# Can we safely assume $\Psi(x,t) = \psi(x)e^{-i\omega t}$ always in QM?

In the particle in a box, harmonic oscillator and in Hydrogen Atom, we can safely assume $$\Psi(x,t) = \psi(x)e^{-i\omega t}.$$ So why not make it a postulate to consider the wave function to be always in the form $$\psi(x)e^{-i\omega t}$$ We can still explain all the foundational experiments, so my question is why not make this amendment to the theory so that we can avoid a lot of nightmares and some peace of mind. Why unnecessarily consider so general thing when we can do away with simple things most of the time?

Oh my actual question is where do we run into problems (practical physical situations/experiments) if we consider such a thing? If we run into problem anywhere, I hope to make any other simple changes to fix them but without going back on this one.

It is not general. For both the harmonic oscillation and the hydrogen atom, we have a Hamitlonian $\hat{H} = -\frac{i\hbar}{2m}\nabla^2 + V$ with a time-independent potential, which implied that the eigenvalue equation $\hat{H}\Psi(x,t) = E\Psi(x,t)$ is separable, and we can therefore write it as a product of a time-independent factor and a space-independent factor.

In other words, $\Psi(x,t) = \Psi(x)e^{-i\omega t}$ is something we get for the energy eigenstates with a time-independent Hamiltonian. A general solution is a superposition of multiple energy eigenstates, and is not going to have that form.

• You say time independent Hamiltonian, and at the same time involve multiple Energy eigen states, So my question is What is the total energy that is conserved in such a Schrodinger equation? – Rajesh Dachiraju Dec 26 '13 at 3:52
• @RajeshD that would be the expectation value of the Hamiltonian. – David Z Dec 26 '13 at 4:07
• @DavidZ : Again these are contrived, and not really demanded from Schrodinger's equation or from conservation of energy. As I understand, these multiple eigen energies are embedded into the wavefunction and they are not generated by the potential V(x) or from the Hamiltonian, right? – Rajesh Dachiraju Dec 26 '13 at 5:00
• What I mean is these are set up and not following directly from te Schrodinger equation. – Rajesh Dachiraju Dec 26 '13 at 5:02
• @RajeshD: No. The energy eigenvalues are the spectrum of the Hamiltonian, and so generated by it directly. In particular, $\hat{H}\Psi_n = E_n\Psi_n$. Since the Schrödinger equation is linear, any superposition of the eigenstates is a solution, and the fact that cover all solutions follows from hermiticity of the Hamiltonian. – Stan Liou Dec 26 '13 at 5:22

Another, as of yet unmentioned, reason is for super-position.

For separation of variables techniques, we find $\Psi(\mathbf{x},t)=\psi(\mathbf{x})f(t)$ to give $$\frac{i\hbar}{f(t)}\frac{df}{dt}=E=\left[\frac{\hbar^2}{2m}\nabla^2+V(\mathbf{x})\right]\psi(\mathbf{x})\tag{1}$$ The term on the left side gives $f\propto\exp\left[iEt/\hbar\right]$, giving us $$\psi\left(\mathbf{x},t\right)=\psi(x)e^{iEt/\hbar}\tag{2}$$ But this is a stationary state, but the particle that Equation (1) describes is not stationary. Thus, we have that the general solution is a linear super-position of states: $$\psi\left(\mathbf{x},t\right)=\sum_ic_i\psi_i(x)e^{iE_it/\hbar}$$

Which cannot be separated into such a form as Equation (2).

• Yes, even with a time-independent Hamiltonian only the eigenfunction are generally time-independent. – dmckee Dec 26 '13 at 3:13
• What is the total energy of such a system, in this case? – Rajesh Dachiraju Dec 26 '13 at 3:39
• @RajeshD: dmckee's statement, if understood as referring to superpositions, follows directly from the Born rule. – Stan Liou Dec 26 '13 at 4:17
• You keep using that word "contrived". What do you think it means? The eigenstates and their spectrum are a consequence of the Hamiltonian. If you have a superposition that means that you don't know the energy of the state. There is no surprise that you don't then know what energy you will get if you measure the energy. This is a tautology. – dmckee Dec 26 '13 at 5:06
• @RajeshD: Schrödinger's equation is linear, so any superposition of solutions is a solution. Thus you almost never have just one eigenstate for a general solution of the SE. Conservation of energy follows from the time-independence of the Hamiltonian, but is otherwise not going to hold. The connection to probabilities of measurement outcomes are and what happens after measurement it not demanded by SE, but by the Born rule. And as dmckee said, the energy eigenvalues are the spectrum of the Hamiltonian. – Stan Liou Dec 26 '13 at 5:16

Only when $V(\mathbf{x},t) = V(\mathbf{x})$, otherwise the time-dependent Schrödinger equation does not separate into time- and space-like parts.

Any explicit time-dependence in the problem and you cannot.

• This Statement and the answer by Stan Liou, should be embossed on every QM text book with a large font size on the cover sheet. Just so that the students are going to relax and also the fact that we are never going for the time dependent potentials. Everybody starts writing $\Psi(x,t)$, which they are never going to use, just to frighten people! – Rajesh Dachiraju Dec 26 '13 at 3:07
• @RajeshD: Bad news: time-dependent perturbation theory is pretty standard for undergrad QM. Though even if $\hat{H}$ is time-independent, you'll still be coming across situations where you need a non-stationary state. – Stan Liou Dec 26 '13 at 3:16

In a different vein from the other answers, an example of physical a problem in which the solution is not separable is one considered in Sakurai 2nd Edition, Chapter 2.1, Pgs. 70 and 71 where we have a spin magnetic moment in the presence of a magnetic field whose intensity can change with time I.e. $B(t)\neq B(0)$, or even more problematically when the magnitude and direction of the magnetic field change with time. In both cases the Hamiltonian is time dependent and so the solution is not separable.