- #1

- 675

- 4

Given an operator [tex]\hat{Q}[/tex] (in the Schrodinger picture) in non-relativistic quantum mechanics and a state [tex]|\psi(t)\rangle[/tex] such that

[tex]\hat{Q} |\psi(t)\rangle=q(t)|\psi(t)\rangle[/tex]

where q(t) is explicitly time-dependent, can we properly say that [tex]|\psi(t)\rangle[/tex] is an eigenstate of Q with a time-dependent eigenvalue. That is, [tex]|\psi(t)\rangle[/tex] remains a eigenstate of Q for all times but its eigenvalue is different depending on when you measure it?

For example, suppose we had a wave function of the form

[tex]\psi(x,t)\propto e^{ixg(t)+h(t))}[/tex]

then applying the momentum operator we find

[tex]-i\frac{\partial}{\partial x}\psi(x,t)=g(t)\psi(x,t)[/tex].

Would you say that [tex]\psi(x,t)[/tex] is momentum eigenstate with momentum = g(t)?

[tex]\hat{Q} |\psi(t)\rangle=q(t)|\psi(t)\rangle[/tex]

where q(t) is explicitly time-dependent, can we properly say that [tex]|\psi(t)\rangle[/tex] is an eigenstate of Q with a time-dependent eigenvalue. That is, [tex]|\psi(t)\rangle[/tex] remains a eigenstate of Q for all times but its eigenvalue is different depending on when you measure it?

For example, suppose we had a wave function of the form

[tex]\psi(x,t)\propto e^{ixg(t)+h(t))}[/tex]

then applying the momentum operator we find

[tex]-i\frac{\partial}{\partial x}\psi(x,t)=g(t)\psi(x,t)[/tex].

Would you say that [tex]\psi(x,t)[/tex] is momentum eigenstate with momentum = g(t)?

Last edited: