Path integral formulation

One common approach to deriving the path integral formula is to divide the time interval into small pieces. Once this is done, the Trotter product formula tells us that the noncommutativity of the kinetic and potential energy operators can be ignored.

For a particle in a smooth potential, the path integral is approximated by zigzag paths, which in one dimension is a product of ordinary integrals. For the motion of the particle from position x_a at time t_a to x_b at time t_b, the time sequence

${\displaystyle t_{a}=t_{0}<t_{1}<\cdots <t_{n-1}<t_{n}<t_{n+1}=t_{b}}$

can be divided up into n + 1 smaller segments t_j − t_{j − 1}, where j = 1, ..., n + 1, of fixed duration

${\displaystyle \varepsilon =\Delta t={\frac {t_{b}-t_{a}}{n+1}}.}$

This process is called time-slicing.

An approximation for the path integral can be computed as proportional to

${\displaystyle \int \limits _{-\infty }^{+\infty }\cdots \int \limits _{-\infty }^{+\infty }\exp \left({\frac {i}{\hbar }}\int _{t_{a}}^{t_{b}}L{\big (}x(t),v(t){\big )}\,dt\right)\,dx_{0}\,\cdots \,dx_{n},}$

where L(x, v) is the Lagrangian of the one-dimensional system with position variable x(t) and velocity v = ẋ(t) considered (see below), and dx_j corresponds to the position at the jth time step, if the time integral is approximated by a sum of n terms.[nb 2]

In the limit n → ∞, this becomes a functional integral, which, apart from a nonessential factor, is directly the product of the probability amplitudes ⟨x_b, t_b|x_a, t_a⟩ (more precisely, since one must work with a continuous spectrum, the respective densities) to find the quantum mechanical particle at t_a in the initial state x_a and at t_b in the final state x_b.

Actually L is the classical Lagrangian of the one-dimensional system considered,

${\displaystyle L(x,{\dot {x}})=T-V={\frac {1}{2}}m|{\dot {x}}|^{2}-V(x)}$

and the abovementioned "zigzagging" corresponds to the appearance of the terms

${\displaystyle \exp \left({\frac {i}{\hbar }}\varepsilon \sum _{j=1}^{n+1}L\left({\tilde {x}}_{j},{\frac {x_{j}-x_{j-1}}{\varepsilon }},j\right)\right)}$

in the Riemann sum approximating the time integral, which are finally integrated over x₁ to x_n with the integration measure dx₁...dx_n, x̃_j is an arbitrary value of the interval corresponding to j, e.g. its center, x_j + x_j−1/2.

Thus, in contrast to classical mechanics, not only does the stationary path contribute, but actually all virtual paths between the initial and the final point also contribute.

In terms of the wave function in the position representation, the path integral formula reads as follows:

${\displaystyle \psi (x,t)={\frac {1}{Z}}\int _{\mathbf {x} (0)=x}{\mathcal {D}}\mathbf {x} \,e^{iS[\mathbf {x} ,{\dot {\mathbf {x} }}]}\psi _{0}(\mathbf {x} (t))\,}$

where ${\displaystyle {\mathcal {D}}\mathbf {x} }$ denotes integration over all paths ${\displaystyle \mathbf {x} }$ with ${\displaystyle \mathbf {x} (0)=x}$ and where ${\displaystyle Z}$ is a normalization factor. Here ${\displaystyle S}$ is the action, given by

${\displaystyle S[\mathbf {x} ,{\dot {\mathbf {x} }}]=\int dt\,L(\mathbf {x} (t),{\dot {\mathbf {x} }}(t))}$

The path integral representation gives the quantum amplitude to go from point x to point y as an integral over all paths. For a free-particle action (for simplicity let m = 1, ħ = 1)

${\displaystyle S=\int {\frac {{\dot {x}}^{2}}{2}}\,dt,}$

the integral can be evaluated explicitly.

To do this, it is convenient to start without the factor i in the exponential, so that large deviations are suppressed by small numbers, not by cancelling oscillatory contributions. The amplitude (or Kernel) reads:

${\displaystyle K(x-y;T)=\int _{x(0)=x}^{x(T)=y}\exp \left(-\int _{0}^{T}{\frac {{\dot {x}}^{2}}{2}}\,dt\right)\,Dx.}$

Splitting the integral into time slices:

${\displaystyle K(x,y;T)=\int _{x(0)=x}^{x(T)=y}\prod _{t}\exp \left(-{\tfrac {1}{2}}\left({\frac {x(t+\varepsilon )-x(t)}{\varepsilon }}\right)^{2}\varepsilon \right)\,Dx,}$

where the Dx is interpreted as a finite collection of integrations at each integer multiple of ε. Each factor in the product is a Gaussian as a function of x(t + ε) centered at x(t) with variance ε. The multiple integrals are a repeated convolution of this Gaussian G_ε with copies of itself at adjacent times:

${\displaystyle K(x-y;T)=G_{\varepsilon }*G_{\varepsilon }*\cdots *G_{\varepsilon },}$

where the number of convolutions is T/ε. The result is easy to evaluate by taking the Fourier transform of both sides, so that the convolutions become multiplications:

${\displaystyle {\tilde {K}}(p;T)={\tilde {G}}_{\varepsilon }(p)^{T/\varepsilon }.}$

The Fourier transform of the Gaussian G is another Gaussian of reciprocal variance:

${\displaystyle {\tilde {G}}_{\varepsilon }(p)=e^{-{\frac {\varepsilon p^{2}}{2}}},}$

and the result is

${\displaystyle {\tilde {K}}(p;T)=e^{-{\frac {Tp^{2}}{2}}}.}$

The Fourier transform gives K, and it is a Gaussian again with reciprocal variance:

${\displaystyle K(x-y;T)\propto e^{-{\frac {(x-y)^{2}}{2T}}}.}$

The proportionality constant is not really determined by the time-slicing approach, only the ratio of values for different endpoint choices is determined. The proportionality constant should be chosen to ensure that between each two time slices the time evolution is quantum-mechanically unitary, but a more illuminating way to fix the normalization is to consider the path integral as a description of a stochastic process.

