# How is the time independent potential term a solution of Schrodinger equation

Consider a time-independent potential: $V(x)$. Then, it is usually stated that $$\Psi(x,t)=\rho(x)\exp{\left(-\frac{i}{\hbar}Et\right)}$$ is the general form of a solution of the Schrodinger equation $$i\hbar \partial_t\Psi=H\Psi$$. Why is this the case?

• There is no $V(x)$ in the formula. (you can render latex-formulae by surrounding them with '$' characters) – M.Herzkamp Jun 19 '14 at 10:59 • Comment to the question (v3): The most general solution to the TDSE is not on product form$\Psi(x,t)=f(x)g(t)$. – Qmechanic Jun 24 '14 at 20:06 ## 1 Answer For a time-independent Hamiltonian$H$, the time-dependent Schrödinger equation$i\hbar\partial_t\Psi(x,t)=H\Psi(x,t)$can be solved by first finding the eigenvalues und eigenstates of the Hamiltonian, that is to say by solving the time-independent Schrödinger equation$H\rho_n(x)=E_n\rho_n(x)$. Then, the time-dependent Schrödinger equation is solved by$\Psi_n(x,t)=\rho_n(x)e^{-iE_nt/\hbar}$: $$i\hbar\partial_t\rho_n(x)e^{-iE_nt/\hbar} = E_n\rho_n(x)e^{-iE_nt/\hbar} = H\rho_n(x)e^{-iE_nt/\hbar}=H\Psi_n(x,t).$$ The general solution of the time-dependent Schrödinger equation is thus a superposition of the solutions$\Psi_n$. Let me show you how to arrive at the above form. Addendum 1: Separation of variables This approach comes from the theory of differential equations, where$H$is seen as a differential operator acting on wave functions. If the Hamiltonian is time-independent, it acts entirely only on the position variable$x$. One can then choose the separation ansatz$\Psi(x,t)=\rho(x)\phi(t)$and substitute it into the Schrödinger equation. Dividing it by$\Psi$gives: $$i\hbar\frac{\partial_t\phi}{\phi} = \frac{H\rho}{\rho}$$ where on the LHS the$\rho$could be canceled as$\partial_t$doesn't act on it and similarly on the RHS the$\phi$could be cancelled as$H$only acts on the position. Now one argues that this equation has to hold for any position$x$, so we require$H\rho/\rho$to be constant. We call the constant$E_n$. There may be any number of solutions, thus the constant is labeled by$n$. It is of course the energy obtained from solving the time-independent Schrödinger equation:$H\rho=E_n\rho$. Replacing the RHS by the constant gives $$i\hbar\partial_t\phi = E_n\phi\qquad\Rightarrow\qquad \phi(t)=e^{-iE_nt/\hbar}.$$ Addendum 2: Translational invariance in time The second approach is more technical but quite elegant and general. We define the time-translation operator$T_{t_0}$whose action on a function of time is defined as$T_{t_0}f(t) = f(t-t_0)$, i.e. that function is shifted in time by$t_0$. Clearly, the translation operator obeys a group-property$T_{t_0}T_{t_1} = T_{t_0+t_1}$. So its eigenvalues can be written in the form$f(t-t_0)=T_{t_0}f(t)=e^{\alpha t_0}f(t)$. So$f$needs to be an exponential function,$f(t)=e^{-\alpha t}$where$\alpha$is arbitrary for now. For a time-independent Hamiltonian$H$, if$\Psi_n$is a wave function with eigenenergy$E_n$, so$H\Psi_n=E_n\Psi_n$, it holds: $$T_{t_0}H\Psi_n(t) = T_{t_0}E_n\Psi_n(t) = E_nT_{t_0}\Psi_n(t) = E_n\Psi_n(t-t_0) = H\Psi_n(t-t_0) = HT_{t_0}\Psi_n(t),$$ so the Hamiltonian and$T_{t_0}$commute which is denoted$[T_{t_0},H]=0$with the commutator$[A,B]:=AB-BA$. When two operators commute, a theorem of linear algebra states that one can find common eigenfunctions. So let's assume$\Psi_n$is an eigenfunction to$H$with eigenvalue$E_n$and to$T_{t_0}$with eigenvalue$e^{\alpha t_0}$, such that the temporal dependence is given by$\Psi\propto e^{-\alpha t}$as reasoned before. Substituting this into the time-dependent Schrödinger equation and you get: $$i\hbar\partial_t\Psi_n(x,t) = -i\hbar\alpha\Psi_n(x,t)\overset{!}{=}E_n\Psi_n(x,t) = H\Psi_n(x,t).$$ Therefore,$\alpha=iE_n/\hbar$, so the time-dependence of$\Psi_n$is given by$e^{-iE_nt/\hbar}\$.

This second method is kind of a Bloch theorem in time. In the usual Bloch theorem, one considers (discrete) translations in space and also finds an exponential dependence.

• Hah, good job realizing what OP was actually asking! You might want to show the process of separation of variables, arriving at the time-independent Schrodinger eqn. from the time-dependent one, to maximize the understanding of OP. – Danu Jun 19 '14 at 12:53