The notion of operators in exponentials is a bit confusing to me. I know that that in some cases one can use the Taylor series of $e^x$, but how do you work with them when that's not the case?

  • 3
    $\begingroup$ Generally if $A | v \rangle = \lambda | v \rangle$, then $f(A) | v \rangle = f(\lambda) |v \rangle$. (Also, I haven't run into a case where you're not allowed to use the Taylor series; that's how we define $e^{\beta H}$ in the first place.) $\endgroup$ – knzhou Oct 17 '17 at 9:31
  • $\begingroup$ The exponential of a self-adjoint operator on an invariant domain from the maximal domain of the operator is defined by means of the so-called "functional calculus", i.e. the exponential is mapped to the von Neumann spectral decomposition of that self-adjoint operator. See also here: mathoverflow.net/questions/95334/… $\endgroup$ – DanielC Oct 17 '17 at 9:52

$\newcommand{\ket}[1]{\left| #1 \right>}$Let's take a step back and look at an arbitrary linear operator $H$ (I don't like putting hats on operators). You probably know from your linear algebra classes how to act on a vector $\ket \psi$. Well it is just $H \ket \psi = \ket \varphi$, which is of course another vector in your vector space, which I have called $\ket \varphi$. Now you probably also know, how $H^2$ acts on a vector; namely you just act with the operator $H$ twice:

$$H^2 \ket \psi = H (H \ket \psi) = H \ket \varphi = \ket \chi$$

where I defined the vector $\ket \chi := H \ket \varphi $

By induction, you also know how $H^n$ acts for $n \in \mathbb N$.

Remember also that you are in a vector space that is you can add two vectors, and in particular two linear operators. For example, let's look at the operator $H^2+H$:

$$ (H^2+H) \ket \psi = H^2 \ket \psi + H \ket \psi = \ket \chi + \ket \varphi $$

Again by induction you know for a polynomial $p(x) = \sum_{k=1}^n a_k x^k$, where $a_k \in \mathbb C$ some constant coefficients, how the operator $p(H)$ acts on states:

$$ p(H) \ket \psi = \sum_{k=1}^n a_k (H^k \ket \psi) $$

Without worrying too much about convergence issues (we are physicists (!)), we can define the exponential map of an operator:

$$ \exp(H) = \sum_{n \in \mathbb N} \frac{1}{n!} \, H^n \implies \exp(H) \ket \psi = \sum_{n \in \mathbb N} \frac{1}{n!} \, (H^n \ket \psi) $$

The upshot is that you have to define the exponential of an operator and there is no way around it. There is however a special case, when for example you have an eigenvector of the operator. Let's assume that the vector $\ket E$ is an eigenvector of $H$ with eigenvalue $E$ i.e. $H \ket E = E \ket E$, then we have:

$$ \exp(H) \ket E = \sum_{n \in \mathbb N} \frac{1}{n!} \, (H^n \ket E ) = \sum_{n \in \mathbb N} \frac{E^n}{n!}\ket E = \exp(E)\ket E$$

In analogy, you can convince yourself that if you can write a function $f$ in taylor series and if $\ket E$ is an eigenvector of $H$ as above then we have $f(H) \ket E = f(E) \ket E $.

I hope that this helps you understand the exponential of an operator. If you have question, don't hesitate to ask!


The exponential of an operator is defined by its series expansion: \begin{align} e^{\beta \hat H}&= \hat 1 +\beta \hat H+\frac{1}{2}\beta^2\hat H^2+\ldots \, ,\\ &=\sum_{k=0}^\infty \frac{\beta^k \hat H^k}{k!}\, .\tag{1} \end{align} If $\vert\psi\rangle$ is an eigenstate of $\hat H$ so that $$ \hat H\vert\psi\rangle=\lambda\vert\psi\rangle\, ,\quad \hat H^2\vert\psi\rangle=\lambda^2\vert\psi\rangle\, ,\quad \hat H^n\vert\psi\rangle=\lambda^n\vert\psi\rangle $$ then $$ e^{\beta \hat H}\vert\psi\rangle =\sum_{k=0}^\infty \frac{\beta^k \hat H^k}{k!}\vert\psi\rangle =\sum_{k=0}^\infty \frac{\beta^k \lambda^k}{k!}\vert\psi\rangle =e^{\beta \lambda}\vert\psi\rangle\, . $$

If $\vert\psi\rangle$ is not an eigenstate, there are fewer options. First, work out in details $$ e^{\beta \hat H}\vert\psi\rangle=\sum_{k=0}^\infty \frac{\beta^k \hat H^k}{k!}\vert\psi\rangle \tag{2} $$ and hope that each $\hat H^k\vert\psi\rangle$ can be simplified. This occurs, for instance, with Pauli matrices where - say - $\hat H=\sigma_j$ and $\sigma_j^2=\hat 1$ for any $j=x,y,z$. It may then be possible to resum the series in (2).

The second alternative is to expand $\vert\psi\rangle$ in terms of eigenstates of $\hat H$. Thus, if $\vert\mu_\alpha\rangle$ is such that \begin{align} \hat H\vert\mu_\alpha\rangle&=\lambda_\alpha\vert\mu_\alpha\rangle\, ,\\ \vert\psi\rangle&=\sum_{\alpha}\vert\mu_\alpha\rangle\langle \mu_\alpha\vert\psi\rangle \end{align} then $$ e^{\beta \hat H}\vert\psi\rangle=\sum_{\alpha}e^{\beta\hat H}\vert\mu_\alpha\rangle\langle \mu_\alpha\vert\psi\rangle =\sum_{\alpha}e^{\beta\lambda_\alpha}\vert\mu_\alpha\rangle\langle \mu_\alpha\vert\psi\rangle $$ which cannot be resummed further in general.

You can still calculate, for instance, \begin{align} \langle\psi\vert e^{\beta \hat H}\vert\psi\rangle &= \sum_{\alpha\kappa}e^{\beta \lambda_\alpha} \langle\psi\vert\mu_\kappa\rangle \langle \mu_\kappa\vert\mu _\alpha\rangle\langle \mu_\alpha\vert\psi\rangle\, ,\\ &= \sum_{\alpha}e^{\beta \lambda_\alpha} \langle\psi\vert\mu_\alpha\rangle \langle \mu_\alpha\vert\psi\rangle \end{align} where $\langle \mu_\kappa\vert\mu _\alpha\rangle=\delta_{\mu\alpha}$ has been used.


The definition of the matrix exponential comes straight from the Taylor expansion of $e^x$. Thus, you can always substitute the matrix exponential for its definition.

You can still use the Taylor expansion in the example you provided. You just take $x$ to be $\beta \hat{H}$ and apply the Taylor expansion as you would normally.


protected by Qmechanic Oct 17 '17 at 13:20

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.