Direct Evaluation of Path Integrals

Abstract

Until now we have always used a trick to calculate the path integral in

$$\displaystyle\begin{array}{rcl} K{\bigl (x_{2},t_{2};x_{1},t_{1}\bigr )}& =& \,\text{e}^{(\text{i}/\hslash )S_{\text{cl}}[x(t)] }\int _{y(t_{1})=0}^{y(t_{2})=0}[dy(t)]\, \\ & & \times \text{exp}\left \{ \frac{\text{i}} {\hslash }\int _{t_{1}}^{t_{2} }dt{\bigl (a(t)y^{2} + b(t)y\dot{y} + c(t)\dot{y}^{2}\bigr )}\right \}\;.{}\end{array}$$

(20.1)

The path integral in (20.1) requires integration over all possible paths y(t) from (0, t ₁) to (0, t ₂) with the associated action

$$\displaystyle{ S{\bigl \{{\bigl (0,t_{2}\bigr )},{\bigl (0,t_{1}\bigr )}\bigr \}} =\int _{ t_{1}}^{t_{2} }dt{\bigl (a(t)y^{2} + b(t)y\dot{y}^{2} + c(t)\dot{y}^{2}\bigr )}\;. }$$

(20.2)

In order to calculate K directly, we divide the time interval T = t ₂ − t ₁ in N steps of width $\varepsilon$: $T = N\varepsilon,\ t_{2}\geqslant t_{1}.$

$$\displaystyle{ \tau _{0}:= t_{1}\,\,\text{(fixed}),\,\tau _{1} = t_{1}+\varepsilon,\ldots \,,\,\tau _{N-1} = t_{1} + (N - 1)\varepsilon,\,\tau _{N}:= t_{2}\,\,\text{(fixed)}\;. }$$

Every time τ _n is assigned a point y _n . We now connect the individual points with a classical path y(τ).  y(τ) is not necessarily the (on-shell trajectory) extremum of the classical action.

These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

Until now we have always used a trick to calculate the path integral in

$$\displaystyle\begin{array}{rcl} K{\bigl (x_{2},t_{2};x_{1},t_{1}\bigr )}& =& \,\text{e}^{(\text{i}/\hslash )S_{\text{cl}}[x(t)] }\int _{y(t_{1})=0}^{y(t_{2})=0}[dy(t)]\, \\ & & \times \text{exp}\left \{ \frac{\text{i}} {\hslash }\int _{t_{1}}^{t_{2} }dt{\bigl (a(t)y^{2} + b(t)y\dot{y} + c(t)\dot{y}^{2}\bigr )}\right \}\;.{}\end{array}$$

(20.1)

The path integral in (20.1) requires integration over all possible paths y(t) from (0, t ₁) to (0, t ₂) with the associated action

$$\displaystyle{ S{\bigl \{{\bigl (0,t_{2}\bigr )},{\bigl (0,t_{1}\bigr )}\bigr \}} =\int _{ t_{1}}^{t_{2} }dt{\bigl (a(t)y^{2} + b(t)y\dot{y}^{2} + c(t)\dot{y}^{2}\bigr )}\;. }$$

(20.2)

In order to calculate K directly, we divide the time interval T = t ₂ − t ₁ in N steps of width $\varepsilon$: $T = N\varepsilon,\ t_{2}\geqslant t_{1}.$

$$\displaystyle{ \tau _{0}:= t_{1}\,\,\text{(fixed}),\,\tau _{1} = t_{1}+\varepsilon,\ldots \,,\,\tau _{N-1} = t_{1} + (N - 1)\varepsilon,\,\tau _{N}:= t_{2}\,\,\text{(fixed)}\;. }$$

Every time τ _n is assigned a point y _n . We now connect the individual points with a classical path y(τ).  y(τ) is not necessarily the (on-shell trajectory) extremum of the classical action. It can be any path between τ _n and τ _n−1 specified by the classical Lagrangian $L(y,\dot{y},t).$ The action along one step of the path from (n) to (n + 1) is thus

$$\displaystyle{ S_{\text{cl}}(n + 1,n) =\int _{ \tau _{n}}^{\tau _{n+1} }d\tau \,L(y,\dot{y};\tau ) }$$

(20.3)

and the action along the entire path, accordingly,

$$\displaystyle{ S_{\text{cl}}(N,0) =\sum _{ n\,=\,0}^{N-1}S_{ \text{cl}}(n + 1,n)\;. }$$

(20.4)

The contribution of each infinitesimal step of the path to the phase integral is then (Dirac, Feynman)

$$\displaystyle{ \phi _{n+1,n} = \text{const.}\,\text{e}^{(\text{i}/\hslash )S_{\mathrm{cl}}(n+1,n)}\;. }$$

(20.5)

The sum over all path contributions is, at last,

$$\displaystyle\begin{array}{rcl} \int _{y(t_{1})\,=\,0}^{y(t_{2})\,=\,0}[dy(t)]\,\text{e}^{(\text{i}/\hslash )S[y(t)]}& =& \,\text{const.}\int _{ -\infty }^{+\infty }dy_{ 1}\ldots \int _{-\infty }^{+\infty }dy_{ N-1} \\ & & \times \text{exp}\left [ \frac{\text{i}} {\hslash }\sum _{n\,=\,0}^{N-1}S_{ \text{cl}}(n + 1,n)\right ]\;.{}\end{array}$$

(20.6)

We do not integrate over $y_{0} = y(\tau _{0}) = y(t_{1}) = 0$ and $y_{N} = y(\tau _{N}) = y(t_{2}) = 0,$ since these are the fixed, chosen initial values of y(τ). In order to obtain all paths between t ₁ and t ₂ in the (y, τ)-plane, we have to form the limit $\varepsilon = (t_{2} - t_{1})/N \rightarrow 0$ or $N \rightarrow \infty$. In order for this limit to exist, the constant in (20.6), i.e., the integration measure, has to be chosen properly.

