Consider the coherent state

$$ |\phi \rangle = \exp \left( \zeta \cdot \sum_\alpha \phi_{\alpha} a_{\alpha}^\dagger \right) | 0 \rangle.$$

For the case of bosons ($\zeta = +1$), the $\phi_\alpha$'s are $c$-numbers and the creation and annihilation operators satisfy some commutation relations. For the case of fermions ($\zeta=-1$), the creation and annihilation operators satisfy some anticommutation relations and the $\phi_\alpha$'s are Grassmann numbers. In both cases the coherent states diagonalize the corresponding annihilation operator:

$$a_\alpha |\phi \rangle = \phi_\alpha |\phi \rangle .$$

It is said that while fermionic coherent states have no classical counterpart, bosonic coherent states are like the classical limit of quantum states. Why? For instance in the answer to this question they say it's because in these states the bosonic creation and annihilation operators commute, which is what "operators" do in classical physics. On the other hand, the exact anticommutation of fermionic annihilation doesn't emulate any classical behaviour. This would explain it. However I don't see that bosonic creation and annihilation operators commute exactly when acting upon coherent states. In the coherent state representation $a_\alpha = \phi_\alpha$ and $a_\alpha^\dagger = \frac{\partial}{\partial \phi_\alpha}$, whence

$$[a_\alpha^\dagger, a_\alpha, ] | \phi\rangle = \frac{\partial}{\partial \phi_\alpha} \left( \phi_ \alpha | \phi \rangle \right) - \phi_\alpha \frac{\partial}{\partial \phi_\alpha} | \phi \rangle = | \phi \rangle$$

which is the usual commutation relation $[a_\alpha^\dagger, a_\beta] = \delta_{\alpha, \beta}$. So the operators don't commute as they should classically. What am I missing? If this is not the reason for saying things like "bosonic coherent states arise when taking the classical limit of quantum states", then what is?


1 Answer 1


In the conventional normalization of the semiclassical correspondence principle between classical physics and quantum physics, the (super) Poisson bracket$^1$ $$\{a,a^{\ast}\}_{SPB}~=~-i\tag{1}$$ turns into the (super) commutation relation$^2$ $$ [\hat{a},\hat{a}^{\dagger}]_{SC}~=~\hbar\hat{\mathbb{1}}. \tag{2}$$ The coherent states are then defined as $$ |z \rangle~:=~e^{\hat{a}^{\dagger}z/\hbar}|0 \rangle, \qquad \hat{a}|z \rangle~=~z|z \rangle ,\qquad \hat{a}^{\dagger}|z \rangle~=~(-1)^{|z|}\hbar\frac{d}{dz}|z \rangle , \tag{3}$$ where $|z|$ denotes the Grassmann-parity of the supernumber $z$. In the classical limit $\hbar\to 0$, the operators (super) commute.


$^1$ One may argue that Grassmann variables are classical variables because they supercommute!

$^2$ Of course, one can obscure the quantum nature of the (super) commutation relation $$[\hat{\alpha},\hat{\alpha}^{\dagger}]_{SC}~=~\hat{\mathbb{1}}\tag{2'}$$ by scaling the quantum operators $\hat{a}=\sqrt{\hbar} \hat{\alpha}$ with a square root power of $\hbar$!

  • $\begingroup$ That is true, but that has nothing to do with coherent states. That equivalence between Poisson brackets and commutators is general. So I don't see why coherent states are chosen to be the "classical limit". $\endgroup$
    – MBolin
    May 21, 2020 at 15:19
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    May 21, 2020 at 16:11
  • $\begingroup$ I think you just added the definition of coherent states right? Maybe it's straightforward, but I don't see how this answers my question. $\endgroup$
    – MBolin
    May 22, 2020 at 11:05

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