In Assa Auerbach's book (Ref. 1), he gave an argument saying that in the normal process of path integral, we lose information about ordering of operators by ignoring the discontinuous path.

What did he want to say? I don't think there is any problem related to the ordering of operators.


  1. Assa Auerbach, Interacting Electrons and Quantum Magnetism, p.102, just below eq. (10.6).

1 Answer 1

  1. Any standard textbook derivation of the correspondence$^1$ between

    $$ \text{Operator formalism}\qquad \longleftrightarrow \qquad \text{Path integral formalism} \tag{1} $$

    is a formal derivation, which discards contributions in the process. This is true whether we work in configuration space (as in Ref. 2), or in phase space; and whether we use position and momentum states, coherent states, or coherent spin states (as in Ref. 3).

    The objects appearing in the formal path integrand are no$^2$ longer non-commutative operators but commutative$^3$ functions a.k.a. symbols. See also this Phys.SE post.

    There is a correspondence/map between

    $$ \text{Operators}\qquad \longleftrightarrow \qquad \text{Functions/Symbols}.\tag{2} $$

    The operator ordering/ambiguity problem is hidden in how to choose this correspondence/map (2).

    Example. The same operator $\frac{\hat{q}\hat{p}+\hat{p}\hat{q}}{2}$ gets translated into the symbol $qp-\frac{ih}{2}$, $qp+\frac{ih}{2}$, or $qp$, depending of whether we choose $\hat{q}\hat{p}$, $\hat{p}\hat{q}$, or Weyl ordering prescription, respectively. Conversely, the same function $qp$ gets translated into the operator $\hat{q}\hat{p}$, $\hat{p}\hat{q}$, or $\frac{\hat{q}\hat{p}+\hat{p}\hat{q}}{2}$, depending of whether we choose $\hat{q}\hat{p}$, $\hat{p}\hat{q}$, or Weyl ordering prescription, respectively.

  2. Let us indicate here where approximations in the correspondence (1) are made in case of the (conceptually simpler) 1D phase space path integral in the Heisenberg picture. The main idea in deriving the path integral is to insert completeness-relations

    $$ \int \!dq ~|q,t \rangle \langle q,t |~=~{\bf 1}, \qquad \text{and} \qquad \int \!dp~ |p,t \rangle \langle p,t |~=~{\bf 1},\tag{3}$$

    of instantaneous$^4$ eigenstates at various times $t$, alternating between position and momentum insertions. The leading contribution leads to a formal path integral

    $$ \langle q_f,t_f|q_i,t_i \rangle~\sim~\int_{q(t_i)=q_i}^{q(t_f)=q_f} \!{\cal D}q~{\cal D}p~\exp\left[ \frac{i}{\hbar}S[q,p]\right],\tag{4}$$

    with formal Hamiltonian action

    $$ S[q,p]~=~\int_{t_i}^{t_f}\!dt~\left[ p\dot{q}- H(q,p)\right],\tag{5}$$

    where $H(q,p)$ denotes the Weyl-symbol for the Hamiltonian operator $\hat{H}$. Weyl-ordering prescription is better than other operator ordering prescriptions, but it is still an approximation.

    Auerbach in Ref.3 is mostly talking about the analogue of the $p\dot{q}$ term for coherent spin states rather than the Hamiltonian term. First recall the $pq$ overlap formula

    $$ \langle p,t \mid q,t \rangle~=~\frac{1}{\sqrt{2\pi\hbar}}\exp\left[\frac{pq}{i\hbar}\right]. \tag{6}$$

    See also this Phys.SE answer.

    Next, two typical neighboring terms in the time slicing procedure are of the form

    $$\begin{align} \langle q_{+},& t+\frac{\epsilon}{2} \mid p,t \rangle \langle p,t \mid q_{-},t- \frac{\epsilon}{2}\rangle \cr ~=~&\langle q_{+},t \mid \exp\left[-\frac{i\epsilon}{2\hbar}\hat{H}\right]\mid p,t \rangle \langle p,t \mid \exp\left[-\frac{i\epsilon}{2\hbar}\hat{H}\right]\mid q_{-},t\rangle\cr ~\approx~&\langle q_{+},t \mid p,t \rangle \langle p,t \mid q_{-},t\rangle \exp\left[-\frac{i\epsilon}{\hbar} H\left(\frac{q_{+}+q_{-}}{2},p\right) \right]\cr ~\stackrel{(6)}{=}~& \frac{1}{2\pi\hbar}\exp\left[\frac{i \epsilon}{\hbar}\left(p\frac{q_{+}-q_{-}}{\epsilon} - H\left(\frac{q_{+}+q_{-}}{2},p\right)\right) \right] \cr ~\approx~& \frac{1}{2\pi\hbar}\exp\left[\frac{i\epsilon}{\hbar}(p\dot{q}-H(q,p)) \right]. \end{align}\tag{7}$$

    We stress that several approximations were made in derivation of eq. (7) by e.g. neglecting differences between different kinds of symbols (corresponding to different kinds of ordering prescriptions). In general, it is not true that such approximations (7) are justified in the limit of infinitesimally fine time slicing $\epsilon\to 0^{+}$.


