Solving the Schrödinger eqn. by commutation of operators

[SemM]

Hi, I noticed that the raising and lowering operators:\begin{equation}
A =\frac{1}{\sqrt{2}}\big(y+\frac{d}{dy}\big)
\end{equation}\begin{equation}
A^{\dagger}=\frac{1}{\sqrt{2}}\big(y-\frac{d}{dy}\big)
\end{equation}can be used to solve the eqn HY = EY

However I am curious about something else. The commutation relation:

\begin{equation}
[A^{\dagger} A] = A^{\dagger} A - AA^{\dagger} = 1
\end{equation}is used to solve the Schrödinger eqn by multiplying

\begin{equation}
AA^{\dagger} Y = EY
\end{equation}with A from the right.

Can this procedure be used if\begin{equation}
[T^{\dagger} T] = T^{\dagger} T - TT^{\dagger} = d/dx
\end{equation}

, if not, how can the commuter (d/dx in this case) be used to solve an ODE composed of $T$ and $T^{\dagger}$?

Thanks!

 

[eljose79]

Hello,

Thank you for your question. The commutation relation you mentioned, [A^{\dagger} A] = A^{\dagger} A - AA^{\dagger} = 1, is a fundamental property of the raising and lowering operators, and it is indeed used in solving the Schrödinger equation. This relation is known as the "commutator" and it represents the difference between the operators when they are applied in different orders. In the case of the raising and lowering operators, this commutator is equal to 1, which is why they are so useful in solving the Schrödinger equation.

In general, the commutator of two operators is not always equal to a constant, and it depends on the specific operators being used. In the case of [T^{\dagger} T], it is not equal to a constant, but rather equal to the operator d/dx. This means that the commutator cannot be used directly in solving the ODE composed of T and T^{\dagger}, as it would not simplify the equation in the same way as in the case of the raising and lowering operators.

However, the commutator can still be useful in solving the ODE. For example, you can use the commutator to derive the eigenvalue equation for T and T^{\dagger}, which can then be used to solve the ODE. Additionally, the commutator can also be used to determine the commutation relations between T and T^{\dagger}, which can provide valuable information for solving the ODE.

I hope this answers your question. If you have any further inquiries, please do not hesitate to ask. Keep up the curiosity and keep exploring the fascinating world of quantum mechanics!