[Physics] Quantum Mechanical Operators in the argument of an exponential

hilbert-spaceoperatorsquantum mechanicsquantum-opticstime evolution

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?

Best Answer

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:

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

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