# Exponential of Hamiltonian-Calculate Probability

jameson2

## Homework Statement

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

## Homework Equations

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

## 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.

## The Attempt at a Solution

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

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