The result has a probability interpretation. The sum over all paths of the exponential factor can be seen as the sum over each path of the probability of selecting that path. The probability is the product over each segment of the probability of selecting that segment, so that each segment is probabilistically independently chosen. The fact that the answer is a Gaussian spreading linearly in time is the central limit theorem, which can be interpreted as the first historical evaluation of a statistical path integral.

The probability interpretation gives a natural normalization choice. The path integral should be defined so that

${\displaystyle \int K(x-y;T)\,dy=1.}$

This condition normalizes the Gaussian and produces a kernel that obeys the diffusion equation:

${\displaystyle {\frac {d}{dt}}K(x;T)={\frac {\nabla ^{2}}{2}}K.}$

For oscillatory path integrals, ones with an i in the numerator, the time slicing produces convolved Gaussians, just as before. Now, however, the convolution product is marginally singular, since it requires careful limits to evaluate the oscillating integrals. To make the factors well defined, the easiest way is to add a small imaginary part to the time increment ε. This is closely related to Wick rotation. Then the same convolution argument as before gives the propagation kernel:

${\displaystyle K(x-y;T)\propto e^{\frac {i(x-y)^{2}}{2T}},}$

which, with the same normalization as before (not the sum-squares normalization – this function has a divergent norm), obeys a free Schrödinger equation:

${\displaystyle {\frac {d}{dt}}K(x;T)=i{\frac {\nabla ^{2}}{2}}K.}$

This means that any superposition of Ks will also obey the same equation, by linearity. Defining

${\displaystyle \psi _{t}(y)=\int \psi _{0}(x)K(x-y;t)\,dx=\int \psi _{0}(x)\int _{x(0)=x}^{x(t)=y}e^{iS}\,Dx,}$

then ψ_t obeys the free Schrödinger equation just as K does:

${\displaystyle i{\frac {\partial }{\partial t}}\psi _{t}=-{\frac {\nabla ^{2}}{2}}\psi _{t}.}$

The Lagrangian for the simple harmonic oscillator is[7]

${\displaystyle {\mathcal {L}}={\tfrac {1}{2}}m{\dot {x}}^{2}-{\tfrac {1}{2}}m\omega ^{2}x^{2}.}$

Write its trajectory x(t) as the classical trajectory plus some perturbation, x(t) = x_c(t) + δx(t) and the action as S = S_c + δS. The classical trajectory can be written as

${\displaystyle x_{\text{c}}(t)=x_{i}{\frac {\sin \omega (t_{f}-t)}{\sin \omega (t_{f}-t_{i})}}+x_{f}{\frac {\sin \omega (t-t_{i})}{\sin \omega (t_{f}-t_{i})}}.}$

This trajectory yields the classical action

${\displaystyle {\begin{aligned}S_{\text{c}}&=\int _{t_{i}}^{t_{f}}{\mathcal {L}}\,dt=\int _{t_{i}}^{t_{f}}\left({\tfrac {1}{2}}m{\dot {x}}^{2}-{\tfrac {1}{2}}m\omega ^{2}x^{2}\right)\,dt\\[6pt]&={\frac {1}{2}}m\omega \left({\frac {(x_{i}^{2}+x_{f}^{2})\cos \omega (t_{f}-t_{i})-2x_{i}x_{f}}{\sin \omega (t_{f}-t_{i})}}\right)~.\end{aligned}}}$

Next, expand the deviation from the classical path as a Fourier series, and calculate the contribution to the action δS, which gives

${\displaystyle S=S_{\text{c}}+\sum _{n=1}^{\infty }{\tfrac {1}{2}}a_{n}^{2}{\frac {m}{2}}\left({\frac {(n\pi )^{2}}{t_{f}-t_{i}}}-\omega ^{2}(t_{f}-t_{i})\right).}$

This means that the propagator is

${\displaystyle {\begin{aligned}K(x_{f},t_{f};x_{i},t_{i})&=Qe^{\frac {iS_{\text{c}}}{\hbar }}\prod _{j=1}^{\infty }{\frac {j\pi }{\sqrt {2}}}\int da_{j}\exp {\left({\frac {i}{2\hbar }}a_{j}^{2}{\frac {m}{2}}\left({\frac {(j\pi )^{2}}{t_{f}-t_{i}}}-\omega ^{2}(t_{f}-t_{i})\right)\right)}\\[6pt]&=e^{\frac {iS_{\text{c}}}{\hbar }}Q\prod _{j=1}^{\infty }\left(1-\left({\frac {\omega (t_{f}-t_{i})}{j\pi }}\right)^{2}\right)^{-{\frac {1}{2}}}\end{aligned}}}$

for some normalization

${\displaystyle Q={\sqrt {\frac {m}{2\pi i\hbar (t_{f}-t_{i})}}}~.}$

Using the infinite-product representation of the sinc function,

${\displaystyle \prod _{j=1}^{\infty }\left(1-{\frac {x^{2}}{j^{2}}}\right)={\frac {\sin \pi x}{\pi x}},}$

the propagator can be written as

${\displaystyle K(x_{f},t_{f};x_{i},t_{i})=Qe^{\frac {iS_{\text{c}}}{\hbar }}{\sqrt {\frac {\omega (t_{f}-t_{i})}{\sin \omega (t_{f}-t_{i})}}}=e^{\frac {iS_{c}}{\hbar }}{\sqrt {\frac {m\omega }{2\pi i\hbar \sin \omega (t_{f}-t_{i})}}}.}$

Let T = t_f − t_i. One may write this propagator in terms of energy eigenstates as

${\displaystyle {\begin{aligned}K(x_{f},t_{f};x_{i},t_{i})&=\left({\frac {m\omega }{2\pi i\hbar \sin \omega T}}\right)^{\frac {1}{2}}\exp {\left({\frac {i}{\hbar }}{\tfrac {1}{2}}m\omega {\frac {(x_{i}^{2}+x_{f}^{2})\cos \omega T-2x_{i}x_{f}}{\sin \omega T}}\right)}\\[6pt]&=\sum _{n=0}^{\infty }\exp {\left(-{\frac {iE_{n}T}{\hbar }}\right)}\psi _{n}(x_{f})\psi _{n}(x_{i})^{*}~.\end{aligned}}}$

