# Why path integral approach may suffer from operator ordering problem?

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.

References:

1. Assa Auerbach, Interacting Electrons and Quantum Magnetism, p.102, just below eq. (10.6).
• For reference, could you please indicate the pages?...Oh, found it myself: Assa Auerbach, Interacting Electrons and Quantum Magnetism, p.102 just below eq. (10.6). – Qmechanic Mar 6 '13 at 16:58
• Any standard textbook derivation from the operator formalism to the path integral formalism (e.g. in Auerbach's book by using coherent states) is a formal derivation in the sense that it e.g. replaces discretized time derivatives with continuous time derivatives neglecting non-commuting contributions in the process. The usual treatment of the Hamiltonian factor is another source of approximations. So naively one would expect operator ordering issues, and further analysis show that they yield in fact non-trivial effects. – Qmechanic Mar 6 '13 at 22:07
• But I still don't understand Where is the non-commuting contributions? – Xiao-Qi Sun Mar 7 '13 at 0:23
• – Qmechanic Mar 14 '13 at 16:16

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

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

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

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

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

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

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

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

with formal Hamiltonian action

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

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

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

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

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

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 $\epsilon\to 0^{+}$.

References:

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).

• Do you know any paper or book where I can read about the path integral measure, how it encodes ordering operator issues, etc? Because the Wey-ordering prescription is connected to the definition of the measure, right? I tried to read Glimm & Jaffe's book but it was too formal for me. I am looking for some reference somehow between the conventional derivation of QM and QFT books and Glimm & Jaffe's. Do you know any reference? A link to some question in this site that partly answers would be welcome as well. Thank you. – Diego Mazón Mar 8 '13 at 18:42
• I know there is approximation. But where is the operator ordering issue? – Xiao-Qi Sun Mar 10 '13 at 8:54
• I updated the answer. – Qmechanic Mar 10 '13 at 16:35
• 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? – Diego Mazón Mar 10 '13 at 17:14
• 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 ${\rm Re}(\epsilon)$. – Qmechanic Mar 10 '16 at 22:16