We know from experience with our examples from Chap. 18 that

$$\displaystyle{ \text{const.} = A^{-N}\;,\quad A(\delta t =\varepsilon ) = \sqrt{\frac{2\pi \text{i} \hslash \varepsilon } {m}} \;. }$$

(20.7)

For this case, the limit exists and we can write

$$\displaystyle\begin{array}{rcl} \int _{y(t_{1})=0}^{y(t_{2})=0}[dy(t)]\,\text{e}^{(\text{i}/\hslash )S[y(t)]}& =& \lim _{\varepsilon \rightarrow \,0} \frac{1} {A}\int _{-\infty }^{+\infty }\frac{dy_{1}} {A} \,\ldots \int _{-\infty }^{+\infty }\frac{dy_{N-1}} {A} \\ & & \times \text{exp}\left [ \frac{\text{i}} {\hslash }\sum _{n\,=\,0}^{N=1}S_{ \text{cl}}(n + 1,n)\right ]\;.{}\end{array}$$

(20.8)

Thus, our formula for calculating the propagator reads

$$\displaystyle\begin{array}{rcl} K{\bigl (x_{2},t_{2};x_{1},t_{1}\bigr )}& =& \,\text{e}^{(\text{i}/\hslash )S_{\text{cl}}(2\vert 1) }\lim _{\varepsilon \rightarrow \,0} \frac{1} {A}\int _{-\infty }^{+\infty }\frac{dy_{1}} {A} \,\ldots \int _{-\infty }^{+\infty }\frac{dy_{N-1}} {A} \\ & & \times \text{exp}\left [ \frac{\text{i}} {\hslash }\sum _{n\,=\,0}^{N-1}S_{ \text{cl}}(n + 1,n)\right ]\;. {}\end{array}$$

(20.9)

We want to now prove that this instruction for the calculation of a propagator indeed leads to this goal, at least for a free particle. The Lagrangian in the (y, τ)-plane is in this case

$$\displaystyle{ L = \frac{m} {2} \dot{y}^{2}(\tau )\;. }$$

(20.10)

The classical action between $n:= (y_{n},\tau _{n})$ and $n + 1:= (y_{n+1},\tau _{n+1})$ is, accordingly,

$$\displaystyle{ S_{\text{cl}}(n + 1,n) = \frac{m} {2} \,\frac{(y_{n+1} - y_{n})^{2}} {\varepsilon } \;. }$$

(20.11)

Then it follows that

$$\displaystyle\begin{array}{rcl} & & \int _{0}^{0}[dy(t)]\,\text{exp}\left [ \frac{\text{i}} {\hslash }\int _{t_{1}}^{t_{2} }dt\frac{m} {2} \dot{y}^{2}(t)\right ] \\ & & \quad =\lim _{\varepsilon \rightarrow 0} \frac{1} {A}\int _{-\infty }^{+\infty }\,\ldots \,\int _{ -\infty }^{+\infty }\frac{dy_{1}} {A} \,\ldots \,\frac{dy_{N-1}} {A} \\ & & \phantom{=} \quad \, \times \text{exp}\left [ \frac{\text{i}} {\hslash }\sum _{n\,=\,0}^{N=1}\frac{m} {2} \,\frac{(y_{n+1} - y_{n})^{2}} {\varepsilon } \right ]\;.{}\end{array}$$

(20.12)

To calculate the first integration, dy ₁, we combine all terms which contain y ₁:

$$\displaystyle\begin{array}{rcl} & & \int _{-\infty }^{+\infty } \frac{1} {A^{2}}dy_{1}\,\text{exp}{\biggl \{ \frac{\text{i}} {\hslash }\,\frac{m} {2\varepsilon } \mathop{\underbrace{{\Bigl [(y_{2} - y_{1})^{2} + (y_{1} -\overbrace{ y_{0}}^{=\,0})^{2}\Bigr ]}}}\limits _{2(y_{1}-y_{2}/2)^{2}+y_{2}^{2}/2}\biggr \}} \\ & & \quad = \frac{1} {A^{2}}\,\text{exp}\left [ \frac{\text{i}} {\hslash }\,\frac{m} {2\varepsilon } \,\frac{y_{2}^{2}} {2} \right ]\mathop{\underbrace{\int _{-\infty }^{+\infty }dy_{1}\,\text{exp}\left [ \frac{\text{i}}{\hslash } \, \frac{m} {\varepsilon } \left (y_{1} -\frac{1} {2}y_{2}\right )^{\!2}\right ]}}\limits _{ =\sqrt{\text{i} \pi \hslash \varepsilon /m}}\;,\quad A^{2} = \frac{2\pi \text{i}\hslash \varepsilon } {m} \\ & & \quad = \sqrt{ \frac{m} {2\pi \text{i}\hslash (2\varepsilon )}}\,\text{exp}\left [ \frac{\text{i}} {2\hslash }\,\frac{m} {2\varepsilon } y_{2}^{2}\right ]\;. {}\end{array}$$

