Finding A Solution Using the Ladder Operators

[Bashyboy]

Hello,

I am reading Griffiths Quantum Mechanics textbook, and am having some difficulty with a derivation on page 56. To me, there seems to be something logically wrong with his arguments, but I can not pin-point precisely what it is.

To provide you with a little background, Griffiths is trying to solve the Schrodinger equation for a quantum harmonic oscillator potential, employing what he refers to as the "algebraic method." He already has defined the raising and lower operators, and has already shown us how to factor the Hamiltonian operator. Here is where I am having trouble:

"Now, here comes the crucial step: I claim that if [itex]\psi[/itex] satisfies the Schrodinger equation with energy E (that is:[itex]H \psi = E \psi[/itex]), then [itex]a_+ \psi[/itex] satisfies the Schrodinger equation with energy [itex]E + \hbar \omega[/itex]..."

The rest involves him going through the proof. As I mentioned above, there seems to be something wrong with the author's reasoning, but I can't quite pin-point it. To me, it seems that I could have chosen absolutely operator I wish, call it [itex]*_1[/itex], and say that the energy is [itex]E+*_2[/itex], and just find a relationship that looks like the Schrodinger equation, which appears to be what he did.

 

[maajdl]

Give a try to your idea.
Take the x operator for example.
You will see that x ψ is not an eigenvector of H even when ψ is an eigenvector (i.e. Hψ=Eψ).

 

[Bashyboy]

Ah, okay. What you have said sort of naturally leads to my next question. Obviously, [itex]\psi[/itex] is a solution to the Schrodinger equation if it satisfies the PDE. But is this equivalent to [itex]\psi[/itex] being an eigen-function of the Hamiltonian; in short, is the following statement true: [itex]\psi[/itex] satisfies the Schrodinger PDE iff [itex]\psi[/itex] is an eigen-function of H?

 

[stevendaryl]

Ah, okay. What you have said sort of naturally leads to my next question. Obviously, [itex]\psi[/itex] is a solution to the Schrodinger equation if it satisfies the PDE. But is this equivalent to [itex]\psi[/itex] being an eigen-function of the Hamiltonian; in short, is the following statement true: [itex]\psi[/itex] satisfies the Schrodinger PDE iff [itex]\psi[/itex] is an eigen-function of H?

Yes.

The time-dependent Schrodinger equation is:

[itex]H \psi(x,t) = i \hbar \frac{d}{dt} \psi(x,t)[/itex]

If [itex]\psi_0(x)[/itex] satisfies [itex]H \psi_0 = E \psi_0[/itex] for some real number [itex]E[/itex], then [itex]\psi(x,t) = e^{\frac{-i E t}{\hbar}} \psi_0(x)[/itex] satisfies the time-dependent Schrodinger equation.

 

[Bashyboy]

This may seem a bit naive, but is there a way of proving that [itex]\psi[/itex] being mapped to a scalar multiple of itself by the Hamiltonian operator is the same as [itex]\psi[/itex] satisfying the PDE? Or does this follow from the fact that the Schrodinger equation can be written, equivalently, as [itex]H \psi = E \psi[/itex]?

 

[maajdl]

What you say is not naive, it is just totally unclear.
Are you talking about the stationary Schrödinger equation or the time-dependent equation?
If ψ is a solution of the stationary equation, then -by definition- it is one of the eigenvectors of H, and it is -also by definition- the spatial part of a "stationary" solution of the full-time-dependent Schrödinger equation (see stevendaryl).

 

[stevendaryl]

This may seem a bit naive, but is there a way of proving that [itex]\psi[/itex] being mapped to a scalar multiple of itself by the Hamiltonian operator is the same as [itex]\psi[/itex] satisfying the PDE? Or does this follow from the fact that the Schrodinger equation can be written, equivalently, as [itex]H \psi = E \psi[/itex]?

Yes. Saying that [itex]\psi[/itex] is mapped to a scalar multiple of itself by the Hamiltonian operator is just saying that there is some [itex]E[/itex] such that [itex]H \psi = E \psi[/itex]. Any [itex]\psi[/itex] that satisfies that equation solves the Schrodinger equation.