How do base kets satisfy Schrödinger's equation in Schrödinger picture and why don't they evolve with time?

Question

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?

Answer 1

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.

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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 $|a\rangle = (1,0,0)$.

Answer 2

Only kets that represent physical systems ("state vectors") satisfy the Schrodinger equation. Basis kets don't represent physical systems, but just a system of coordinates, so they don't.

Your question is analogous to asking why the coordinates of a random point in space don't satisfy Hamilton's or the Euler-Lagrange equations. There's just nothing there to time-evolve.

Answer 3

It's true that kets do not evolve with time (kets are not functions of time). However, the state vector $|\psi(t)\rangle$ does evolve with time (since we're in the Schrodinger picture).

That is, I think you're failing to distinguish between the eigenket $|a'\rangle$ of the operator $A$ and the state vector $|\psi(t)\rangle$ which is a ket valued function of time that satisfies the Schrodinger equation.

Assuming the Hamiltonian $H$ is time independent and stipulating that the state vector at time $t = 0$ is the ket $|\psi_0\rangle$, the state vector at any other time $t$ is given by

$$|\psi(t)\rangle = e^{-\frac{it}{\hbar}H}|\psi_0\rangle = |\psi_0\rangle - \frac{it}{\hbar}H|\psi_0\rangle + \cdots$$

Since, in general, $H|\psi_0\rangle$ is not proportional to $|\psi_0\rangle$, the state vector at time $t \ne 0$ is in a different ray than $|\psi_0\rangle$. However, in the case that $H|\psi_0\rangle = E|\psi_0\rangle$, we have

$$|\psi(t)\rangle = e^{-\frac{it}{\hbar}H}|\psi_0\rangle = e^{-\frac{iEt}{\hbar}}|\psi_0\rangle$$

and so the state vector remains in the initial ray (since states are rays, all $e^{i\theta}|\psi_0\rangle$ represent the same state).

In summary, if $|\psi_0\rangle = |a'\rangle$ then, unless $A$ commutes with $H$, the state vector $|\psi(t)\rangle$ will evolve with time to a ket in a different ray. Again, $H$ does not evolve the ket $|a'\rangle$ but, rather, $H$ evolves the state vector $|\psi(t)\rangle$

Answer 4

I think there is a lot of confusion in what you wrote. The Schrodinger equation is satisfied,of course not by the eigenkets themselves, but by the expansion coefficients of any state expanded on the basis of eigenkets. The eigenkets are defined as being time independent exactly because, for a general state, the time dependence of each expansion coefficient on the eigenket basis is really simple. To make this clear, for any state $|\psi\rangle$, take a complete set of eigenkets $(|a_1\rangle,|a_2\rangle,|a_3\rangle,...)$, then the state $|\psi\rangle$ can be expressed as: $$|\psi\rangle=\sum_n c_n(t)*|a_n\rangle$$

So $|\psi\rangle$ is actually $|\psi(t)\rangle$, and it does satisfy Schrodingers equation. So if the $|a_n\rangle$ are the eigenkets of the Hamiltonian, then the time dependence of the expansion coefficients is just the usual $e^{-i\frac {E_n}{\hbar}t}$, and the expansion of the state is just: $$|\psi(t)\rangle=\sum_n c_n(0)*e^{-i\frac {E_n}{\hbar}t}*|a_n\rangle$$