The one-dimensional free-particle propagator is
K 0 ( x f , t f ; x i , t i ) = [ m 2 π i ℏ ( t f − t i ) ] 1 / 2 exp [ i m ( x f − x i ) 2 2 ℏ ( t f − t i ) ] , K_0(x_f,t_f;x_i,t_i)
=
\left[
\frac{m}{2\pi i\hbar (t_f-t_i)}
\right]^{1/2}
\exp\left[
\frac{im(x_f-x_i)^2}{2\hbar(t_f-t_i)}
\right], K 0 ( x f , t f ; x i , t i ) = [ 2 π i ℏ ( t f − t i ) m ] 1/2 exp [ 2ℏ ( t f − t i ) im ( x f − x i ) 2 ] , for T = t f − t i ≠ 0 T=t_f-t_i\ne0 T = t f − t i = 0 m / ( 2 π ℏ ∣ T ∣ ) e − i π sgn ( T ) / 4 \sqrt{m/(2\pi\hbar|T|)}e^{-i\pi\operatorname{sgn}(T)/4} m / ( 2 π ℏ∣ T ∣ ) e − iπ sgn ( T ) /4 K 0 ( x f , − T ; x i , 0 ) = K 0 ∗ ( x i , T ; x f , 0 ) K_0(x_f,-T;x_i,0)=K_0^*(x_i,T;x_f,0) K 0 ( x f , − T ; x i , 0 ) = K 0 ∗ ( x i , T ; x f , 0 )
U 0 ( t f , t i ) = exp [ − i ℏ p 2 2 m ( t f − t i ) ] . U_0(t_f,t_i)
=
\exp\left[
-\frac{i}{\hbar}\frac{p^2}{2m}(t_f-t_i)
\right]. U 0 ( t f , t i ) = exp [ − ℏ i 2 m p 2 ( t f − t i ) ] . This page derives the kernel, checks its normalization and composition law, and explains why it is the basic short-time building block for path integrals.
Let
T = t f − t i , Δ x = x f − x i . T=t_f-t_i,
\qquad
\Delta x=x_f-x_i. T = t f − t i , Δ x = x f − x i . Insert the momentum resolution of identity into
K 0 ( x f , t f ; x i , t i ) = ⟨ x f ∣ U 0 ( T ) ∣ x i ⟩ . K_0(x_f,t_f;x_i,t_i)
=
\langle x_f|U_0(T)|x_i\rangle. K 0 ( x f , t f ; x i , t i ) = ⟨ x f ∣ U 0 ( T ) ∣ x i ⟩ . With
⟨ x ∣ p ⟩ = 1 2 π ℏ e i p x / ℏ , \langle x|p\rangle
=
\frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar}, ⟨ x ∣ p ⟩ = 2 π ℏ 1 e i p x /ℏ , one obtains
K 0 ( x f , t f ; x i , t i ) = ∫ − ∞ ∞ d p 2 π ℏ exp [ i p Δ x ℏ − i p 2 T 2 m ℏ ] . \begin{aligned}
K_0(x_f,t_f;x_i,t_i)
&=
\int_{-\infty}^{\infty}
\frac{dp}{2\pi\hbar}\,
\exp\left[
\frac{ip\Delta x}{\hbar}
-
\frac{ip^2T}{2m\hbar}
\right].