(20.13)

One can show by induction that the n-th step of integration yields the following result:

$$\displaystyle{ \sqrt{ \frac{m} {2\pi \text{i}\hslash (n + 1)\varepsilon }}\,\text{exp}\left [ \frac{\text{i}} {2\hslash }\, \frac{m} {(n + 1)\varepsilon }y_{n+1}^{2}\right ]\;. }$$

(20.14)

After (N − 1) integrations one obtains

$$\displaystyle{ \lim _{\varepsilon \rightarrow 0}\sqrt{ \frac{m} {2\pi \text{i}\hslash N\varepsilon }}\,\text{exp}\left \{ \frac{\text{i}} {2\hslash }\,\frac{m} {N\varepsilon }{\bigl (\mathop{\underbrace{y_{N}}}\limits _{=\,0} -\mathop{\underbrace{ y_{0}}}\limits _{=\,0}\bigr )}^{2}\right \}\mathop{ =}\limits _{ N\varepsilon \,=T}\sqrt{ \frac{m} {2\pi \text{i}\hslash T}}\;. }$$

So we again find for the propagator of a free particle

$$\displaystyle{ K{\bigl (x_{2},t_{2};x_{1},t_{1}\bigr )} = \sqrt{ \frac{m} {2\pi \text{i}\hslash T}}\,\text{exp}\left [ \frac{\text{i}} {\hslash }\,\frac{m} {2} \frac{(x_{2} - x_{1})^{2}} {T} \right ]\;. }$$

(20.15)

For the linear harmonic oscillator we write in the same fashion

$$\displaystyle\begin{array}{rcl} K{\bigl (x_{2},t_{2};x_{1},t_{1}\bigr )}& =& \,\text{e}^{(\text{i}/\hslash )S_{\text{cl}}}\int _{ 0}^{0}[dy(t)] \\ & & \times \text{exp}\left [ \frac{\text{i}} {\hslash }\int _{t_{1}}^{t_{2} }dt\frac{m} {2} {\bigl (\dot{y}^{2} -\omega ^{2}y^{2}\bigr )}\right ]\;.{}\end{array}$$

(20.16)

The value of the path integral before the first caustic is repeated here:

$$\displaystyle{ \int _{0}^{0}[dy(t)]\,\text{exp}\left [ \frac{\text{i}} {\hslash }\int _{t_{1}}^{t_{2} }dt\frac{m} {2} {\bigl (\dot{y}^{2} -\omega ^{2}y^{2}\bigr )}\right ] = \sqrt{ \frac{m\omega } {2\pi \text{i}\hslash \,\sin [\omega (t_{2} - t_{1})]}}\;. }$$

(20.17)

At this point we want to establish contact with the conventional approach to quantum mechanics, which is based on the Schrödinger equation.

We found a dynamical equation (integral equation) in (17.10) that describes the time development of the Schrödinger amplitude. As promised, we now wish to show that this equation is equivalent to the Schrödinger equation. To that end, we again consider the one-dimensional motion of a particle in a potential V (x). The Lagrangian is then

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

(20.18)

with the propagator

$$\displaystyle{ K{\bigl (x,t_{2};y,t_{1}\bigr )} =\int _{ x(t_{1})\,=\,y}^{x(t_{2})\,=\,x}[dx(t)]\,\text{exp}\left \{ \frac{\text{i}} {\hslash }\int _{t_{1}}^{t_{2} }dt\left [\frac{m} {2} \dot{x}^{2} - V (x)\right ]\right \}\;. }$$

(20.19)

Writing t ₁ = t and $t_{2} = t+\varepsilon,$ our integral equation reads

$$\displaystyle{ \psi (x,t+\varepsilon ) =\int _{ -\infty }^{+\infty }dy\,K(x,t+\varepsilon;y,t)\psi (y,t)\;. }$$

(20.20)

Since $\varepsilon$ is assumed very small, we can use the following approximation for K in (20.20):

$$\displaystyle{ K(x,t+\varepsilon;y,t) \simeq \frac{1} {A}\text{exp}\left \{ \frac{\text{i}} {\hslash }\left [\frac{m} {2} \,\frac{(x - y)^{2}} {\varepsilon ^{2}} - V (y)\right ]\varepsilon \right \}\;. }$$

(20.21)

The constant A is still to be determined. Using (20.21) in (20.20), we obtain

$$\displaystyle{ \psi (x,t+\varepsilon ) \simeq \frac{1} {A}\int _{-\infty }^{+\infty }dy\,\text{exp}\left \{ \frac{\text{i}} {\hslash }\left [\frac{m} {2} \frac{(x - y)^{2}} {\varepsilon } - V (y)\varepsilon \right ]\right \}\psi (y,t) }$$

$$\displaystyle{ y = x+\xi;\,= \frac{1} {A}\int _{-\infty }^{+\infty }d\xi \,\text{exp}\left \{ \frac{\text{i}} {\hslash }\left [\frac{m} {2} \,\frac{\xi ^{2}} {\varepsilon } -\varepsilon V (x+\xi )\right ]\right \}\psi (x+\xi,t)\;. }$$

