Suppose that we have a hermitian operator. Let's take for example the Identical operator $I: I\Psi=\Psi$. What does the operator $e^{I}$ do? does it result in $e^{\Psi}$ or something else? What happens if we replace it with the momentum operator $p$?
Asked
Active
Viewed 365 times
2 Answers
9
You can Taylor-expand it just like you would a polynomial. Given the definition of the exponential function as a Taylor series, $$ e^{ax} = 1 + ax + \frac{(ax)^2}{2!} + \frac{(ax)^3}{3!} + \cdots,$$
you can generalize it to operators. Accordingly, the exponentiation of an operator, say the Hamiltonian $\hat{H},$ can be written as an infinite series of operators: $$ e^{\hat{H}} = \mathbb{I} + \hat{H} + \frac{\hat{H}^2}{2!} + \frac{\hat{H}^3}{3!} + \cdots$$ where the first operator on the right-hand-side is the identity operator: $\mathbb{I}: \mathbb{I} \psi = \psi.$
Of course, the choice of the operator is arbitrary, and as far as I know, it does not need to be Hermitian or unitary. In Quantum Mechanics, exponentiated operators serve the role of generators of motion in time or in a Hilbert space. For instance, the momentum operator $\hat{p}$ exponentiated as $e^{- i \, \textbf{x} \cdot \hat{p}/\hbar}$ represents a translation in position when applied to a wavefunction $\psi(x).$
Yejus
- 2,136
3
Functions of operators are understood as Taylor expansions. Thus, in the case of an exponent: $$ e^{\hat{A}} = \sum_{n=0}^{+\infty} \frac{\hat{A}^n}{n!} $$
Roger V.
- 68,984