- #1
TheCanadian
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- 13
Homework Statement
## H ## is the Hamiltonian of an electron and is a Hermitian operator. It satisfies the following equation:
##H |\phi_n\rangle = E_n |\phi_n\rangle ##
Let ## U = e^{\frac {iHt}{\hbar}} ##. Find the expectation value of U in state ##|\phi_n\rangle##
Homework Equations
## \langle U\rangle = \langle\phi_n| U |\phi_n\rangle ##
The Attempt at a Solution
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I have tried to use the general expression for H, which is: ## H = i\hbar \frac{\partial}{\partial t} ##
Applying this to U, I have: ## U = e^{\frac {iH\frac{\partial t}{\partial t}}{\hbar}} = e^{-1} ##.
This implies:
## \langle U\rangle = \langle\phi_n| e^{-1} |\phi_n\rangle = e^{-1} \langle\phi_n| \phi_n \rangle = e^{-1} ##
But I've been told this is wrong, although I have not been told exactly why. Instead I have been told the solution is:
## \langle U\rangle= \langle\phi_n| e^{\frac {iHt}{\hbar}} |\phi_n\rangle = \langle\phi_n| e^{\frac {iE_nt}{\hbar}} \phi_n \rangle
= e^{\frac {iE_nt}{\hbar}} \langle\phi_n |\phi_n \rangle= e^{\frac {iE_nt}{\hbar}} ##
Although the explanation for this alternate solution was lacking. Namely, why is it true that:
## e^{\frac {iHt}{\hbar}} | \phi_n \rangle = e^{\frac {iE_nt}{\hbar}} |\phi_n\rangle ## ?
Isn't it only true if the Hamiltonian, H, is directly acting on ## |\phi_n\rangle ##? Why is it that when H is in the exponent of the operator that it is still acting on the state function to yield ## E_n## as the eigenvalue? Any explanation on this and why my approach is wrong would be great!
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