According to Sakurai, eigenvalue equation for an operator $A$, $A|a'\rangle=a'|a'\rangle$. In the Schrodinger 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.

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

Schrodinger 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 Schrodinger equation not satisfied?

Question 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 Schrodinger Equation is now satisfied.

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

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?

According to Sakurai, eigenvalue equation for an operator A, $A|a'\rangle=a'|a'\rangle$. In the Schrodinger 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.

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

Schrodinger equation $i\hbar\frac{\partial |a',t\rangle}{\partial t}=H|a',t\rangle$

LHS is zero and RHS is non-zero. Why is Schrodinger equation not satisfied?

Question 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 Schrodinger Equation is now satisfied.

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

According to Sakurai, eigenvalue equation for an operator $A$, $A|a'\rangle=a'|a'\rangle$. In the Schrodinger 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.

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

Schrodinger 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 Schrodinger equation not satisfied?

Question 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 Schrodinger Equation is now satisfied.

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