\frac{\phi_n(x)+\phi_n^*(x)}{2}$ (which is real, as for any complex number or function z, $\frac{z+z^*}{2}$ is the real part of z, where the star represents the complex conjugate) also solves the TISE with energy $E_n$. But, if we are in a one dimensional bound state where $\lim_{x\rightarrow \pm \infty}\phi_n(x)=0$, since both solve the TISE with the same energy, they are physically the same solution (i.e. any non-real solution is just the real solution $\phi_{R,n}(x)$ multiplied by $e^{is}$). Therefore, we can pick whichever one we want to use, since they mean the same thing. We like to pick the real solutions for obvious reasons.

TL,DR: if we are in one dimension, in a bound state with the condition $\lim_{x\rightarrow \pm \infty}\phi_n(x)=0$, then we can restrict ourselves to real $\phi_n(x)$.

 

[PeroK]

Summary:: Energy Eigenstates Coefficients Real-valued?

Given any system with discreet energy eigenstates, φ_n(x)e^-iEnt . The φ_n are functions only of position. But are they also almost always real-valued?Thanks in advance.

The energy eigenstates can be taken to be real-valued functions. There is a simple proof of this. Basically, all you do is move any complex factors the the coefficients. You could, for example, have a system in the initial state:
$$\psi_0(x) = a \phi_n(x)$$
Where $a$ is any complex number with unit modulus. Of course, you could equally take this function, with the $a$ included, to be your basis eigenstate.