(20.22)

Because of the smallness of $\varepsilon,\ \xi ^{2}/\varepsilon$ is the dominant term in the exponential. For large $\xi,$ the integrand oscillates very rapidly and makes no contributions. The main contribution to the integral comes from values $\xi \sim \sqrt{\varepsilon },$ or, better: $-\sqrt{\varepsilon \hslash /m}\leqslant \xi \leqslant \sqrt{\varepsilon \hslash /m}.$ In the case of expansion of the integral up to linear terms in $\varepsilon,$ the integrand must be expanded up to quadratic terms in $\xi.$ Thus, with

$$\displaystyle{ \psi (x+\xi,t) \simeq \psi (x,t) +\xi \frac{\partial \psi (x,t)} {\partial x} + \frac{1} {2}\xi ^{2}\frac{\partial ^{2}\psi (x,t)} {\partial x^{2}} }$$

we get in (20.22)

$$\displaystyle\begin{array}{rcl} \psi (x,t+\varepsilon )& \simeq & \frac{1} {A}\int _{-\infty }^{+\infty }d\xi \,\text{exp}\left \{ \frac{\text{i}} {\hslash }\,\frac{m} {2} \,\frac{\xi ^{2}} {\varepsilon } \left (1 - \frac{\text{i}} {\hslash }\varepsilon V (x)+\,\ldots \right )\right \} {}\\ & & \times \left [\psi (x,t) +\mathop{ \xi \frac{\partial \psi } {\partial x}}\limits _{\text{odd}\,\,\text{in}\,\,\xi } + \frac{1} {2}\xi ^{2} \frac{\partial ^{2}} {\partial x^{2}}\psi (x,t)+\,\ldots \right ]\;. {}\\ \end{array}$$

Using the integrals

$$\displaystyle\begin{array}{rcl} & & \int _{-\infty }^{+\infty }\text{exp}\left [ \frac{\text{i}} {\hslash }\,\frac{m} {2} \,\frac{\xi ^{2}} {\varepsilon } \right ]d\xi = \left (\frac{2\pi \text{i}\hslash \varepsilon } {m} \right )^{\!1/2} =: A^{{\prime}}\;, {}\\ & & \int _{-\infty }^{+\infty }\text{exp}\left [ \frac{\text{i}} {\hslash }\,\frac{m} {2} \,\frac{\xi ^{2}} {\varepsilon } \right ]\xi d\xi = 0\;, {}\\ & & \int _{-\infty }^{+\infty }\text{exp}\left [ \frac{\text{i}} {\hslash }\,\frac{m} {2} \,\frac{\xi ^{2}} {2}\right ]\xi ^{2}d\xi = \sqrt{2\pi }\left (\frac{\text{i}\hslash \varepsilon } {m}\right )^{\!3/2} {}\\ \end{array}$$

it then follows that

$$\displaystyle\begin{array}{rcl} \psi (x,t) +\varepsilon \frac{\partial \psi (x,t)} {\partial t} +\,\ldots & =& \,\psi (x,t+\varepsilon )\mathop{\cong} \frac{1} {A}A^{{\prime}}\psi (x,t) {}\\ & & +\varepsilon \left [A^{{\prime}}\frac{1} {A}\,\frac{\text{i}\hslash } {m}\,\frac{1} {2}\,\frac{\partial ^{2}\psi (x,t)} {\partial x^{2}} - A^{{\prime}}\frac{1} {A}\, \frac{\text{i}} {\hslash }V (x)\psi (x,t)\right ]\;. {}\\ \end{array}$$

If we take the limit $\varepsilon \rightarrow 0,$ then we can identify

$$\displaystyle{ A = A^{{\prime}} = \left (\frac{2\pi \text{i}\hslash \varepsilon } {m} \right )^{\!1/2}\;. }$$

A comparison of the linear terms in $\varepsilon$ yields, finally,

$$\displaystyle{ \frac{\partial } {\partial t}\psi (x,t) = \frac{\text{i}\hslash } {2m}\,\frac{\partial ^{2}\psi (x,t)} {\partial x^{2}} - \frac{\text{i}} {\hslash }V (x)\psi (x,t) }$$

or

$$\displaystyle{ \text{i}\hslash \frac{\partial \psi (x,t)} {\partial t} = -\frac{\hslash ^{2}} {2m}\, \frac{\partial ^{2}} {\partial x^{2}}\psi (x,t) + V (x)\psi (x,t)\;. }$$

(20.23)

This is the well-known Schrödinger equation of a particle in the potential V (x). 

