I have been working through the derivation of the partition function expressed as a path integral in terms of coherent states, following the many-body condensed-matter field theory books of Altland & Simons and Piers Coleman, and while I can follow the mathematical arguments, I am struggling with some of the concepts involved. I would very much appreciate some help with several points, which I have been unable to clarify to my satisfaction, despite referring to numerous other textbooks. I apologize in advance if these questions seem too basic.

Question: Starting from the very first step of the derivation, where the partition function is expressed as the trace of the operator $\exp[-\beta(\hat{H}-\mu\hat{N})]$, firstly in terms of a complete set of Fock space states and then coherent states, is it necessary for the basis to be chosen such that the Hamiltonian is diagonal in this basis? I know that the trace, and hence the partition function, is the sum of the diagonal matrix elements, but does the matrix need to be diagonal for this to be evaluated and make sense?

  • $\begingroup$ I'm currently working on the mathematics of Herman-Kluk propagators and initial value representations for coherent state propagation in semiclassical mechanics. They're NOT basic at all. There's a series of misunderstandings and wrong articles that make things even worse. I wish you good luck! $\endgroup$ – QuantumBrick Jun 8 '16 at 14:06
  • $\begingroup$ Does this question boil down to asking whether the trace entering the partition function is basis-independent? If so, then yes, it is basis-independent. Otherwise the coherent state basis would not work for the trace, since no physical Hamiltonian is diagonal in this basis. $\endgroup$ – Mark Mitchison Jun 8 '16 at 14:48

While the trace is invariant under a transform to another basis, you need to take into account here that the coherent state basis is not an orthogonal basis and it is overcomplete. We can evaluate the trace of an operator $A$ by inserting identity operators in front of and after the operator and then using resolution of identity in terms of the coherent basis vectors. We then need to use that the resolution of identity is now given as:

$$I = \frac{1}{\pi}\int d^2\alpha \left|\alpha\right\rangle\left\langle\alpha\right|$$

So, there is an extra factor of $\frac{1}{\pi}$ due to overcompleteness. We can thus write:

$$\operatorname{Tr}A = \sum_n\left\langle n\left|A\right|n\right\rangle = \frac{1}{\pi^2}\sum_{n}\int d^2\alpha d^2\beta\left\langle n\right|\left.\alpha\right\rangle\left\langle\alpha\right|A\left|\beta\right\rangle\left\langle\beta\right|\left.n\right\rangle$$

If you now sum over the (complete, orthonormal) basis $\left|n\right\rangle$, you get using completeness in this basis:

$$\operatorname{Tr}A =\frac{1}{\pi^2}\int d^2\alpha d^2\beta\left\langle\beta\right|\left.\alpha\right\rangle\left\langle\alpha\right|A\left|\beta\right\rangle$$

You can then evaluate this by using the fact that the overlap between two coherent basis states is given by:

$$\left\langle\beta\right|\left.\alpha\right\rangle = \exp\left[-\frac{1}{2}\left(\left|\alpha\right|^2+\left|\beta\right|^2-2\beta^*\alpha\right)\right]$$


No, you don't need to work in the basis where the Hamiltonian is diagonal. It's a fact of linear algebra that the sum of the diagonal elements of a matrix is the same no matter what basis you're in, so you can easily evaluate the trace in whichever basis is convenient.

  • $\begingroup$ That makes sense. I knew that the coherent states in quantum mechanics are not eigenstates of the Hamiltonian, so I did not see why they should be in quantum field theory. One statement from the textbook about coherent states puzzles me though: "The reason for emphasizing normal ordering is that such an operator can be readily diagonalized by means of coherent states". How should this be interpreted? $\endgroup$ – dgwp Jun 10 '16 at 12:01
  • $\begingroup$ @dgwp Operators don't necessarily commute, so in the operator Hamiltonian formalism, the order of the terms matters. But in the coherent state path integral, the fields are just c-numbers and therefore commute. So if you just naively replaced all operators with their corresponding coherent state fields, you'd lose important information about the operators' ordering. The way you preserve the information held in the operator ordering is by requiring that the operators be normal ordered before you replace them with coherent state fields. $\endgroup$ – tparker Jun 10 '16 at 17:40
  • $\begingroup$ @dgwp So for example, the operator term $\hat{a}^\dagger \hat{a} \hat{a}^\dagger \hat{a}$ would correspond to the coherent state terms $\bar{a} \bar{a} a a + \bar{a} a$ - the normal-ordering would result in a shift in the mass/chemical potential term. $\endgroup$ – tparker Jun 10 '16 at 17:43
  • $\begingroup$ Thanks, but that wasn't really what I was wanting to know. I was really interested in how an operator can be diagonalized by coherent states. Sorry, I should have made myself clearer. $\endgroup$ – dgwp Jun 11 '16 at 18:59
  • $\begingroup$ @dgwp Well, a coherent state ket is an eigenstate of the annihilation operator, and a coherent state bra is a (left-)eigenstate of the creation operator. So if you take a normal-ordered operator made up of creation and annihilation operators (in the formalism of second quantization, pretty much all operators in a Hamiltonian have this form) and sandwich it in between coherent states, then all the creation and annihilation operator can be replaced by their eigenvalues, which are much more convenient to deal with because they're just c-numbers. $\endgroup$ – tparker Jun 11 '16 at 21:22

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.