Exponential of Hamiltonian-Calculate Probability

[jameson2]

Homework Statement

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

Homework Equations

Eigenvalues of H are E_\pm =\hbar \omega (n +\frac{1}{2}) \pm \hbar \lambda\sqrt{n+1} and eigenstates are |E_\pm&gt; =\frac{1}{\sqrt{2}}(|+,n&gt;\pm|-,n+1&gt;).

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 P(t)=cos^2(t\labda \sqrt{n+1}) but I've no idea how to start.

 

[sgd37]

if you are familiar with the taylor expansion of the exponential then you can interpret the exponential of the Hamiltonian as

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

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

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