The infinitesimal propagator obtained in this manner (20.21) can be factorized as follows:

$$\displaystyle\begin{array}{rcl} K(x,t+\varepsilon;y,t)& \simeq & \sqrt{ \frac{m} {2\pi \text{i}\hslash \varepsilon }}\text{exp}\left [ \frac{\text{i}} {\hslash }\,\frac{m} {2} \,\frac{(x - y)^{2}} {\varepsilon } \right ]\,\text{exp}\left [-\frac{\text{i}} {\hslash }V (x)\varepsilon \right ] \\ & =& K_{0}(x,t+\varepsilon;y,t)\phi _{C}^{I}(x,t+\varepsilon;y,t)\;, {}\end{array}$$

(20.24)

where K ₀ is the infinitesimal free particle propagator and ϕ _C ^I is the phase factor that corresponds to the interaction

$$\displaystyle{ \phi _{C}^{I} = \text{exp}\left [-\frac{\text{i}} {\hslash }\int _{C}dt\,V (x)\right ]\;. }$$

(20.25)

The methods just described are now to be applied to the direct calculation of the path integral (with potential):

$$\displaystyle{ K{\bigl (x_{f},t_{f};x_{i},t_{i}\bigr )} =\int _{ x(t_{i})\,=\,x_{i}}^{x(t_{f})\,=\,x_{f} }[dx(t)]\,\text{exp}\left [ \frac{\text{i}} {\hslash }\int _{t_{i}}^{t_{f} }dt\,L(x,\dot{x};t)\right ]\;. }$$

(20.26)

To this end, we again divide the time interval between t _i and t _f in N equal parts, $\varepsilon = (t_{f} - t_{i})/N$:

$$\displaystyle{ t_{0} = t_{i}\,\,(\text{fixed})\;,\,\,\,t_{1} = t_{i}+\varepsilon \;,\ldots \;,\,\,\,t_{N-1} = t_{i} +\varepsilon (N - 1)\;,\,\,\,t_{N} = t_{f}\,\,(\text{fixed})\;. }$$

Now we recall the definition of the path integral:

$$\displaystyle\begin{array}{rcl} K(f\vert i)& =& \int _{x(t_{i})=x_{i}}^{x(t_{f})=x_{f} }[dx(t)]\,\text{exp}\left \{ \frac{\text{i}} {\hslash }S[x(t)]\right \} {}\\ & =& \int _{-\infty }^{+\infty }\,\ldots \,\int _{ -\infty }^{+\infty }dx_{ 1}\,\ldots \,dx_{N-1}K\mathop{\underbrace{{\bigl (x_{N}}}\limits _{=x_{f}},\,\mathop{\underbrace{t_{N}}}\limits _{t_{f}};\,x_{N-1},\,t_{N-1}\bigr )}\,\ldots {}\\ & & \times \,K{\bigl (x_{1},\,t_{1};\,\mathop{\underbrace{x_{0}}}\limits _{=x_{i}},\,\mathop{\underbrace{t_{0}}}\limits _{t_{i}}\,\bigr )}\;. {}\\ \end{array}$$

For $N \rightarrow \infty,$ or $\varepsilon \rightarrow 0,$ we again use infinitesimal propagators:

$$\displaystyle\begin{array}{rcl} K(y,t+\varepsilon;x,t)& \simeq & \left ( \frac{m} {2\pi \text{i}\hslash \varepsilon }\right )^{\!1/2}\,\text{exp}\left [ \frac{\text{i}} {\hslash }\,\frac{m} {2} \,\frac{(x - y)^{2}} {\varepsilon } \right ] \\ & & \times \text{exp}\left [-\frac{\text{i}} {\hslash }\varepsilon V \left (\frac{x + y} {2} \right )\right ]\;. {}\end{array}$$

(20.27)

Then we get for the propagator $K(f\vert i)$:

$$\displaystyle\begin{array}{rcl} & & \int _{x(t_{i})=x_{i}}^{x(t_{f})=x_{f} }[dx(t)]\,\text{exp}\left \{ \frac{\text{i}} {\hslash }\int _{t_{i}}^{t_{f} }dt\left [\frac{m} {2} \dot{x}^{2} - V (x)\right ]\right \} \\ & & \quad =\lim _{N\rightarrow \infty }\left ( \frac{m} {2\pi \text{i}\hslash \varepsilon }\right )^{\!N/2}\int dx_{ 1}\,\ldots \,\int dx_{N-1} \\ & & \quad \phantom{ =}\, \times \text{exp}\left [ \frac{\text{i}} {\hslash }\sum _{k=1}^{N}\frac{m} {2} \,\frac{(x_{k} - x_{k-1})^{2}} {\varepsilon } - V \left (\frac{x_{k} + x_{k-1}} {2} \right )\varepsilon \right ]\;.{}\end{array}$$

(20.28)

For the path integral, one often simply finds the expression

$$\displaystyle\begin{array}{rcl} & & \sim \int dx_{1}\,\ldots \,\int dx_{N-1} \\ & & \quad \times \text{exp}\left [ \frac{\text{i}} {\hslash }\sum _{k=1}^{N}\frac{m} {2} \,\frac{(x_{k} - x_{k-1})^{2}} {\varepsilon } - V \left (\frac{x_{k} + x_{k-1}} {2} \right )\varepsilon \right ]\;.{}\end{array}$$

(20.29)