Using the identities i sin ωT = 1/2e^iωT (1 − e^−2iωT) and cos ωT = 1/2e^iωT (1 + e^−2iωT), this amounts to

${\displaystyle K(x_{f},t_{f};x_{i},t_{i})=\left({\frac {m\omega }{\pi \hbar }}\right)^{\frac {1}{2}}e^{\frac {-i\omega T}{2}}\left(1-e^{-2i\omega T}\right)^{-{\frac {1}{2}}}\exp {\left(-{\frac {m\omega }{2\hbar }}\left(\left(x_{i}^{2}+x_{f}^{2}\right){\frac {1+e^{-2i\omega T}}{1-e^{-2i\omega T}}}-{\frac {4x_{i}x_{f}e^{-i\omega T}}{1-e^{-2i\omega T}}}\right)\right)}.}$

One may absorb all terms after the first e^−iωT/2 into R(T), thereby obtaining

${\displaystyle K(x_{f},t_{f};x_{i},t_{i})=\left({\frac {m\omega }{\pi \hbar }}\right)^{\frac {1}{2}}e^{\frac {-i\omega T}{2}}\cdot R(T).}$

One may finally expand R(T) in powers of e^−iωT: All terms in this expansion get multiplied by the e^−iωT/2 factor in the front, yielding terms of the form

${\displaystyle e^{\frac {-i\omega T}{2}}e^{-in\omega T}=e^{-i\omega T\left({\frac {1}{2}}+n\right)}\quad {\text{for }}n=0,1,2,\ldots .}$

Comparison to the above eigenstate expansion yields the standard energy spectrum for the simple harmonic oscillator,

${\displaystyle E_{n}=\left(n+{\tfrac {1}{2}}\right)\hbar \omega ~.}$

Feynman's time-sliced approximation does not, however, exist for the most important quantum-mechanical path integrals of atoms, due to the singularity of the Coulomb potential e²/r at the origin. Only after replacing the time t by another path-dependent pseudo-time parameter

${\displaystyle s=\int {\frac {dt}{r(t)}}}$

the singularity is removed and a time-sliced approximation exists, which is exactly integrable, since it can be made harmonic by a simple coordinate transformation, as discovered in 1979 by İsmail Hakkı Duru and Hagen Kleinert.[8] The combination of a path-dependent time transformation and a coordinate transformation is an important tool to solve many path integrals and is called generically the Duru–Kleinert transformation.

The path integral reproduces the Schrödinger equation for the initial and final state even when a potential is present. This is easiest to see by taking a path-integral over infinitesimally separated times.

${\displaystyle \psi (y;t+\varepsilon )=\int _{-\infty }^{\infty }\psi (x;t)\int _{x(t)=x}^{x(t+\varepsilon )=y}e^{i\int _{t}^{t+\varepsilon }{\bigl (}{\frac {1}{2}}{\dot {x}}^{2}-V(x){\bigr )}dt}Dx(t)\,dx\qquad (1)}$

Since the time separation is infinitesimal and the cancelling oscillations become severe for large values of ẋ, the path integral has most weight for y close to x. In this case, to lowest order the potential energy is constant, and only the kinetic energy contribution is nontrivial. (This separation of the kinetic and potential energy terms in the exponent is essentially the Trotter product formula.) The exponential of the action is

${\displaystyle e^{-i\varepsilon V(x)}e^{i{\frac {{\dot {x}}^{2}}{2}}\varepsilon }}$

The first term rotates the phase of ψ(x) locally by an amount proportional to the potential energy. The second term is the free particle propagator, corresponding to i times a diffusion process. To lowest order in ε they are additive; in any case one has with (1):

${\displaystyle \psi (y;t+\varepsilon )\approx \int \psi (x;t)e^{-i\varepsilon V(x)}e^{\frac {i(x-y)^{2}}{2\varepsilon }}\,dx\,.}$

As mentioned, the spread in ψ is diffusive from the free particle propagation, with an extra infinitesimal rotation in phase which slowly varies from point to point from the potential:

${\displaystyle {\frac {\partial \psi }{\partial t}}=i\cdot \left({\tfrac {1}{2}}\nabla ^{2}-V(x)\right)\psi \,}$

and this is the Schrödinger equation. The normalization of the path integral needs to be fixed in exactly the same way as in the free particle case. An arbitrary continuous potential does not affect the normalization, although singular potentials require careful treatment.

Since the states obey the Schrödinger equation, the path integral must reproduce the Heisenberg equations of motion for the averages of x and ẋ variables, but it is instructive to see this directly. The direct approach shows that the expectation values calculated from the path integral reproduce the usual ones of quantum mechanics.

Start by considering the path integral with some fixed initial state

${\displaystyle \int \psi _{0}(x)\int _{x(0)=x}e^{iS(x,{\dot {x}})}\,Dx\,}$

Now x(t) at each separate time is a separate integration variable. So it is legitimate to change variables in the integral by shifting: x(t) = u(t) + ε(t) where ε(t) is a different shift at each time but ε(0) = ε(T) = 0, since the endpoints are not integrated:

${\displaystyle \int \psi _{0}(x)\int _{u(0)=x}e^{iS(u+\varepsilon ,{\dot {u}}+{\dot {\varepsilon }})}\,Du\,}$

The change in the integral from the shift is, to first infinitesimal order in ε:

${\displaystyle \int \psi _{0}(x)\int _{u(0)=x}\left(\int {\frac {\partial S}{\partial u}}\varepsilon +{\frac {\partial S}{\partial {\dot {u}}}}{\dot {\varepsilon }}\,dt\right)e^{iS}\,Du\,}$

which, integrating by parts in t, gives:

${\displaystyle \int \psi _{0}(x)\int _{u(0)=x}-\left(\int \left({\frac {d}{dt}}{\frac {\partial S}{\partial {\dot {u}}}}-{\frac {\partial S}{\partial u}}\right)\varepsilon (t)\,dt\right)e^{iS}\,Du\,}$

But this was just a shift of integration variables, which doesn't change the value of the integral for any choice of ε(t). The conclusion is that this first order variation is zero for an arbitrary initial state and at any arbitrary point in time:

${\displaystyle \left\langle \psi _{0}\left|{\frac {\delta S}{\delta x}}(t)\right|\psi _{0}\right\rangle =0}$

this is the Heisenberg equation of motion.

