# What is the definition of variance for a non-Hermitian operator?

I am trying to understand what is the correct way to compute the variance of a non-Hermitian operator.

I was thinking that it was simply something that:

$$\langle (\Delta a)^2 \rangle = |\langle \psi| \hat{a}^2 |\psi\rangle| -| \langle \psi |\hat{a}|\psi\rangle|^2$$

But now I have read on Wikipedia that the variance for a random complex variable can be written as:

$$Var[Z] = \mathbb{E}[|Z|^2] - |\mathbb{E}[Z]|^2$$

In the first term, the absolute value is computed before the expectation value, so I think the formula I have written before may be wrong.

Now I am thinking that it may be

$$\langle (\Delta a)^2 \rangle = |\langle \psi| \hat{a}^\dagger \hat{a} |\psi\rangle| -| \langle \psi |\hat{a}|\psi\rangle|^2$$

Is this correct?

• $\uparrow$ Yes. Oct 4, 2018 at 13:03
• Compare to the question of how you would define the variance of a complex variable, without knowledge of quantum mechanics.
– J.G.
Oct 4, 2018 at 15:36

Since is not possible to associate an observable to a non-Hermitian operator $$\hat{a}$$, is impossible to give a physical meaningful definition of variance since the measure process is undefined. Let us ignore this problem and suppose that an hypothetical measure of the operator $$\hat{a}$$ has as results the (complex) eigenvalues of $$\hat{a}$$ with probability distribution given by the squared of the projection of the state on the eigenvectors, just like the Hermitian operators. Here it is an additional problem. If an operator is not Hermitian their eigenvectors may not form complete set of the Hilbert space.

Let suppose the operator $$\hat{a}$$ admit a complete set of eigenvectors to avoid this problem.

\par Since the outcome of the measure is a complex number, the measure process should be modelled like a complex random variable. The definition of variance for a complex random variable is $$$$\mathrm{Var}[Z] = \mathbb{E}[|Z|^2] - |\mathbb{E}[Z]|^2$$$$

Is clear that the second term in the expression is equal to the modulus squared of the expected value, which can be written like \begin{align} \mathbb{E}[Z] &= \sum_\alpha \alpha | \langle \alpha| \psi \rangle |^2 = \\ &= \sum_\alpha \alpha \langle \psi| \alpha \rangle \langle \alpha|\psi\rangle = \\ &= \sum_\alpha \langle \psi|\alpha\rangle\langle \alpha|\hat{a} |\psi\rangle = \\ &= \langle\psi|\hat{a}|\psi\rangle \end{align}

For the first term, instead \begin{align*} \mathbb{E}[|Z|^2] &= \sum_\alpha |\alpha|^2 |\langle \alpha| \psi \rangle |^2 = \\ &= \sum_\alpha \alpha\alpha^* \langle \psi|\alpha\rangle \langle \alpha|\psi\rangle = \\ &= \sum_\alpha \langle \psi|\hat{a}^\dagger| \alpha\rangle \langle \alpha|\hat{a}|\psi\rangle = \\ &= \langle \psi |\hat{a}^\dagger \hat{a} | \psi\rangle \end{align*}

Finally,

$$$$\langle \left( \Delta a \right)^2 \rangle = \langle\psi|\hat{a}^\dagger \hat{a} | \psi\rangle - \langle \psi|\hat{a}^\dagger |\psi\rangle\langle \psi|\hat{a} |\psi\rangle$$$$

This definition is consistent with the usual one for Hermitian operator because in case $$\hat{a}=\hat{a}^\dagger$$

$$$$\langle \left( \Delta a \right)^2 \rangle = \langle \psi|\hat{a}^\dagger \hat{a}| \psi\rangle - \langle \psi|\hat{a}^\dagger|\psi\rangle \langle\psi|\hat{a} |\psi\langle = \langle \psi|\hat{a}^2 |\psi\rangle - |\langle \psi|\hat{a} |\psi\rangle|^2$$$$

In both these definition we have used the hypothesis that, despite being a non-Hermitian operator, $$\hat{a}$$ admits a complete set of eigenvectors. Therefore this assumption is fundamental as it guarantee to have a normalized distribution. In fact, if $$\left\{|\alpha\rangle \right\}$$ is not a complete set, the sum of the squared projection can be less than one. If instead $$\left\{|\alpha\rangle\right\}$$ is an overcomplete set, the sum of the probabilities will be greater than one.