schrodinger picture derivation

[skateboarding]

I had a math question about the following steps.

The Shroedinger's equation can be written as follows.

\LARGE i\hbar \frac{d}{dt}U(t) |\psi(0)> = HU(t)|\psi(0)>

Where H is the hamiltonian and U is the time evolution operator.
So U satisfies the schrodinger equation.

\LARGE i\hbar\frac{d}{dt}U(t) = HU(t)

This is the part that I don't quite understand.

If the hamiltonian H is independent of time, the solution to the above equation is

\LARGE U(t) = e^{-iHt/\hbar}

This is a solution of integration with respect to U to solve for the differential equation.
My problem is that i can't seem to justify the expression:

\LARGE \int i\hbar \frac{dU(t)}{U(t)} = \int Hdt

Am I allowed to divide operators and shuffle them around? Furthermore, when I think of operators in their matrix representation, I am even more confused, since I do not know of matrix division, only inverse operations on matrix.
Any help would be appreciated. Thanks.

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1. The problem statement, all variables and given/known data



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3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

[Tangent87]

If you're not happy with separating variables you can always use an integrating factor:


\LARGE \frac{d}{dt}\left(U(t)e^{iHt/{\hbar}}\right) = 0


So then:


\LARGE U(t)e^{iHt/{\hbar}} = U(0)


U(0) can be normalised to 1 therefore
\LARGE U(t) = e^{-iHt/\hbar}