Here, the integration measure has been “forgotten”. This can sometimes be justified by calculating, instead of a single path integral, the ratio of two path integrals, both of which describe a particle of mass m. Then the measure drops out and we have simply

$$\displaystyle\begin{array}{rcl} & & \int _{x(t_{i})=x_{i}}^{x(t_{f})=x_{f} }[dx(t)]\,\text{exp}\left \{ \frac{\text{i}} {\hslash }S[x(t)]\right \}\bigg/\int _{x(t_{i})=x_{i}}^{x(t_{f})=x_{f} }[dx(t)]\text{exp}\left \{ \frac{\text{i}} {\hslash }S_{0}[x(t)]\right \} \\ & =& \lim _{N\rightarrow \infty } \frac{\int _{-\infty }^{+\infty }\ldots \int _{-\infty }^{+\infty }dx_{1}\,\ldots \,dx_{N-1}\,\text{exp}\left \{ \frac{\text{i}}{\hslash } \sum _{k=1}^{N}\frac{m} {2} \,\frac{(x_{k}-x_{k-1})^{2}} {\varepsilon } - V \left (\frac{x_{k}+x_{k-1}} {2} \right )\varepsilon \right \}} {\int _{-\infty }^{+\infty }\ldots \int _{-\infty }^{+\infty }dx_{1}\ldots dx_{N-1}\,\text{exp}\left \{ \frac{\text{i}}{\hslash } \sum _{k=1}^{N}\frac{m} {2} \,\frac{(x_{k}-x_{k-1})^{2}} {\varepsilon } - V _{0}\left (\frac{x_{k}+x_{k-1}} {2} \right )\varepsilon \right \}}\;.{}\end{array}$$

(20.30)

Here, S ₀ refers to the free propagator, so that we indeed are calculating the ratio of the interacting propagator to the free propagator.

If we want to calculate a path integral, we need not necessarily sum piecewise over straight lines in (x, t)-space. Other complete classes of paths can equally well be used. For example, if we wish to calculate the propagator

$$\displaystyle{ K(0,T;0,0) }$$

we normally begin by dividing an arbitrary path x(t) in segments which connect the intermediate points x ₁ = x(t ₁) up to x _N−1 = x(t _N−1). Now, however, we shall approximate the path x(t) by using a “Fourier path”, i.e., a path of the form

$$\displaystyle{ \tilde{x}(t) =\sum _{ k=1}^{N-1}a_{ k}\,\sin \left ( \frac{\pi k} {T}t\right )\;. }$$

(20.31)

If we then choose the coefficients a _k in such a manner that

$$\displaystyle{ x_{j} =\sum _{ k=1}^{N-1}a_{ k}\,\sin \left ( \frac{\pi k} {N}j\right ) }$$

(20.32)

then the Fourier paths obviously go through the same intermediary points of (x, t)-space.

The approximated path is then once again completely characterized by the vertex coordinates $(x_{1},\,\ldots,\,x_{N-1}),$ and we again can sum up the contributions of the various paths by integrating over the vertex coordinates. In practice, however, it is more convenient to integrate over the Fourier components $(a_{1},\,\ldots,\,a_{N-1}).$ Since the relation (20.32) is one-to-one between the $(x_{1},\,\ldots,\,x_{N-1})$ and the $(a_{1},\,\ldots,\,a_{N-1}),$ it is immediately evident that

$$\displaystyle{ \int \ldots \int \{./.\}dx_{1}\,\ldots \,dx_{N-1} =\int \ldots \int \{./.\}\Big\vert \left (\frac{\partial x_{i}} {\partial a_{j}}\right )\Big\vert da_{1}\,\ldots \,da_{N-1}\;. }$$

Since the transformation (20.32) is linear, the Jacobi determinant is independent of $(a_{1},\,\ldots,\,a_{N-1})$! Its value is thus unimportant (drops out) when calculating the ratio again:

$$\displaystyle\begin{array}{rcl} & & \int _{x(0)=0}^{x(T)=0}[dx(t)]\,\text{exp}\left \{ \frac{\text{i}} {\hslash }S[x(t)]\right \}\bigg/\int _{x(0)=0}^{x(T)=0}[dx(t)]\,\text{exp}\left \{ \frac{\text{i}} {\hslash }S_{0}[x(t)]\right \} \\ & & =\lim _{N\rightarrow \infty } \frac{\int da_{1}\ldots da_{N-1}\,\text{exp}\left \{ \frac{\text{i}}{\hslash } S\left [\sum _{k=1}^{N-1}a_{k}\,\sin \left ( \frac{\pi k} {T}t\right )\right ]\right \}} {\int da_{1}\ldots da_{N-1}\,\text{exp}\left \{ \frac{\text{i}}{\hslash } S_{0}\left [\sum _{k=1}^{N-1}a_{k}\,\sin \left ( \frac{\pi k} {T}t\right )\right ]\right \}}\;. {}\end{array}$$

(20.33)

In order to illustrate how this procedure works, we again return to the example of the harmonic oscillator, which we all treasure:

$$\displaystyle\begin{array}{rcl} K{\bigl (x_{2},T;x_{1},0\bigr )}& =& \,\text{exp}\left \{ \frac{\text{i}} {\hslash }S{\bigl [x_{\text{cl}}\bigr ]}\right \}\int _{x(0)\,=\,0}^{x(T)\,=\,0}[dx(t)] \\ & & \times \,\text{exp}\left \{ \frac{\text{i}} {\hslash }\int _{0}^{T}dt\left [\frac{m} {2} \dot{x}^{2} -\frac{m} {2} \omega ^{2}x^{2}\right ]\right \}\;.{}\end{array}$$

(20.34)

