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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 $p$?

Qmechanic
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Disjoint Union
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  • See https://physics.stackexchange.com/q/574621/50583, https://physics.stackexchange.com/q/133758/50583 for more discussions of exponential operators – ACuriousMind Sep 30 '20 at 18:11

2 Answers2

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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
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    The reason that this is a sensical definition is that a lot of identities from the exponential function can be proven using the Taylor series. So this carries over nicely to the operator definition. Try to 'proof' $\frac{d}{dx}e^{ax}=ae^{ax}$ using the Taylor series for example. A notable exception is that $e^{\hat A} e^{\hat B}\neq e^{\hat A\hat B}$. The Baker-Campbell-Hausdorff formula describes why and how this fails. https://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula – AccidentalTaylorExpansion Sep 30 '20 at 08:49
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    Another important exception: if $\hat H(t)$ depends on a parameter $t$ in such a way that $\left[\hat H(t), \tfrac{\mathrm d\hat H}{\mathrm dt}(t)\right]\neq 0$, then $ \tfrac{\mathrm d}{\mathrm dt}\left[e^{\hat H(t)} \right]\neq \tfrac{\mathrm d\hat H}{\mathrm dt}(t) e^{\hat H(t)} $. This is why time-ordered exponentials need to be used for the formal solution of the Schrödinger equation. – Emilio Pisanty Sep 30 '20 at 09:35
  • The series definition really only works for bounded exponents. For the general proper definition see https://physics.stackexchange.com/q/574621/50583. – ACuriousMind Sep 30 '20 at 18:13
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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.
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    Though in the case where the operator is diagonalizable, we can also take the more direct approach $f(A)\vert\psi\rangle:=f(a)\vert\psi\rangle$ for eigenstates $\vert\psi\rangle$ of $A$ with eigenvalue $a$ and then extending linearly to all other states. This approach also works for functions which do not have a series representation. – Vercassivelaunos Sep 30 '20 at 08:52
  • This is the first time I see this approach, but it does make sense. – Roger V. Sep 30 '20 at 08:55
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    The series definition really only works for bounded exponents. For the general proper definition see https://physics.stackexchange.com/q/574621/50583. – ACuriousMind Sep 30 '20 at 18:13