\end{aligned} K 0 ( x f , t f ; x i , t i ) = ∫ − ∞ ∞ 2 π ℏ d p exp [ ℏ i p Δ x − 2 m ℏ i p 2 T ] . This is a Fourier transform of the free-particle energy phase e − i p 2 T / ( 2 m ℏ ) e^{-ip^2T/(2m\hbar)} e − i p 2 T / ( 2 m ℏ )
Complete the square:
p Δ x ℏ − p 2 T 2 m ℏ = − T 2 m ℏ ( p − m Δ x T ) 2 + m ( Δ x ) 2 2 ℏ T . \frac{p\Delta x}{\hbar}
-
\frac{p^2T}{2m\hbar}
=
-
\frac{T}{2m\hbar}
\left(
p-\frac{m\Delta x}{T}
\right)^2
+
\frac{m(\Delta x)^2}{2\hbar T}. ℏ p Δ x − 2 m ℏ p 2 T = − 2 m ℏ T ( p − T m Δ x ) 2 + 2ℏ T m ( Δ x ) 2 . Including the factor of i i i
K 0 = e i m ( Δ x ) 2 / ( 2 ℏ T ) ∫ − ∞ ∞ d p 2 π ℏ exp [ − i T 2 m ℏ ( p − m Δ x T ) 2 ] . K_0
=
e^{im(\Delta x)^2/(2\hbar T)}
\int_{-\infty}^{\infty}
\frac{dp}{2\pi\hbar}\,
\exp\left[
-\frac{iT}{2m\hbar}
\left(
p-\frac{m\Delta x}{T}
\right)^2
\right]. K 0 = e im ( Δ x ) 2 / ( 2ℏ T ) ∫ − ∞ ∞ 2 π ℏ d p exp [ − 2 m ℏ i T ( p − T m Δ x ) 2 ] . The remaining Fresnel integral is defined by a convergence prescription, for example T → T − i 0 + T\to T-i0^+ T → T − i 0 + T > 0 T\gt0 T > 0
K 0 = ( m 2 π i ℏ T ) 1 / 2 exp [ i m ( Δ x ) 2 2 ℏ T ] . K_0
=
\left(
\frac{m}{2\pi i\hbar T}
\right)^{1/2}
\exp\left[
\frac{im(\Delta x)^2}{2\hbar T}
\right]. K 0 = ( 2 π i ℏ T m ) 1/2 exp [ 2ℏ T im ( Δ x ) 2 ] . The prefactor enforces the delta-function initial condition and unitarity. The small negative imaginary time is a convergence regulator; by itself it does not impose relativistic causal support.
At equal times, the propagator must become the identity kernel:
lim T → 0 K 0 ( x f , t i + T ; x i , t i ) = δ ( x f − x i ) \lim_{T\to0}
K_0(x_f,t_i+T;x_i,t_i)
=
\delta(x_f-x_i) T → 0 lim K 0 ( x f , t i + T ; x i , t i ) = δ ( x f − x i ) in the distributional sense. This does not mean the expression has an ordinary pointwise limit. Instead, for a sufficiently nice test wavefunction,
lim T → 0 ∫ − ∞ ∞ d x i K 0 ( x f , t i + T ; x i , t i ) ψ ( x i ) = ψ ( x f ) . \lim_{T\to0}
\int_{-\infty}^{\infty}dx_i\,
K_0(x_f,t_i+T;x_i,t_i)\psi(x_i)
=
\psi(x_f). T → 0 lim ∫ − ∞ ∞ d x i K 0 ( x f , t i + T ; x i , t i ) ψ ( x i ) = ψ ( x f ) . The rapidly oscillating phase away from x i = x f x_i=x_f x i = x f
The kernel satisfies the free Schrödinger equation in the final coordinate:
i ℏ ∂ K 0 ∂ t f = − ℏ 2 2 m ∂ 2 K 0 ∂ x f 2 . i\hbar\frac{\partial K_0}{\partial t_f}
=
-\frac{\hbar^2}{2m}
\frac{\partial^2K_0}{\partial x_f^2}. i ℏ ∂ t f ∂ K 0 = − 2 m ℏ 2 ∂ x f 2 ∂ 2 K 0 . Together with the delta-function initial condition, this characterizes the free-particle kernel. The same expression also satisfies the adjoint equation in the initial variables, as required by unitarity.
For t i < t < t f t_i\lt t\lt t_f t i < t < t f
K 0 ( x f , t f ; x i , t i ) = ∫ − ∞ ∞ d x K 0 ( x f , t f ; x , t ) K 0 ( x , t ; x i , t i ) . K_0(x_f,t_f;x_i,t_i)
=
\int_{-\infty}^{\infty}dx\,
K_0(x_f,t_f;x,t)
K_0(x,t;x_i,t_i). K 0 ( x f , t f ; x i , t i ) = ∫ − ∞ ∞ d x K 0 ( x f , t f ; x , t ) K 0 ( x , t ; x i , t i ) . The cleanest verification uses the momentum representation. Each short kernel is an integral over a momentum phase; the intermediate position integral gives a delta distribution equating the two momenta. The result is the single momentum integral for the total time t f − t i t_f-t_i t f − t i
This composition law is the starting point for the time-sliced construction in From Propagators to Path Integrals .
