[Physics] How do base kets satisfy Schrödinger’s equation in Schrödinger picture and why don’t they evolve with time

hilbert-spaceoperatorsquantum mechanicsschroedinger equationtime evolution

According to Sakurai, eigenvalue equation for an operator $A$, $A|a'\rangle=a'|a'\rangle$. In the Schrödinger picture, $A$ does not change, so the base kets, obtained as the solutions to this eigenvalue equation at t=0, for instance, must remain unchanged.

  1. Since base kets do not evolve with time $|a',t\rangle=|a'\rangle$ and is independent of t.

    Schrödinger equation
    $$i\hbar\frac{\partial |a',t\rangle}{\partial t}=H|a',t\rangle,$$
    the LHS is zero and RHS is non-zero. Why is the Schrödinger equation not satisfied?

  2. Suppose $A$ commutes with $H$ (Hamiltonian).

    $A|a'\rangle=a'|a'\rangle$ and evolution operator is $U(t,0)=\exp(-\frac{iHt}{\hbar})$

    $$UA|a'\rangle=Ua'|a'\rangle$$

    Since $H$ and $A$ commute, $U$ and $A$ also commute.

    $$AU|a'\rangle=a'U|a'\rangle$$

    So the eigenvalue remains same and eigenket is now $U|a'\rangle$ and evolves with time, which reduces to $|a'\rangle$ at t=0.

    So, I can conclude that base kets evolve with time when $A$ commutes with Hamiltonian. This has an additional advantage that Schrödinger Equation is now satisfied.

As stated in the book, the base kets do not change in the Schrödinger picture. Is this statement wrong in the above case?

Best Answer

Since base kets do not evolve with time $|a',t\rangle=|a'\rangle$

This is a misreading of that statement. When we say that the energy-eigenbasis kets of a time-independent hamiltonian to not evolve with time, what we mean is that if $H|E\rangle = E|E\rangle$ then $$ |\Psi_E(t)\rangle = e^{-iEt/\hbar}|E\rangle $$ is a solution of the Schrödinger equation, which starts off at $|\Psi_E(0)\rangle = |E\rangle$ and which maintains a unit inner product with its initial condition, $$ |\langle \Psi(0)|\Psi(t)\rangle| = |\langle E |\Psi(t)\rangle| = 1. $$ The phase factor $e^{-iEt/\hbar}$ is absolutely crucial for the Schrödinger equation to be satisfied. On the other hand, at any given time $t$ it acts as a global phase factor, which is irrelevant for any physical observable, which is why it is justified to say that the base ket does not 'really' evolve with time.

For your second question, your initial manipulations are correct, and they are best understood in the form $$ A\ U|a\rangle = U\ A|a\rangle = U\ a|a\rangle = a\ U|a\rangle $$ (where I've dropped the primes, to $A|a\rangle = a|a\rangle$, because yours didn't make any sense). In other words, if you start off in an eigenvector $|a\rangle$ of $A$ with eigenvalue $a$ and $A$ commutes with $H$, then the time-evolved state $U(t)|a\rangle$ will always be an eigenstate of $A$.

Now, if you know that $A$ is non-degenerate on that eigenspace, then that combination allows you to conclude that $|a\rangle$ is also an eigenstate of $H$ and the time evolution will keep $U(t)|a\rangle$ as a multiple of $|a\rangle$.

On the other hand, it is perfectly possible for $A$ to have a degenerate eigenspace there, in which case $U(t)|a\rangle$ can have a nontrivial time dependence. If you want an explicit example, try $$ A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix} \ \text{under} \ H = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 3 \end{pmatrix} $$ with $|a\rangle = (1,0,0)$.

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