# Quantum Mechanical Operators in the argument of an exponential

In Quantum Optics and Quantum Mechanics, the time evolution operator

$$U(t,t_i) = \exp\left[\frac{-i}{\hbar}H(t-t_i)\right]$$

is used quite a lot.

Suppose $t_i =0$ for simplicity, and say the eigenvalue and eigenvectors of the hamiltionian are $\lambda_i, \left|\lambda_i\right>$. Now, nearly every book i have read and in my lecture courses the following result is given with very little or no explanation:

$$U(t,0) = \sum\limits_i \exp\left[-\frac{i}{\hbar}\lambda_it\right]\left|\lambda_i\right>\left<\lambda_i\right|$$

This is quite a logical jump and I can't see where it comes from, could anyone enlighten me?

-

Starting with:

$$U(t,t_i) = e^{\frac{-i}{\hbar }H(t-t_i)}$$

If $t_i=0$:

$$U(t,0) = e^{\frac{-i}{\hbar }Ht}$$

Using the identity: $\sum\limits_i \left|\lambda_i\right>\left<\lambda_i\right|=\mathbb{I}$

$$U(t,0) = \sum\limits_i e^{\frac{-i}{\hbar }Ht}\left|\lambda_i\right>\left<\lambda_i\right|$$

Since the exponential of an operator is (by Taylor expanding): $e^H=\mathbb{I}+H+\frac{1}{2}H^2+\dots$

And: $H\left| \lambda_i \right> =\lambda_i \left| \lambda_i \right>$

You should be able to see that:

$$U(t,0) = \sum\limits_i e^{\frac{-i}{\hbar }\lambda_it}\left|\lambda_i\right>\left<\lambda_i\right|$$

-
I always forget about the series expansion, thanks very much –  StickyCube May 18 '13 at 17:06
Without loss of generality, let's take the $|\lambda_i\rangle$ to be orthonormal. Notice that, by the spectral theorem, the hamiltonian can be written as follows: $$H = \sum_i \lambda_i P_i, \qquad P_i = |\lambda_i\rangle\langle \lambda_i|$$ Each operator $P_i$ is a projectors onto the subspace spanned by $|\lambda_i\rangle$. Notice, in particular, that $$P_i^2 = P_i, \qquad P_iP_j = P_jP_i = 0$$ and a mathematical induction argument gives $$P_i^n = P_i$$ for all $n\geq 1$. Now, for notational simplicity let $$\mu = -\frac{i}{\hbar}t$$ Then we have $$U(t,0) = e^{\mu H} = \sum_{n=0}^\infty \frac{1}{n!}\mu^nH^n$$ but notice that using the properties of projection operators written above, we have $$H^n = \sum_{i_1, \dots, i_n}\lambda_{i_1}\cdots\lambda_{i_n}P_{i_1}\cdots P_{i_n} = \sum_i\lambda_i^nP_i$$ and therefore $$U(t,0) = \sum_i\sum_n\frac{1}{n!}(\mu\lambda_i)^nP_i = \sum_ie^{\mu\lambda_i}P_i$$ as desired.