The classical path from ( x i , t i ) (x_i,t_i) ( x i , t i ) ( x f , t f ) (x_f,t_f) ( x f , t f )
v c l = x f − x i t f − t i . v_{\rm cl}
=
\frac{x_f-x_i}{t_f-t_i}. v cl = t f − t i x f − x i . The classical action is
S c l = ∫ t i t f 1 2 m v c l 2 d t = m ( x f − x i ) 2 2 ( t f − t i ) . S_{\rm cl}
=
\int_{t_i}^{t_f}
\frac12m v_{\rm cl}^2\,dt
=
\frac{m(x_f-x_i)^2}{2(t_f-t_i)}. S cl = ∫ t i t f 2 1 m v cl 2 d t = 2 ( t f − t i ) m ( x f − x i ) 2 . Therefore
K 0 ( x f , t f ; x i , t i ) = [ 1 2 π i ℏ ( − ∂ 2 S c l ∂ x f ∂ x i ) ] 1 / 2 e i S c l / ℏ . K_0(x_f,t_f;x_i,t_i)
=
\left[
\frac{1}{2\pi i\hbar}
\left(
-\frac{\partial^2 S_{\rm cl}}
{\partial x_f\,\partial x_i}
\right)
\right]^{1/2}
e^{iS_{\rm cl}/\hbar}. K 0 ( x f , t f ; x i , t i ) = [ 2 π i ℏ 1 ( − ∂ x f ∂ x i ∂ 2 S cl ) ] 1/2 e i S cl /ℏ . For the free particle this semiclassical-looking expression is exact because the action is quadratic.
The propagator evolves an initial wavefunction by
ψ ( x f , t f ) = ∫ − ∞ ∞ d x i K 0 ( x f , t f ; x i , t i ) ψ ( x i , t i ) . \psi(x_f,t_f)
=
\int_{-\infty}^{\infty}dx_i\,
K_0(x_f,t_f;x_i,t_i)\psi(x_i,t_i). ψ ( x f , t f ) = ∫ − ∞ ∞ d x i K 0 ( x f , t f ; x i , t i ) ψ ( x i , t i ) . For an initial Gaussian packet
ψ ( x , 0 ) = 1 ( 2 π σ 0 2 ) 1 / 4 exp [ − ( x − x 0 ) 2 4 σ 0 2 + i p 0 ( x − x 0 ) ℏ ] , \psi(x,0)
=
\frac{1}{(2\pi\sigma_0^2)^{1/4}}
\exp\left[
-\frac{(x-x_0)^2}{4\sigma_0^2}
+
\frac{ip_0(x-x_0)}{\hbar}
\right], ψ ( x , 0 ) = ( 2 π σ 0 2 ) 1/4 1 exp [ − 4 σ 0 2 ( x − x 0 ) 2 + ℏ i p 0 ( x − x 0 ) ] , the probability density remains Gaussian. Its center follows the classical trajectory
x c ( t ) = x 0 + p 0 m t , x_c(t)=x_0+\frac{p_0}{m}t, x c ( t ) = x 0 + m p 0 t , while its width becomes
σ t = σ 0 [ 1 + ( ℏ t 2 m σ 0 2 ) 2 ] 1 / 2 . \sigma_t
=
\sigma_0
\left[
1+
\left(
\frac{\hbar t}{2m\sigma_0^2}
\right)^2
\right]^{1/2}. σ t = σ 0 [ 1 + ( 2 m σ 0 2 ℏ t ) 2 ] 1/2 . The packet spreads because the free-particle dispersion relation ω = ℏ k 2 / ( 2 m ) \omega=\hbar k^2/(2m) ω = ℏ k 2 / ( 2 m ) Gaussian Wave Packets and Wave-Packet Spreading .
For a short time step ϵ \epsilon ϵ
K 0 ( x j + 1 , t j + ϵ ; x j , t j ) = ( m 2 π i ℏ ϵ ) 1 / 2 exp [ i m ( x j + 1 − x j ) 2 2 ℏ ϵ ] . K_0(x_{j+1},t_j+\epsilon;x_j,t_j)
=
\left(
\frac{m}{2\pi i\hbar\epsilon}
\right)^{1/2}
\exp\left[
\frac{im(x_{j+1}-x_j)^2}{2\hbar\epsilon}
\right]. K 0 ( x j + 1 , t j + ϵ ; x j , t j ) = ( 2 π i ℏ ϵ m ) 1/2 exp [ 2ℏ ϵ im ( x j + 1 − x j ) 2 ] . When a potential is present, a short-time kinetic-potential splitting gives the approximate kernel
K ( x j + 1 , t j + ϵ ; x j , t j ) ≈ K 0 ( x j + 1 , t j + ϵ ; x j , t j ) e − i ϵ V ( x j ) / ℏ . K(x_{j+1},t_j+\epsilon;x_j,t_j)
\approx
K_0(x_{j+1},t_j+\epsilon;x_j,t_j)
e^{-i\epsilon V(x_j)/\hbar}. K ( x j + 1 , t j + ϵ ; x j , t j ) ≈ K 0 ( x j + 1 , t j + ϵ ; x j , t j ) e − i ϵ V ( x j ) /ℏ . Multiplying many such short-time kernels and integrating over intermediate positions is the formal route to the real-time path integral.