[The path integral is the same as in (20.17), where we used y(t) instead of x(t). ] According to (20.33), we can rewrite (20.34) as

$$\displaystyle\begin{array}{rcl} K{\bigl (x_{2},T;x_{1},0\bigr )}& =& \,\text{exp}\left \{ \frac{\text{i}} {\hslash }S{\bigl [x_{\text{cl}}\bigr ]}\right \}K_{0}(0,T;0,0) \\ & & \times \,\lim _{N\rightarrow \infty } \frac{\int \ldots \int da_{1}\ldots da_{N-1}\,\text{exp}\left \{ \frac{\text{i}}{\hslash } S[./.]\right \}} {\int \ldots \int da_{1}\ldots da_{N-1}\,\text{exp}\left \{ \frac{\text{i}}{\hslash } S_{0}[./.]\right \}}\;.{}\end{array}$$

(20.35)

The free propagator (ω = 0) is known to be given by

$$\displaystyle{ K_{0}(0,T;0,0) = \left ( \frac{m} {2\pi \text{i}\hslash T}\right )^{\!1/2}\;. }$$

So now we have to calculate the remaining multiple integrals over the a _i . First we want to show that

$$\displaystyle{ S\left [\sum _{k=1}^{N-1}a_{ k}\,\sin \left ( \frac{\pi k} {T}t\right )\right ] = \frac{mT} {4} \sum _{k=1}^{N-1}a_{ k}^{2}\left (\frac{k^{2}\pi ^{2}} {T^{2}} -\omega ^{2}\right )\;. }$$

(20.36)

FormalPara Proof

$$\displaystyle\begin{array}{rcl} & & x(t)\mathop{=}\limits _{(31)}\sum _{k=1}^{N-1}a_{ k}\,\sin \left ( \frac{\pi k} {T}t\right ) =\sum _{ k=1}^{N-1}a_{ k}\,\sin {\bigl (\omega _{k}t\bigr )}\;;\quad \omega _{k} = \frac{k\pi } {T} {}\\ & & \dot{x} =\sum _{ k=1}^{N-1}a_{ k}\omega _{k}\,\cos {\bigl (\omega _{k}t\bigr )}\;,\quad \dot{x}^{2} =\sum _{ k=1}^{N-1}\sum _{ j=1}^{N-1}a_{ k}a_{j}\omega _{k}\omega _{j}\,\cos {\bigl (\omega _{k}t\bigr )}\,\mathrm{cos}{\bigl (\omega _{j}t\bigr )}\;. {}\\ \end{array}$$

We need

$$\displaystyle{ \int _{0}^{T}dt\,\dot{x}^{2} =\sum _{ k,j}a_{k}a_{j}\omega _{k}\omega _{j}\int _{0}^{T}dt\,\cos \left ( \frac{k\pi } {T}t\right )\cos \left ( \frac{j\pi } {T}t\right ) = \frac{T} {2} \sum _{k=1}^{N-1}a_{ k}^{2}\omega _{ k}^{2}\;. }$$

Similarly, it follows that

$$\displaystyle{ \int _{0}^{T}dt\,x^{2} = \frac{T} {2} \sum _{k=1}^{N-1}a_{ k}^{2}\;. }$$

Therefore, we find indeed

$$\displaystyle{ \text{exp}\left [ \frac{\text{i}} {\hslash }\int _{0}^{T}dt\frac{m} {2} {\bigl (\dot{x}^{2} -\omega ^{2}x^{2}\bigr )}\right ] = \text{exp}\left [ \frac{\text{i}} {\hslash }\,\frac{mT} {4} \sum _{k=1}^{N-1}\left (\frac{k^{2}\pi ^{2}} {T^{2}} -\omega ^{2}\right )a_{ k}^{2}\right ]\;. }$$

Now the integration over the a _i is simple to perform, since the exponential is not only quadratic but also diagonal in $(a_{1},\,\ldots,\,a_{N-1}).$

$$\displaystyle\begin{array}{rcl} & & \int \ldots \int da_{1}\,\ldots \,da_{N-1}\,\text{exp}\left \{ \frac{\text{i}} {\hslash }S\left [\sum _{k=1}^{N-1}a_{ k}\sin \left ( \frac{k\pi } {T}t\right )\right ]\right \} \\ & & =\int \ldots \int da_{1}\,\ldots \,da_{N-1}\,\text{exp}\left [ \frac{\text{i}} {\hslash }\,\frac{mT} {4} \sum _{k=1}^{N-1}\left (\frac{k^{2}\pi ^{2}} {T^{2}} -\omega ^{2}\right )a_{ k}^{2}\right ]\;.{}\end{array}$$

(20.37)

Using

$$\displaystyle\begin{array}{rcl} & & \int _{-\infty }^{+\infty }\text{e}^{\text{i}\alpha x^{2} }\,dx = \sqrt{\frac{\pi \text{i} } {\alpha }} \quad: {}\\ & & \int _{-\infty }^{+\infty }da_{ k}\,\text{exp}\left [ \frac{\text{i}} {\hslash }\,\frac{mT} {4} \left (\omega _{k}^{2} -\omega ^{2}\right )a_{ k}^{2}\right ] = \sqrt{ \frac{4\pi \text{i} \hslash } {mT(\omega _{k}^{2} -\omega ^{2})}} {}\\ \end{array}$$

we obtain in (20.37)

$$\displaystyle{ \left (\sqrt{ \frac{4\pi \text{i} \hslash } {mT}}\right )^{\!N-1}\prod _{ k=1}^{N-1}\left (\frac{k^{2}\pi ^{2}} {T^{2}} -\omega ^{2}\right )^{\!-1/2}\;. }$$

The free particle propagator is obtained simply by setting ω = 0. The forefactor $(\sqrt{4\pi \text{i} \hslash /mT})^{N-1}$ drops out when constructing the ratio in (20.35).

