I am a beginner in Quantum mechanics and as I understand,the superposition of stationary states is also a solution of time-independent Schrödinger equation (TISE). The wave functions that are the solutions of TISE are stationary states. So this should mean the superposition of stationary state is also a stationary state. But it turns out it is not as the probability density varies with time if the energies of each state are different. I am terribly stuck after reading a lot of books so here's my question:
Is the superposition of stationary states a stationary state? If no, how can the solution to a TISE not be a stationary state (as I see in all the physical problems we find solutions to TISE and they are stationary)?
Also, I read somewhere that the expected value of Hamiltonian stays constant with time for such superposed wavefunction. If the expectation of Hamiltonian is constant then why isn't the superposed state stationary?
Best Answer
A superposition of stationary states is not a stationary state. Suppose we have two kets, $| E_1 \rangle$ and $|E_2 \rangle$, which solve the TISE like so: $$ \hat{H} | E_i \rangle = E_i | E_i \rangle $$ Here, the left hand side is the Hamiltonian operator and and the right hand side just shows that the stationary state picks up an eigenvalue. This is the definition of a solution of the TISE, although you may be more familiar with the differential operator version which is equivalent to left-multiplying by a position eigenstate $\langle x |$. Now, suppose we create a new superposition state, $| \psi \rangle$, defined as: $$ | \psi \rangle = \frac{1}{\sqrt{2}} \left( | E_1 \rangle + | E_2 \rangle \right) $$ If this is a solution to the TISE (that is, if it is a stationary state/eigenstate of the Hamiltonian) then it should follow that $\hat{H} | \psi \rangle = c | \psi \rangle$, where $c$ is a constant. But we can use linearity to write: $$ H | \psi \rangle = \frac{1}{\sqrt{2}} \left( \hat{H} | E_1 \rangle + \hat{H} | E_2 \rangle \right) \\ = \frac{1}{\sqrt{2}} \left( E_1 | E_1 \rangle + E_2 | E_2 \rangle \right) $$ This is not a constant $c$ multiplied by the vector $| \psi \rangle$ unless $E_1 = E_2$. So we can see that two stationary states cannot be combined into another stationary state unless they share the same eigenvalue.
As for the expectation value of the Hamiltonian, let's evaluate that real quick: $$ \langle \hat{H} \rangle = \langle \psi | \hat{H} | \psi \rangle \\ = \frac{1}{2} \left( \langle E_1 | + \langle E_2 | \right) \hat{H} \left( | E_1 \rangle + | E_2 \rangle \right) \\ = \frac{1}{2} \left( \langle E_1 | \hat{H} | E_1 \rangle + \langle E_2 | \hat{H} | E_1 \rangle + \langle E_2 | \hat{H} | E_2 \rangle + \langle E_1 | \hat{H} | E_2 \rangle \right)\\ = \frac{1}{2} \left(E_1 + E_2 \right) $$ Nothing about time dependence here at all. Basically, a combination of energy eigenkets will have a Hamiltonian expectation value which is a weighted average of the energies. Since the relative weights of energy eigenkets won't change, it stays constant--and we can always decompose a ket into a sum of energy eigenkets, so this is always true. (Assuming a time-independent Hamiltonian.) Note that this doesn't apply for other operators necessarily, since if it doesn't share eigenkets with the Hamiltonian you might find that $\langle E_1 | \hat{A} | E_2 \rangle$ does not evaluate to zero for an arbitrary operator $A$.