Many times in quantum physics we face a problem of calculating expectation values of normal product operators. Assume we have a two level system with creation and annihilation operators $\hat{a}$ and $\hat{b}$. We would like to find the time evolution of expectation values of the form

$$\langle (\hat{a}^{\dagger})^{k}(\hat{b})^{l}\rangle$$ $$\langle (\hat{b}^{\dagger})^{k}(\hat{a})^{l}\rangle$$

We can do that using generating function method. For example, if we take Hamiltonian $H = S_{z}^2 = \frac{1}{4}(\hat{a}^{\dagger}\hat{a} - \hat{b}^{\dagger}\hat{b})^2$, equation for density matrix coefficients expanded in the Fock state basis become: $$\hat{\rho}(t) = \sum\limits_{k_1,k_2,n_1,n_2} \rho^{k_1,k_2}_{n_1,n_2}(t) |k_1,k_2\rangle\langle n_1,n_2|.$$ using von Neumann equation we arrive at: $$i\dot{\rho}^{k_1,k_2}_{n_1,n_2}(t) = \frac{1}{4}\rho^{k_1,k_2}_{n_1,n_2}(t)\left[ (k_1-k_2)^2 - (n_1-n_2)^2\right]$$ Of course we can solve this equation easily, but the idea is different. Let's say we wan to calculate expectation value: $$\langle \hat{a}^{\dagger}\hat{b}\rangle = \sum\limits_{k_1,k_2}\rho^{k_1-1,k_2+1}_{k_1,k_2}\sqrt{k_1(k_2+1)}$$ We introduce generating function of the form: $$G(x,y,t) = \sum\limits_{k_1,k_2}x^{k_1}y^{k_2}\rho^{k_1-1,k_2+1}_{k_1,k_2}(t)\sqrt{k_1(k_2+1)}$$ such that $G(1,1,t) = \langle \hat{a}^{\dagger}\hat{b}\rangle$. Using recurrence relation we can write $$i\partial_{t}G(x,y,t) = (1-x\partial_{x} + y\partial_{y})G(x,y,t)$$ We can solve the differential equation and get the result.

This was a simple example just to show the idea. What about more complicated Hamiltonians like: $H = S_{x} = \frac{1}{2}(\hat{a}^{\dagger}\hat{b} + \hat{b}^{\dagger}\hat{a})$ or $H = S_{z}S_{y} + S_{y}S_{z}$, where $S_y = \frac{1}{2i}(\hat{a}^{\dagger}\hat{b} - \hat{b}^{\dagger}\hat{a})$.

Is it a general trick, or it works only for diagonal Hamiltonians? Maybe some of you know other methods that can be used for calculating expectation values of operator products?


1 Answer 1


You might take a look at the quasiprobability distributions and optical equivalence theorem. The general idea is that you apply a transformation to the density matrix $\rho$:

$\chi(\xi,\xi^*) = Tr[D(\xi,\xi^*) \rho]$

where $D(\xi,\xi^*) = \exp[\xi \hat{a}^{\dagger} - \xi^* \hat{a}] $ is a displacement operator, and get a characteristic function $\chi$. Now you can calculate the expectation values as:

$\langle \hat{a}^{\dagger k}\hat{a}^{m}\rangle = \frac{\partial^{m+n}\chi(\xi,\xi^*)}{\partial(\xi^*)^k\partial(\xi)^m}$.

You can define other characteristic functions in order to calculate the expectation for different ordering of the operators.

  • $\begingroup$ I know about the method of characteristic functions. They were very useful in quantum optics and I think you can find them in any book. In my case it maybe useful for $H = S_x$, but quadratic Hamiltonians cause a lot of trouble and I couldn't find the characteristic function useful. $\endgroup$
    – WoofDoggy
    Feb 17, 2016 at 22:07
  • $\begingroup$ @Ilya your last expression the quantity $\langle(\hat{a}_{}^{\dagger})_{}^{k}(\hat{a}_{}^{})_{}^{m}\rangle$ should be replaced by symmetric or Weyl ordered average. $\endgroup$
    – Sunyam
    Aug 5, 2018 at 21:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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