With

$$\displaystyle{ \frac{\prod _{k=1}^{N-1}(k^{2}\pi ^{2}/T^{2} -\omega ^{2})^{-1/2}} {\prod _{k=1}^{N-1}(k^{2}\pi ^{2}/T^{2})^{-1/2}} =\prod _{ k=1}^{N-1}\left (1 -\frac{\omega ^{2}T^{2}} {k^{2}\pi ^{2}} \right )^{\!-1/2} }$$

we find

$$\displaystyle\begin{array}{rcl} K{\bigl (x_{2},T;x_{1},0\bigr )}& =& \,\text{exp}\left \{ \frac{\text{i}} {\hslash }S{\bigl [x_{\text{cl}}\bigr ]}\right \} \\ & & \,\times \sqrt{ \frac{m} {2\pi \text{i}\hslash T}}\lim _{N\rightarrow \infty }\prod _{k=1}^{N-1}\left (1 -\frac{\omega ^{2}T^{2}} {k^{2}\pi ^{2}} \right )^{\!-1/2}\;.{}\end{array}$$

(20.38)

Euler’s famous product formula for the sine function

$$\displaystyle{ \left (\frac{z} {\sin \,z}\right )^{\!1/2} =\prod _{ k=1}^{\infty }\left (1 - \frac{z^{2}} {k^{2}\pi ^{2}}\right )^{\!-1/2}\;,\quad z \equiv \omega T }$$

therefore yields for $N \rightarrow \infty$ or $\varepsilon \rightarrow 0$ in (20.38):

$$\displaystyle{ K{\bigl (x_{2},T;x_{1},0\bigr )} = \sqrt{ \frac{m\omega } {2\pi \text{i}\hslash \,\sin (\omega T)}}\,\text{exp}\left \{ \frac{\text{i}} {\hslash }S_{\text{cl}}^{\text{H.O.}}[x]\right \}\;. }$$

We already know from our earlier considerations that the phase dependence is not yet correctly described. Let us recall, however,

$$\displaystyle\begin{array}{rcl} \int _{-\infty }^{+\infty }dx\,\text{e}^{\text{i}\lambda x^{2} } = \sqrt{\frac{\pi } {\vert \lambda \vert }}\,\text{e}^{(\text{i}\pi /4)\text{sign}\,\lambda } = \left \{\begin{array}{*{10}c} \sqrt{\frac{\pi } {\vert \lambda \vert }}& \text{e}^{\text{i}\pi /4}\;, &\lambda > 0\;, \\ \sqrt{\frac{\pi }{\vert \lambda \vert }}&\text{e}^{-\text{i}\pi /4}\;,&\lambda < 0\;, \end{array} \right.& & {}\\ \end{array}$$

and return to the action expressed in the Fourier path,

$$\displaystyle{ \text{exp}\left [ \frac{\text{i}} {\hslash }S\right ] = \text{exp}\left [ \frac{\text{i}} {\hslash }\,\frac{mT} {4} \sum _{k\,=\,1}^{N-1}a_{ k}^{2}\left (\frac{\pi ^{2}k^{2}} {T^{2}} -\omega ^{2}\right )\right ]\;. }$$

Then we see that the analogue of $\lambda$ is negative if k < ω T∕π, and positive if k > ω T∕π. Once again, the additional phase factor exp{ − (iπ∕2)[ω T∕π]} has to be included, so that our multiple integral is given more precisely by

$$\displaystyle\begin{array}{rcl} & & \int \ldots \int da_{1}\,\ldots \,da_{N-1}\,\text{exp}\left [ \frac{\text{i}} {\hslash }\,\frac{mT} {4} \sum _{k\,=\,1}^{N-1}a_{ k}^{2}\left (\frac{\pi ^{2}k^{2}} {T^{2}} -\omega ^{2}\right )\right ] \\ & & \quad = \text{exp}\left \{-\text{i} \frac{\pi } {2}\left [\frac{\omega T} {\pi } \right ]\right \}\left (\text{e}^{\text{i}\pi /4}\sqrt{ \frac{4\pi \hslash } {mT}}\right )^{\!N-1}\prod _{ k=1}^{N-1}\bigg\vert \frac{\pi ^{2}k^{2}} {T^{2}} -\omega ^{2}\bigg\vert ^{-1/2}\;.{}\end{array}$$

(20.39)

(For the free propagator this ambiguity does not appear.) The correct formula for the propagator of a particle in the potential of the harmonic oscillator well is therefore

$$\displaystyle\begin{array}{rcl} K{\bigl (x_{2},T;x_{1},0\bigr )}& =& \,\text{exp}\left [-\text{i}\left ( \frac{\pi } {4} + \frac{\pi } {2}\left [\frac{\omega T} {\pi } \right ]\right )\right ]\sqrt{ \frac{m\omega } {2\pi \hslash \vert \mathrm{sin}(\omega T)\vert }} \\ & & \times \,\text{exp}\left \{ \frac{\text{i}} {\hslash }S_{\mathrm{cl}}^{\text{H.O.}}[x]\right \}\;. {}\end{array}$$

(20.40)

This is precisely the Feynman–Soriau formula  (19.71).