If the action contains terms which multiply ẋ and x, at the same moment in time, the manipulations above are only heuristic, because the multiplication rules for these quantities is just as noncommuting in the path integral as it is in the operator formalism.

If the variation in the action exceeds ħ by many orders of magnitude, we typically have destructive interference other than in the vicinity of those trajectories satisfying the Euler–Lagrange equation, which is now reinterpreted as the condition for constructive interference. This can be shown using the method of stationary phase applied to the propagator. As ħ decreases, the exponential in the integral oscillates rapidly in the complex domain for any change in the action. Thus, in the limit that ħ goes to zero, only points where the classical action does not vary contribute to the propagator.

The formulation of the path integral does not make it clear at first sight that the quantities x and p do not commute. In the path integral, these are just integration variables and they have no obvious ordering. Feynman discovered that the non-commutativity is still present.[9]

To see this, consider the simplest path integral, the brownian walk. This is not yet quantum mechanics, so in the path-integral the action is not multiplied by i:

${\displaystyle S=\int \left({\frac {dx}{dt}}\right)^{2}\,dt}$

The quantity x(t) is fluctuating, and the derivative is defined as the limit of a discrete difference.

${\displaystyle {\frac {dx}{dt}}={\frac {x(t+\varepsilon )-x(t)}{\varepsilon }}}$

The distance that a random walk moves is proportional to √t, so that:

${\displaystyle x(t+\varepsilon )-x(t)\approx {\sqrt {\varepsilon }}}$

This shows that the random walk is not differentiable, since the ratio that defines the derivative diverges with probability one.

The quantity xẋ is ambiguous, with two possible meanings:

${\displaystyle [1]=x{\frac {dx}{dt}}=x(t){\frac {x(t+\varepsilon )-x(t)}{\varepsilon }}}$

${\displaystyle [2]=x{\frac {dx}{dt}}=x(t+\varepsilon ){\frac {x(t+\varepsilon )-x(t)}{\varepsilon }}}$

In elementary calculus, the two are only different by an amount which goes to 0 as ε goes to 0. But in this case, the difference between the two is not 0:

${\displaystyle [2]-[1]={\frac {{\big (}x(t+\varepsilon )-x(t){\big )}^{2}}{\varepsilon }}\approx {\frac {\varepsilon }{\varepsilon }}}$

Let

${\displaystyle f(t)={\frac {{\big (}x(t+\varepsilon )-x(t){\big )}^{2}}{\varepsilon }}}$

Then f(t) is a rapidly fluctuating statistical quantity, whose average value is 1, i.e. a normalized "Gaussian process". The fluctuations of such a quantity can be described by a statistical Lagrangian

${\displaystyle {\mathcal {L}}=(f(t)-1)^{2}\,,}$

and the equations of motion for f derived from extremizing the action S corresponding to L just set it equal to 1. In physics, such a quantity is "equal to 1 as an operator identity". In mathematics, it "weakly converges to 1". In either case, it is 1 in any expectation value, or when averaged over any interval, or for all practical purpose.

Defining the time order to be the operator order:

${\displaystyle [x,{\dot {x}}]=x{\frac {dx}{dt}}-{\frac {dx}{dt}}x=1}$

This is called the Itō lemma in stochastic calculus, and the (euclideanized) canonical commutation relations in physics.

For a general statistical action, a similar argument shows that

${\displaystyle \left[x,{\frac {\partial S}{\partial {\dot {x}}}}\right]=1}$

and in quantum mechanics, the extra imaginary unit in the action converts this to the canonical commutation relation,

${\displaystyle [x,p]=i}$

For a particle in curved space the kinetic term depends on the position, and the above time slicing cannot be applied, this being a manifestation of the notorious operator ordering problem in Schrödinger quantum mechanics. One may, however, solve this problem by transforming the time-sliced flat-space path integral to curved space using a multivalued coordinate transformation (nonholonomic mapping explained here).

Sometimes (e.g. a particle moving in curved space) we also have measure-theoretic factors in the functional integral:

${\displaystyle \int \mu [x]e^{iS[x]}\,{\mathcal {D}}x.}$

This factor is needed to restore unitarity.

For instance, if

${\displaystyle S=\int \left({\frac {m}{2}}g_{ij}{\dot {x}}^{i}{\dot {x}}^{j}-V(x)\right)\,dt,}$

then it means that each spatial slice is multiplied by the measure √g. This measure cannot be expressed as a functional multiplying the Dx measure because they belong to entirely different classes.

Matrix elements of the kind ${\displaystyle \langle x_{f}|e^{-{\frac {i}{\hbar }}{\hat {H}}(t-t')}F({\hat {x}})e^{-{\frac {i}{\hbar }}{\hat {H}}(t')}|x_{i}\rangle }$ take the form

${\displaystyle \int _{x(0)=x_{i}}^{x(t)=x_{f}}{\mathcal {D}}[x]F(x(t'))e^{{\frac {i}{\hbar }}\int dtL(x(t),{\dot {x}}(t))}}$.

