I My problem with time-dependent Schrodinger equation

[wirefree]

TL;DR

David A. B. Miller's book titled 'Quantum Mechanics for Scientists and Engineers' has an incocsistency

Section 3.3 titled 'Solutions of the time-dependent Schrödinger equation' states in its 1st line that the time-dependent solution is not an eigenvalue equation:

The same section ends with a comment on eigenstates:

How do you reconcile this: are solutions to the time-dependent equation eigenvalue solutions or not?

 

[Nugatory]

are solutions to the time-dependent equation eigenvalue solutions or not?

They are not.
The book is saying that the time-independent Schrodinger equation is an eigenvalue equation: We are solving for functions $\psi(x)$ that satisfy $\hat{H}\psi=E\psi$ where $E$ is a constant.

The time-dependent Schrodinger equation is a differential equation: We are solving for functions $\psi(x,t)$ that satisfy $\hat{H}\psi=\hat{E}\psi$ where $\hat{E}$ is the operator $\hat{E}=i\hbar\frac{\partial}{\partial t}$.

The connection between the two is that if $\psi(x)$ is a eigenfunction solution of the time-independent equation with eigenvalue $E$, then $\psi(x,t)=\psi(x)e^{iEt/\hbar}$ is a solution of the time-dependent equation.

 

[wirefree]

The connection between the two is that if $\psi(x)$ is a eigenfunction solution of the time-independent equation with eigenvalue $E$, then $\psi(x,t)=\psi(x)e^{iEt/\hbar}$ is a solution of the time-dependent equation.

The operator E in the time-dependent version, as well as the operator H in the time-independent version, is a differential operator: former of time, and latter of space.
So, they both clearly result in differential equations.

The question becomes: what makes one an eigenvalue problem?

Regards & Best Wishes,
wirefree

 

[Nugatory]

The question becomes: what makes one an eigenvalue problem?

The eigenvalue problem for a given operator is finding the vectors that are mapped to a scalar multiple of themselves by that operator. So the time-independent Schrodinger equation is an eigenvalue equation because $E$ is a scalar.

 

[wirefree]

The eigenvalue problem for a given operator is finding the vectors that are mapped to a scalar multiple of themselves by that operator.

That is indeed the definition of an eigenvalue problem.

Why does one type of differential equation lead to such a situation, while the other doesn't?

 

[PeroK]

That is indeed the definition of an eigenvalue problem.

Why does one type of differential equation lead to such a situation, while the other doesn't?

The "value" in eigenvalue refers to a number. Like the $\lambda$ in $M\vec v = \lambda \vec v$. This linear algebra extends to function spaces, where the matrix ($M$) becomes an operator, as in $\hat T \psi(x) = \lambda \psi(x)$.

The TDSE is of the form: $\frac {\partial} {\partial t} \psi(x, t) = \hat T \psi(x, t)$. This is not an eigenvalue problem, but a more general operator equation. The technique of separation of variables, however, reduces the problem to an eigenvalue problem in the spatial variable.

 

[MichPod]

are solutions to the time-dependent equation eigenvalue solutions or not?

Some solutions of time-dependent Schrodinger equation may in fact be eigenvalue solutions of time-independent Schrodinger equation as well, if you allow such an informality as to substitute $\Psi(r,t)$ into an equation which normally expects $\psi(r)$.

Others are not, but may be expressed as linear combinations of those solutions which are eigenvalues (in the sense I discussed in the above paragraph).

PS: I like QMSE book of Miller and IMO he is a great teacher.