  1. F. Bastianelli and P. van Nieuwenhuizen, Path Integrals and Anomalies in Curved Space, 2006.

  2. J.J. Sakurai, Modern Quantum Mechanics, 1994, Section 2.5.

  3. A. Auerbach, Interacting Electrons and Quantum Magnetism, 1994, p.102 just below eq. (10.6).

$^1$ The operator-path integral correspondence (1) is in general highly non-trivial. For instance, for quantization of a non-relativistic point particle in a classical curved background, the Hamiltonians on the two sides of the correspondence (1) differ by curvature corrections at second order in $\hbar$. See. e.g. Ref. 1. To keep the discussion simple, we do not address regularization/renormalization issues of the correspondence (1) in this answer.

$^2$ Strictly speaking, time derivatives inside the formal path integrand is a remaining source of non-commutative objects as time derivatives should be understood in a time-ordered fashion to reflect the underlying time slicing procedure. See e.g. this and this Phys.SE answer.

$^3$ The standard point-wise multiplication $fg=gf$ of functions/symbols is commutative. There also exists a so-called star-product $f\star g$ of functions/symbols, which is non-commutative, since it reflects the non-commutativity of the corresponding operator composition $\hat{f}\circ \hat{g}$. The star-product $\star$ itself depends on the choice of ordering prescription.

$^4$ Instantaneous eigenstates are often introduced in textbooks of quantum mechanics to derive the path-integral formalism from the operator formalism in the simplest cases, see e.g. Ref. 2. Note that the instantaneous eigenstates $\mid q,t \rangle $ and $\mid p,t \rangle $ are time-independent states (as they should be in the Heisenberg picture).

  • 1
    $\begingroup$ Thank you for the update. I though that there was a correspondence between the ordering prescription (Weyl's, $q$ to the left or whatever) and the definition on the path integral measure. And that this correspondence made the path integral unambiguous. But I do not know in what cases this is right and how the ordering-prescription and the measure's ambiguities cancel each other. Do you have an answer? $\endgroup$ Commented Mar 10, 2013 at 17:14
  • $\begingroup$ Note to self for later: In order for a possible $p$-integration to be be convergent, we must demand that $0<{\rm Re}(i\epsilon)=-{\rm Im}(\epsilon)$, irrespectively of the sign of ${\rm Re}(\epsilon)$. $\endgroup$
    – Qmechanic
    Commented Mar 10, 2016 at 22:16
  • $\begingroup$ 1. Am i correct in thinking that $\langle q_{+}, t+\frac{\epsilon}{2} | p,t \rangle \langle p,t | q_{-},t- \frac{\epsilon}{2}\rangle$ represents $\langle q_{+}, t+\frac{\epsilon}{2} | q_{-},t- \frac{\epsilon}{2}\rangle$? 2. (In Sakurai they say that this is equivalent to $e^{i\mathcal{S}}$) 3. If so, I don't understand why these statements are equivalent. 4. If this is a completeness relation then do we not need $\int dp |p,t\rangle\langle p,t|$? $\endgroup$ Commented Sep 7, 2022 at 11:28
  • 1
    $\begingroup$ Hi Adrien Amour. Thanks for the feed-back. 1. No. 2. Note that Sakurai's $S$ is the action, while my $S$ is the Hamiltonian action (5). 4. Yes, we eventually need the integration $\int dp$. $\endgroup$
    – Qmechanic
    Commented Sep 7, 2022 at 12:34
  • $\begingroup$ Thanks for your response, to me the final expression in (7) is already the $e^{\frac{i}{\hbar}\mathcal{S}}$ we were looking for, since we have the Lagrangian multiplied by an infinitesimal timestep. It feels like integrating over every possible momentum value will change this. Am I correct? In that case what is the correct final expression for $\langle q_{+}, t+\frac{\epsilon}{2} \mid p,t \rangle \langle p,t \mid q_{-},t- \frac{\epsilon}{2}\rangle$ $\endgroup$ Commented Sep 7, 2022 at 13:04

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