This generalizes to multiple operators, for example

${\displaystyle \langle x_{f}|e^{-{\frac {i}{\hbar }}{\hat {H}}(t-t_{1})}F_{1}({\hat {x}})e^{-{\frac {i}{\hbar }}{\hat {H}}(t_{1}-t_{2})}F_{2}({\hat {x}})e^{-{\frac {i}{\hbar }}{\hat {H}}(t_{2})}|x_{i}\rangle =\int _{x(0)=x_{i}}^{x(t)=x_{f}}{\mathcal {D}}[x]F_{1}(x(t_{1}))F_{2}(x(t_{2}))e^{{\frac {i}{\hbar }}\int dtL(x(t),{\dot {x}}(t))}}$,

and to the general expectation value

${\displaystyle \langle F\rangle ={\frac {\int {\mathcal {D}}[\phi ]F(\phi )e^{{\frac {i}{\hbar }}S[\phi ]}}{\int {\mathcal {D}}[\phi ]e^{{\frac {i}{\hbar }}S[\phi ]}}}}$.

If we replace ${\displaystyle t}$ by ${\displaystyle -it}$, the time-evolution operator ${\displaystyle e^{-it{\hat {H}}/\hbar }}$ is replaced by ${\displaystyle e^{-t{\hat {H}}/\hbar }}$. (This change is known as a Wick rotation.) If we repeat the derivation of the path-integral formula in this setting, we obtain[10]

${\displaystyle \psi (x,t)={\frac {1}{Z}}\int _{\mathbf {x} (0)=x}e^{-S_{\mathrm {Euclidean} }(\mathbf {x} ,{\dot {\mathbf {x} }})/\hbar }\psi _{0}(\mathbf {x} (t))\,{\mathcal {D}}\mathbf {x} \,}$,

where ${\displaystyle S_{\mathrm {Euclidean} }}$ is the Euclidean action, given by

${\displaystyle S_{\mathrm {Euclidean} }(\mathbf {x} ,{\dot {\mathbf {x} }})=\int \left[{\frac {m}{2}}|{\dot {\mathbf {x} }}(t)|^{2}+V(\mathbf {x} (t))\right]\,dt}$.

Note the sign change between this and the normal action, where the potential energy term is negative. (The term Euclidean is from the context of quantum field theory, where the change from real to imaginary time changes the space-time geometry from Lorentzian to Euclidean.)

Now, the contribution of the kinetic energy to the path integral is as follows:

${\displaystyle {\frac {1}{Z}}\int _{\mathbf {x} (0)=x}f(\mathbf {x} )e^{-{\frac {m}{2}}\int |{\dot {\mathbf {x} }}|^{2}dt}\,{\mathcal {D}}\mathbf {x} \,}$

where ${\displaystyle f(\mathbf {x} )}$ includes all the remaining dependence of the integrand on the path. This integral has a rigorous mathematical interpretation as integration against the Wiener measure, denoted ${\displaystyle \mu _{x}}$. The Wiener measure, constructed by Norbert Wiener gives a rigorous foundation to Einstein's mathematical model of Brownian motion. The subscript ${\displaystyle x}$ indicates that the measure ${\displaystyle \mu _{x}}$ is supported on paths ${\displaystyle \mathbf {x} }$ with ${\displaystyle \mathbf {x} (0)=x}$.

We then have a rigorous version of the Feynman path integral, known as the Feynman–Kac formula:[11]

${\displaystyle \psi (x,t)=\int e^{-\int V(\mathbf {x} (t)\,dt/\hbar }\,\psi _{0}(\mathbf {x} (t))\,d\mu _{x}(\mathbf {x} )}$,

where now ${\displaystyle \psi (x,t)}$ satisfies the Wick-rotated version of the Schrödinger equation,

${\displaystyle \hbar {\frac {\partial }{\partial t}}\psi (x,t)=-{\hat {H}}\psi (x,t)}$.

Although the Wick-rotated Schrödinger equation does not have a direct physical meaning, interesting properties of the Schrödinger operator ${\displaystyle {\hat {H}}}$ can be extracted by studying it.[12]

Much of the study of quantum field theories from the path-integral perspective, in both the mathematics and physics literatures, is done in the Euclidean setting, that is, after a Wick rotation. In particular, there are various results showing that if a Euclidean field theory with suitable properties can be constructed, one can then undo the Wick rotation to recover the physical, Lorentzian theory.[13] On the other hand, it is much more difficult to give a meaning to path integrals (even Euclidean path integrals) in quantum field theory than in quantum mechanics.[nb 3]

The path integral is just the generalization of the integral above to all quantum mechanical problems—

${\displaystyle Z=\int e^{\frac {i{\mathcal {S}}[\mathbf {x} ]}{\hbar }}\,{\mathcal {D}}\mathbf {x} \quad {\text{where }}{\mathcal {S}}[\mathbf {x} ]=\int _{0}^{t_{f}}L[\mathbf {x} (t),{\dot {\mathbf {x} }}(t)]\,dt}$

is the action of the classical problem in which one investigates the path starting at time t = 0 and ending at time t = t_f, and ${\displaystyle {\mathcal {D}}\mathbf {x} }$ denotes the integration measure over all paths. In the classical limit, ${\displaystyle {\mathcal {S}}[\mathbf {x} ]\gg \hbar }$, the path of minimum action dominates the integral, because the phase of any path away from this fluctuates rapidly and different contributions cancel.[14]

The connection with statistical mechanics follows. Considering only paths which begin and end in the same configuration, perform the Wick rotation it = ħβ, i.e., make time imaginary, and integrate over all possible beginning-ending configurations. The Wick-rotated path integral—described in the previous subsection, with the ordinary action replaced by its "Euclidean" counterpart—now resembles the partition function of statistical mechanics defined in a canonical ensemble with inverse temperature proportional to imaginary time, 1/T = ik_Bt/ħ. Strictly speaking, though, this is the partition function for a statistical field theory.

Clearly, such a deep analogy between quantum mechanics and statistical mechanics cannot be dependent on the formulation. In the canonical formulation, one sees that the unitary evolution operator of a state is given by

${\displaystyle |\alpha ;t\rangle =e^{-{\frac {iHt}{\hbar }}}|\alpha ;0\rangle }$

where the state α is evolved from time t = 0. If one makes a Wick rotation here, and finds the amplitude to go from any state, back to the same state in (imaginary) time iβ is given by

${\displaystyle Z=\operatorname {Tr} \left[e^{-H\beta }\right]}$

which is precisely the partition function of statistical mechanics for the same system at the temperature quoted earlier. One aspect of this equivalence was also known to Erwin Schrödinger who remarked that the equation named after him looked like the diffusion equation after Wick rotation. Note, however, that the Euclidean path integral is actually in the form of a classical statistical mechanics model.

