Consider the holomorphic representation of the path integral (for a single degree of freedom):

$$ U(a^{*}, a, t'', t') = \int e^{\alpha^{*}(t'') \alpha(t'')} \exp\left\{\intop_{t'}^{t''} dt \left( -a^{*} \dot{a} - i h(a, a^{*}) \right) \right\} \prod_t \frac{da^{*}(t) da(t)}{2\pi i}. $$

The proper boundary conditions are of form

$$ a(t') = a; \quad a^{*}(t'') = a^{*}. $$

My question is: how are $a$ and $a^{*}$ related and why?

One observation is that we treat them as independent variables in the path integral, so they can't be complex-conjugate ad hoc.

Another observation is that we should impose the reality condition on the boundary (which is analogous to the $\text{Im} \left(x(t',t'')\right) = 0$ condition in the coordinate representation). But how (and why) should we relate $a(t')$ to $a^{*}(t'')$ which are taken at different instants of time?

UPDATE: my original idea was that they are not related at all. We simply constrain our description to holomorphic wavefunctions $\Psi(a^{*}(t''))$ and $\Phi(a(t'))$ which is analogous to constraining it to the real-variable wavefunctions in the coordinate basis. But my professor keeps insisting otherwise (he doesn't actually care to give a convincing argument though, just keeps saying "no").


1 Answer 1


OP's question is essentially pondering (in the context of the holomorphic/coherent state path integral) if a pair of variables is a complex conjugate pair or$^1$ truly independent variables?

TL;DR: Well, it depends.

Notation in this answer: In this answer, let $z,z^{\ast}\in \mathbb{C}$ denote two independent complex numbers. Let $\overline{z}$ denote the complex conjugate of $z$.

Recall that the coherent ket state is

$$ |z \rangle~:=~e^{\hat{a}^{\dagger}z/\hbar}|0 \rangle, \qquad \hat{a}|z \rangle~=~z|z \rangle , \qquad [\hat{a},\hat{a}^{\dagger}]~=~\hbar \hat{\bf 1}.\tag{1}$$

It is customary$^2$ to define the coherent bra state

$$ \langle z | ~:=~ |\bar{z} \rangle^{\dagger} ~\stackrel{(1)}{=}~\langle 0 |e^{z\hat{a}/\hbar}\tag{2}$$

in terms of the coherent ket state (1) by including a complex conjugation, cf. e.g. Ref. 1. In other words, we have the convenient rule that

$$ \langle z^{\ast} | ~\stackrel{(2)}{=}~ \langle 0 |e^{z^{\ast}\hat{a}/\hbar}, \qquad \langle z^{\ast} |\hat{a}^{\dagger} ~=~z^{\ast} \langle z^{\ast} | . \tag{3}$$

With this convention (2), the completeness relation reads$^3$

$$ \int_{\mathbb{C}} \frac{d\bar{z}~dz}{2 \pi i\hbar} e^{-\bar{z}z/\hbar} |z \rangle\langle \bar{z} |~=~\hat{\bf 1}.\tag{4} $$

It is important to realize that the coherent states are an overcomplete set of states

$$ \langle z^{\ast}|z \rangle~=~e^{z^{\ast} z/\hbar} \tag{5}$$

with non-orthogonal overlaps. The coherent state path integral reads

$$\begin{align} \langle z_f^{\ast}, t_f | z_i, t_i \rangle ~=~& \int_{z(t_i)=z_i}^{\bar{z}(t_f)=z^{\ast}_f} \! {\cal D}\bar{z}~{\cal D}z ~e^{iS[z,\bar{z}]/\hbar}, \cr {\cal D}\bar{z}~{\cal D}z~:=~& \prod_{n=1}^N \frac{d\bar{z}_n~dz_n}{2 \pi i\hbar}, \end{align}\tag{6} $$

$$\begin{align} iS[z,z^{\ast}]~:=&~ (1-\lambda)z^{\ast}(t_f)~z(t_f) + \lambda z^{\ast}(t_i) z(t_i) \cr +& \int_{t_i}^{t_f}\! dt \left[\lambda \dot{z}^{\ast} z -(1-\lambda) z^{\ast} \dot{z}- iH_N(z^{\ast},z) \right],\end{align}\tag{7} $$

where $\lambda\in \mathbb{R}$ is a real constant which the action (7) does not actually depend on, due to the fundamental theorem of calculus. The Hamiltonian function

$$ H_N(z^{\ast},z)~:=~\frac{\langle z^{\ast}|\hat{H}(\hat{a}^{\dagger},\hat{a})|z \rangle}{\langle z^{\ast}|z \rangle}\tag{8}$$

is the normal/Wick-ordered function/symbol corresponding to the quantum Hamiltonian operator

$$\hat{H}(\hat{a}^{\dagger},\hat{a})~=~\left. e^{\hat{a}^{\dagger}\frac{\partial}{\partial z^{\ast}}}e^{\hat{a}\frac{\partial}{\partial z}}H_N(z^{\ast},z)\right|_{z=0=z^{\ast}}.\tag{9}$$

Concerning operator ordering in the path integral, see also e.g. this Phys.SE post.

In the standard Feynman path integral there are 2 real boundary conditions (BCs), typically Dirichlet BCs

$$ q(t_i)~=~q_i \qquad\text{and}\qquad q(t_f)~=~q_f.\tag{10} $$

The position $\hat{q}/\sqrt{2}$ and the momentum $\hat{p}/\sqrt{2}$ operators are related to

$$ {\rm Re}(\hat{a})~:=~\frac{\hat{a}+\hat{a}^{\dagger}}{2} \qquad\text{and}\qquad{\rm Im}(\hat{a})~:=~\frac{\hat{a}-\hat{a}^{\dagger}}{2i},\tag{11}$$

respectively. In the coherent state path integral (6), there are 2 complex (= 4 real) BCs

$$ z(t_i)~=~z_i \qquad\text{and}\qquad \bar{z}(t_f)~=~z^{\ast}_f. \tag{12}$$

In other words, we specify both initial position and initial momentum, naively violating the HUP. Similar for the final state. This is related to the overcompleteness (5) of the coherent states.

