Exponential of Hamiltonian-Calculate Probability

jameson2
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Homework Statement


I have to evaluate [tex]P(t)=|<+,n|\exp{\frac{-iHt}{\hbar}}|+,n>|^2[/tex] where [tex]H=\hbar \omega_0 S_z + \hbar \omega a^+a+\hbar \lambda(a^+S_-+aS_+)[/tex] and [tex]|+,n>=\left( \begin{array}{c}<br /> 1\\0 \end{array} \right)[/tex]

Homework Equations


Eigenvalues of H are [tex]E_\pm =\hbar \omega (n +\frac{1}{2}) \pm \hbar \lambda\sqrt{n+1}[/tex] and eigenstates are [tex]|E_\pm> =\frac{1}{\sqrt{2}}(|+,n>\pm|-,n+1>)[/tex].


The Attempt at a Solution


Basically, I don't know how to treat the hamiltonian when it's in the exponential like that. The answer is given as [tex]P(t)=cos^2(t\labda \sqrt{n+1})[/tex] but I've no idea how to start.
 
on Phys.org
if you are familiar with the taylor expansion of the exponential then you can interpret the exponential of the Hamiltonian as

[tex]e^{\frac{-iHt}{\hbar}} = 1+ \frac{-iHt}{\hbar} + \frac{1}{2} (\frac{-iHt}{\hbar})^2 + \cdots[/tex]

for a state satisfying [tex]H \left| \psi \right\rangle = \lambda \left| \psi \right\rangle[/tex] the action of the exponentiated hamiltonian on such a state is given by

[tex]e^{\frac{-iHt}{\hbar}} \left| \psi \right\rangle = e^{\frac{-i \lambda t}{\hbar}} \left| \psi \right\rangle[/tex]
 

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