In relativistic theories, there is both a particle and field representation for every theory. The field representation is a sum over all field configurations, and the particle representation is a sum over different particle paths.

The nonrelativistic formulation is traditionally given in terms of particle paths, not fields. There, the path integral in the usual variables, with fixed boundary conditions, gives the probability amplitude for a particle to go from point x to point y in time T:

${\displaystyle K(x,y;T)=\langle y;T\mid x;0\rangle =\int _{x(0)=x}^{x(T)=y}e^{iS[x]}\,Dx.}$

This is called the propagator. Superposing different values of the initial position x with an arbitrary initial state ψ₀(x) constructs the final state:

${\displaystyle \psi _{T}(y)=\int _{x}\psi _{0}(x)K(x,y;T)\,dx=\int ^{x(T)=y}\psi _{0}(x(0))e^{iS[x]}\,Dx.}$

For a spatially homogeneous system, where K(x, y) is only a function of (x − y), the integral is a convolution, the final state is the initial state convolved with the propagator:

${\displaystyle \psi _{T}=\psi _{0}*K(;T).}$

For a free particle of mass m, the propagator can be evaluated either explicitly from the path integral or by noting that the Schrödinger equation is a diffusion equation in imaginary time, and the solution must be a normalized Gaussian:

${\displaystyle K(x,y;T)\propto e^{\frac {im(x-y)^{2}}{2T}}.}$

Taking the Fourier transform in (x − y) produces another Gaussian:

${\displaystyle K(p;T)=e^{\frac {iTp^{2}}{2m}},}$

and in p-space the proportionality factor here is constant in time, as will be verified in a moment. The Fourier transform in time, extending K(p; T) to be zero for negative times, gives Green's function, or the frequency-space propagator:

${\displaystyle G_{\text{F}}(p,E)={\frac {-i}{E-{\frac {{\vec {p}}^{2}}{2m}}+i\varepsilon }},}$

which is the reciprocal of the operator that annihilates the wavefunction in the Schrödinger equation, which wouldn't have come out right if the proportionality factor weren't constant in the p-space representation.

The infinitesimal term in the denominator is a small positive number, which guarantees that the inverse Fourier transform in E will be nonzero only for future times. For past times, the inverse Fourier transform contour closes toward values of E where there is no singularity. This guarantees that K propagates the particle into the future and is the reason for the subscript "F" on G. The infinitesimal term can be interpreted as an infinitesimal rotation toward imaginary time.

It is also possible to reexpress the nonrelativistic time evolution in terms of propagators going toward the past, since the Schrödinger equation is time-reversible. The past propagator is the same as the future propagator except for the obvious difference that it vanishes in the future, and in the Gaussian t is replaced by −t. In this case, the interpretation is that these are the quantities to convolve the final wavefunction so as to get the initial wavefunction:

${\displaystyle G_{\text{B}}(p,E)={\frac {-i}{-E-{\frac {i{\vec {p}}^{2}}{2m}}+i\varepsilon }}.}$

Given the nearly identical only change is the sign of E and ε, the parameter E in Green's function can either be the energy if the paths are going toward the future, or the negative of the energy if the paths are going toward the past.

For a nonrelativistic theory, the time as measured along the path of a moving particle and the time as measured by an outside observer are the same. In relativity, this is no longer true. For a relativistic theory the propagator should be defined as the sum over all paths that travel between two points in a fixed proper time, as measured along the path (these paths describe the trajectory of a particle in space and in time):

${\displaystyle K(x-y,\mathrm {T} )=\int _{x(0)=x}^{x(\mathrm {T} )=y}e^{i\int _{0}^{\mathrm {T} }{\sqrt {{\dot {x}}^{2}}}-\alpha \,d\tau }.}$

The integral above is not trivial to interpret because of the square root. Fortunately, there is a heuristic trick. The sum is over the relativistic arc length of the path of an oscillating quantity, and like the nonrelativistic path integral should be interpreted as slightly rotated into imaginary time. The function K(x − y, τ) can be evaluated when the sum is over paths in Euclidean space:

${\displaystyle K(x-y,\mathrm {T} )=e^{-\alpha \mathrm {T} }\int _{x(0)=x}^{x(\mathrm {T} )=y}e^{-L}.}$

This describes a sum over all paths of length Τ of the exponential of minus the length. This can be given a probability interpretation. The sum over all paths is a probability average over a path constructed step by step. The total number of steps is proportional to Τ, and each step is less likely the longer it is. By the central limit theorem, the result of many independent steps is a Gaussian of variance proportional to Τ:

${\displaystyle K(x-y,\mathrm {T} )=e^{-\alpha \mathrm {T} }e^{-{\frac {(x-y)^{2}}{\mathrm {T} }}}.}$

The usual definition of the relativistic propagator only asks for the amplitude is to travel from x to y, after summing over all the possible proper times it could take:

${\displaystyle K(x-y)=\int _{0}^{\infty }K(x-y,\mathrm {T} )W(\mathrm {T} )\,d\mathrm {T} ,}$

where W(Τ) is a weight factor, the relative importance of paths of different proper time. By the translation symmetry in proper time, this weight can only be an exponential factor and can be absorbed into the constant α:

${\displaystyle K(x-y)=\int _{0}^{\infty }e^{-{\frac {(x-y)^{2}}{\mathrm {T} }}-\alpha \mathrm {T} }\,d\mathrm {T} .}$

This is the Schwinger representation. Taking a Fourier transform over the variable (x − y) can be done for each value of Τ separately, and because each separate Τ contribution is a Gaussian, gives whose Fourier transform is another Gaussian with reciprocal width. So in p-space, the propagator can be reexpressed simply:

${\displaystyle K(p)=\int _{0}^{\infty }e^{-\mathrm {T} p^{2}-\mathrm {T} \alpha }\,d\mathrm {T} ={\frac {1}{p^{2}+\alpha }},}$

which is the Euclidean propagator for a scalar particle. Rotating p₀ to be imaginary gives the usual relativistic propagator, up to a factor of −i and an ambiguity, which will be clarified below:

${\displaystyle K(p)={\frac {i}{p_{0}^{2}-{\vec {p}}^{2}-m^{2}}}.}$

This expression can be interpreted in the nonrelativistic limit, where it is convenient to split it by partial fractions:

${\displaystyle 2p_{0}K(p)={\frac {i}{p_{0}-{\sqrt {{\vec {p}}^{2}+m^{2}}}}}+{\frac {i}{p_{0}+{\sqrt {{\vec {p}}^{2}+m^{2}}}}}.}$

For states where one nonrelativistic particle is present, the initial wavefunction has a frequency distribution concentrated near p₀ = m. When convolving with the propagator, which in p space just means multiplying by the propagator, the second term is suppressed and the first term is enhanced. For frequencies near p₀ = m, the dominant first term has the form

${\displaystyle 2mK_{\text{NR}}(p)={\frac {i}{(p_{0}-m)-{\frac {{\vec {p}}^{2}}{2m}}}}.}$

This is the expression for the nonrelativistic Green's function of a free Schrödinger particle.

The second term has a nonrelativistic limit also, but this limit is concentrated on frequencies that are negative. The second pole is dominated by contributions from paths where the proper time and the coordinate time are ticking in an opposite sense, which means that the second term is to be interpreted as the antiparticle. The nonrelativistic analysis shows that with this form the antiparticle still has positive energy.

The proper way to express this mathematically is that, adding a small suppression factor in proper time, the limit where t → −∞ of the first term must vanish, while the t → +∞ limit of the second term must vanish. In the Fourier transform, this means shifting the pole in p₀ slightly, so that the inverse Fourier transform will pick up a small decay factor in one of the time directions:

${\displaystyle K(p)={\frac {i}{p_{0}-{\sqrt {{\vec {p}}^{2}+m^{2}}}+i\varepsilon }}+{\frac {i}{p_{0}-{\sqrt {{\vec {p}}^{2}+m^{2}}}-i\varepsilon }}.}$

Without these terms, the pole contribution could not be unambiguously evaluated when taking the inverse Fourier transform of p₀. The terms can be recombined:

${\displaystyle K(p)={\frac {i}{p^{2}-m^{2}+i\varepsilon }},}$

which when factored, produces opposite-sign infinitesimal terms in each factor. This is the mathematically precise form of the relativistic particle propagator, free of any ambiguities. The ε term introduces a small imaginary part to the α = m², which in the Minkowski version is a small exponential suppression of long paths.

So in the relativistic case, the Feynman path-integral representation of the propagator includes paths going backwards in time, which describe antiparticles. The paths that contribute to the relativistic propagator go forward and backwards in time, and the interpretation of this is that the amplitude for a free particle to travel between two points includes amplitudes for the particle to fluctuate into an antiparticle, travel back in time, then forward again.

Unlike the nonrelativistic case, it is impossible to produce a relativistic theory of local particle propagation without including antiparticles. All local differential operators have inverses that are nonzero outside the light cone, meaning that it is impossible to keep a particle from travelling faster than light. Such a particle cannot have a Green's function which is only nonzero in the future in a relativistically invariant theory.

However, the path integral formulation is also extremely important in direct application to quantum field theory, in which the "paths" or histories being considered are not the motions of a single particle, but the possible time evolutions of a field over all space. The action is referred to technically as a functional of the field: S[ϕ], where the field ϕ(x^μ) is itself a function of space and time, and the square brackets are a reminder that the action depends on all the field's values everywhere, not just some particular value. One such given function ϕ(x^μ) of spacetime is called a field configuration. In principle, one integrates Feynman's amplitude over the class of all possible field configurations.

Much of the formal study of QFT is devoted to the properties of the resulting functional integral, and much effort (not yet entirely successful) has been made toward making these functional integrals mathematically precise.

Such a functional integral is extremely similar to the partition function in statistical mechanics. Indeed, it is sometimes called a partition function, and the two are essentially mathematically identical except for the factor of i in the exponent in Feynman's postulate 3. Analytically continuing the integral to an imaginary time variable (called a Wick rotation) makes the functional integral even more like a statistical partition function and also tames some of the mathematical difficulties of working with these integrals.

In quantum field theory, if the action is given by the functional S of field configurations (which only depends locally on the fields), then the time-ordered vacuum expectation value of polynomially bounded functional F, ⟨F⟩, is given by

${\displaystyle \langle F\rangle ={\frac {\int {\mathcal {D}}\varphi F[\varphi ]e^{i{\mathcal {S}}[\varphi ]}}{\int {\mathcal {D}}\varphi e^{i{\mathcal {S}}[\varphi ]}}}.}$

The symbol ∫Dϕ here is a concise way to represent the infinite-dimensional integral over all possible field configurations on all of space-time. As stated above, the unadorned path integral in the denominator ensures proper normalization.

Strictly speaking, the only question that can be asked in physics is: What fraction of states satisfying condition A also satisfy condition B? The answer to this is a number between 0 and 1, which can be interpreted as a conditional probability, written as P(B|A). In terms of path integration, since P(B|A) = P(A∩B) / P(A), this means

${\displaystyle \operatorname {P} (B\mid A)={\frac {\sum _{F\subset A\cap B}\left|\int {\mathcal {D}}\varphi O_{\text{in}}[\varphi ]e^{i{\mathcal {S}}[\varphi ]}F[\varphi ]\right|^{2}}{\sum _{F\subset A}\left|\int {\mathcal {D}}\varphi O_{\text{in}}[\varphi ]e^{i{\mathcal {S}}[\varphi ]}F[\varphi ]\right|^{2}}},}$

where the functional O_in[ϕ] is the superposition of all incoming states that could lead to the states we are interested in. In particular, this could be a state corresponding to the state of the Universe just after the Big Bang, although for actual calculation this can be simplified using heuristic methods. Since this expression is a quotient of path integrals, it is naturally normalised.