The overcomplete BCs (12) means that there typically is not an underlying physical classical path with


that fulfills [besides the Euler-Lagrange (EL) equations

$$ \dot{z}~\approx~-i\frac{\partial H(z,z^{\ast})}{\partial z^{\ast}} \quad\text{and}\quad \dot{z}^{\ast}~\approx~i\frac{\partial H(z,z^{\ast})}{\partial z}, \tag{14} $$

i.e. the Hamilton's equations] all the BCs (12) simultaneously unless we tune the BCs (12) appropriately, cf. e.g. Ref. 1. The precise tuning depends on the theory at hand.

Nevertheless, we may in principle always find a classical path $t\mapsto (z_0(t),z_0^{\ast}(t))$ in twice as many variables that satisfies eqs. (12) and (14) but not necessarily eq. (13). We may then use complex function theory to deform$^4$ the integration contour in the coherent state path integral in order to only integrate over quantum fluctuations $\eta$

$$ \langle z_f^{\ast}, t_f | z_i, t_i \rangle ~\stackrel{(6)}{=}~ \int_{\eta(t_i)=0}^{\bar{\eta}(t_f)=0} \! {\cal D}\bar{\eta}~{\cal D}\eta ~e^{iS[z_0+\eta,z_0^{\ast}+\bar{\eta}]/\hbar}, \tag{15} $$

and in this way still achieve a WKB/stationary phase approximation around a classical path $(z_0,z_0^{\ast})$.


  1. L.S. Brown, QFT; Section 1.8.


$^1$ For more on complex conjugation and independence of variables, see also e.g. this Phys.SE post.

$^2$ Notabene: Some authors do not include a complex conjugation in definition (2), cf. e.g. Wikipedia!

$^3$ With the displayed order, the formulas in this answer also work for the Grassmann-odd/fermionic coherent state path integral, except one should leave out the normalization factor $2\pi i$ in eqs. (4) & (6).

$^4$ The deformation of the integration contour is easiest to justify using position and momentum variables

$$q~=~\frac{z+z^{\ast}}{\sqrt{2}} \qquad\text{and}\qquad p~=~\frac{z-z^{\ast}}{\sqrt{2}i} ,\tag{16}$$

that might be complex.

  • 1
    $\begingroup$ Is like a sum over histories but not with high precision paths over space or momentum space, is something in between (trajectories of Gaussians in phase space) $\endgroup$
    – Nogueira
    Commented Nov 12, 2015 at 1:41
  • $\begingroup$ Notes for later: $\quad \{z,z^{\ast}\}=-i$; $\quad \omega=i\mathrm{d}z^{\ast}\wedge \mathrm{d}z$; $\quad \star_N=\exp\left(\hbar \frac{\stackrel{\longleftarrow}{\partial}}{\partial z} \frac{\stackrel{\longrightarrow}{\partial}}{\partial z^{\ast}} \right)$; $\endgroup$
    – Qmechanic
    Commented Nov 29, 2022 at 14:26
  • $\begingroup$ Heisenberg instantaneous eigenstate: $\quad |z,t\rangle_H:=e^{\hat{a}_H^{\dagger}(t)z/\hbar}|0 \rangle$; $\quad \hat{a}_H(t)|z,t\rangle_H~=~z|z,t\rangle_H$; $\quad \hat{a}^{\dagger}_H(t)|z,t\rangle_H~=~\hbar\frac{\partial}{\partial z}|z,t\rangle_H$; $\quad |z^{\ast},t\rangle_H:=\int\! dz~e^{-z^{\ast}z/\hbar} |z,t\rangle_H$; $\quad \hat{a}^{\dagger}_H(t)|z^{\ast},t\rangle_H~=~z^{\ast}|z^{\ast},t\rangle_H$; We assume implicitly that the vacuum $|0 \rangle$ is invariant (up to a phase) under time-evolution. $\endgroup$
    – Qmechanic
    Commented Feb 8, 2023 at 10:22
  • $\begingroup$ $\quad |q,t\rangle_H:=\int_{i\mathbb{R}}\! dz~e^{(z^2/2-\sqrt{2}qz)\hbar} |z,t\rangle_H$; $\quad \hat{q}_H(t)|q,t\rangle_H~=~q|q,t\rangle_H$; $\quad |p,t\rangle_H:=\int_{\mathbb{R}}\! dz~e^{(\sqrt{2}ipz-z^2/2)\hbar} |z,t\rangle_H$; $\quad \hat{p}_H(t)|p,t\rangle_H~=~p|p,t\rangle_H$; $\endgroup$
    – Qmechanic
    Commented Feb 8, 2023 at 14:30
  • $\begingroup$ Notes for later: Polar coordinates: $\quad z=re^{i\theta}$; $\quad q=\sqrt{2}r\cos\theta$; $\quad p=\sqrt{2}r\sin\theta$; $\quad\omega=\mathrm{d}\theta\wedge\mathrm{d}(r^2)$; $\quad\{r^2,\theta\}=1$; The symplectic vector field $X=\frac{\partial}{\partial r^2}=-\{\theta,\cdot\}$ corresponds to a translation $r^2\to r^2 +{\rm const}$. $\endgroup$
    – Qmechanic
    Commented Jan 10 at 8:56

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