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?

  • $\begingroup$ There is no $V(x)$ in the formula. (you can render latex-formulae by surrounding them with '$' characters) $\endgroup$
    – M.Herzkamp
    Commented Jun 19, 2014 at 10:59
  • $\begingroup$ Comment to the question (v3): The most general solution to the TDSE is not on product form $\Psi(x,t)=f(x)g(t)$. $\endgroup$
    – Qmechanic
    Commented Jun 24, 2014 at 20:06

1 Answer 1


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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Danu
    Commented Jun 19, 2014 at 12:53

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