Since this formulation of quantum mechanics is analogous to classical action principle, one might expect that identities concerning the action in classical mechanics would have quantum counterparts derivable from a functional integral. This is often the case.

In the language of functional analysis, we can write the Euler–Lagrange equations as

${\displaystyle {\frac {\delta {\mathcal {S}}[\varphi ]}{\delta \varphi }}=0}$

(the left-hand side is a functional derivative; the equation means that the action is stationary under small changes in the field configuration). The quantum analogues of these equations are called the Schwinger–Dyson equations.

If the functional measure Dϕ turns out to be translationally invariant (we'll assume this for the rest of this article, although this does not hold for, let's say nonlinear sigma models), and if we assume that after a Wick rotation

${\displaystyle e^{i{\mathcal {S}}[\varphi ]},}$

which now becomes

${\displaystyle e^{-H[\varphi ]}}$

for some H, it goes to zero faster than a reciprocal of any polynomial for large values of φ, then we can integrate by parts (after a Wick rotation, followed by a Wick rotation back) to get the following Schwinger–Dyson equations for the expectation:

${\displaystyle \left\langle {\frac {\delta F[\varphi ]}{\delta \varphi }}\right\rangle =-i\left\langle F[\varphi ]{\frac {\delta {\mathcal {S}}[\varphi ]}{\delta \varphi }}\right\rangle }$

for any polynomially-bounded functional F. In the deWitt notation this looks like[15]

${\displaystyle \left\langle F_{,i}\right\rangle =-i\left\langle F{\mathcal {S}}_{,i}\right\rangle .}$

These equations are the analog of the on-shell EL equations. The time ordering is taken before the time derivatives inside the S_,i.

If J (called the source field) is an element of the dual space of the field configurations (which has at least an affine structure because of the assumption of the translational invariance for the functional measure), then the generating functional Z of the source fields is defined to be

${\displaystyle Z[J]=\int {\mathcal {D}}\varphi e^{i\left({\mathcal {S}}[\varphi ]+\langle J,\varphi \rangle \right)}.}$

Note that

${\displaystyle {\frac {\delta ^{n}Z}{\delta J(x_{1})\cdots \delta J(x_{n})}}[J]=i^{n}\,Z[J]\,\left\langle \varphi (x_{1})\cdots \varphi (x_{n})\right\rangle _{J},}$

or

${\displaystyle Z^{,i_{1}\cdots i_{n}}[J]=i^{n}Z[J]\left\langle \varphi ^{i_{1}}\cdots \varphi ^{i_{n}}\right\rangle _{J},}$

where

${\displaystyle \langle F\rangle _{J}={\frac {\int {\mathcal {D}}\varphi F[\varphi ]e^{i\left({\mathcal {S}}[\varphi ]+\langle J,\varphi \rangle \right)}}{\int {\mathcal {D}}\varphi e^{i\left({\mathcal {S}}[\varphi ]+\langle J,\varphi \rangle \right)}}}.}$

Basically, if Dφ e^iS[φ] is viewed as a functional distribution (this shouldn't be taken too literally as an interpretation of QFT, unlike its Wick-rotated statistical mechanics analogue, because we have time ordering complications here!), then ⟨φ(x₁) ... φ(x_n)⟩ are its moments, and Z is its Fourier transform.

If F is a functional of φ, then for an operator K, F[K] is defined to be the operator that substitutes K for φ. For example, if

${\displaystyle F[\varphi ]={\frac {\partial ^{k_{1}}}{\partial x_{1}^{k_{1}}}}\varphi (x_{1})\cdots {\frac {\partial ^{k_{n}}}{\partial x_{n}^{k_{n}}}}\varphi (x_{n}),}$

and G is a functional of J, then

${\displaystyle F\left[-i{\frac {\delta }{\delta J}}\right]G[J]=(-i)^{n}{\frac {\partial ^{k_{1}}}{\partial x_{1}^{k_{1}}}}{\frac {\delta }{\delta J(x_{1})}}\cdots {\frac {\partial ^{k_{n}}}{\partial x_{n}^{k_{n}}}}{\frac {\delta }{\delta J(x_{n})}}G[J].}$

Then, from the properties of the functional integrals

${\displaystyle \left\langle {\frac {\delta {\mathcal {S}}}{\delta \varphi (x)}}[\varphi ]+J(x)\right\rangle _{J}=0}$

we get the "master" Schwinger–Dyson equation:

${\displaystyle {\frac {\delta {\mathcal {S}}}{\delta \varphi (x)}}\left[-i{\frac {\delta }{\delta J}}\right]Z[J]+J(x)Z[J]=0,}$

or

${\displaystyle {\mathcal {S}}_{,i}[-i\partial ]Z+J_{i}Z=0.}$

If the functional measure is not translationally invariant, it might be possible to express it as the product M[φ] Dφ, where M is a functional and Dφ is a translationally invariant measure. This is true, for example, for nonlinear sigma models where the target space is diffeomorphic to Rⁿ. However, if the target manifold is some topologically nontrivial space, the concept of a translation does not even make any sense.

In that case, we would have to replace the S in this equation by another functional

${\displaystyle {\hat {\mathcal {S}}}={\mathcal {S}}-i\ln M.}$

If we expand this equation as a Taylor series about J = 0, we get the entire set of Schwinger–Dyson equations.

Now how about the on shell Noether's theorem for the classical case? Does it have a quantum analog as well? Yes, but with a caveat. The functional measure would have to be invariant under the one parameter group of symmetry transformation as well.

Let's just assume for simplicity here that the symmetry in question is local (not local in the sense of a gauge symmetry, but in the sense that the transformed value of the field at any given point under an infinitesimal transformation would only depend on the field configuration over an arbitrarily small neighborhood of the point in question). Let's also assume that the action is local in the sense that it is the integral over spacetime of a Lagrangian, and that

${\displaystyle Q[{\mathcal {L}}(x)]=\partial _{\mu }f^{\mu }(x)}$