Treating K 0 K_0 K 0 x f x_f x f x i x_i x i
Dropping the phase of the square-root prefactor.
Forgetting that the equal-time limit is a distribution, not an ordinary function.
Reusing the free kernel in a box, on a half-line, or on a ring without enforcing the correct boundary conditions.
Confusing the free-particle propagator with a relativistic or QFT propagator.
Ignoring Fourier-transform conventions when deriving the prefactor.
R. P. Feynman and A. R. Hibbs, Quantum Mechanics and Path Integrals , McGraw-Hill, 1965.
H. Kleinert, Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets , 5th ed., World Scientific, 2009.
J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics , 3rd ed., Cambridge University Press, 2020.
L. S. Schulman, Techniques and Applications of Path Integration , Wiley, 1981.
R. Shankar, Principles of Quantum Mechanics , 2nd ed., Springer, 1994.
Starting from the momentum integral for K 0 K_0 K 0
Solution Start with
K 0 = ∫ − ∞ ∞ d p 2 π ℏ exp [ i p Δ x ℏ − i p 2 T 2 m ℏ ] . K_0
=
\int_{-\infty}^{\infty}
\frac{dp}{2\pi\hbar}
\exp\left[
\frac{ip\Delta x}{\hbar}
-
\frac{ip^2T}{2m\hbar}
\right]. K 0 = ∫ − ∞ ∞ 2 π ℏ d p exp [ ℏ i p Δ x − 2 m ℏ i p 2 T ] . Completing the square gives
i p Δ x ℏ − i p 2 T 2 m ℏ = i m ( Δ x ) 2 2 ℏ T − i T 2 m ℏ ( p − m Δ x T ) 2 . \frac{ip\Delta x}{\hbar}
-
\frac{ip^2T}{2m\hbar}
=
\frac{im(\Delta x)^2}{2\hbar T}
-
\frac{iT}{2m\hbar}
\left(
p-\frac{m\Delta x}{T}
\right)^2. ℏ i p Δ x − 2 m ℏ i p 2 T = 2ℏ T im ( Δ x ) 2 − 2 m ℏ i T ( p − T m Δ x ) 2 . The shifted Fresnel integral is
∫ − ∞ ∞ d q 2 π ℏ exp ( − i T q 2 2 m ℏ ) = ( m 2 π i ℏ T ) 1 / 2 , \int_{-\infty}^{\infty}
\frac{dq}{2\pi\hbar}
\exp\left(
-\frac{iTq^2}{2m\hbar}
\right)
=
\left(
\frac{m}{2\pi i\hbar T}
\right)^{1/2}, ∫ − ∞ ∞ 2 π ℏ d q exp ( − 2 m ℏ i T q 2 ) = ( 2 π i ℏ T m ) 1/2 , with the standard convergence prescription. Combining factors gives the result.
Verify the free Schrödinger equation for the kernel by differentiating with respect to T T T Δ x \Delta x Δ x
Solution Write
K 0 = C T − 1 / 2 exp [ i m ( Δ x ) 2 2 ℏ T ] . K_0=C T^{-1/2}
\exp\left[
\frac{im(\Delta x)^2}{2\hbar T}
\right]. K 0 = C T − 1/2 exp [ 2ℏ T im ( Δ x ) 2 ] . Then
∂ K 0 ∂ T = [ − 1 2 T − i m ( Δ x ) 2 2 ℏ T 2 ] K 0 , \frac{\partial K_0}{\partial T}
=
\left[
-\frac{1}{2T}
-
\frac{im(\Delta x)^2}{2\hbar T^2}
\right]K_0, ∂ T ∂ K 0 = [ − 2 T 1 − 2ℏ T 2 im ( Δ x ) 2 ] K 0 , so
i ℏ ∂ K 0 ∂ T = [ − i ℏ 2 T + m ( Δ x ) 2 2 T 2 ] K 0 . i\hbar\frac{\partial K_0}{\partial T}
=
\left[
-\frac{i\hbar}{2T}
+
\frac{m(\Delta x)^2}{2T^2}
\right]K_0. i ℏ ∂ T ∂ K 0 = [ − 2 T i ℏ + 2 T 2 m ( Δ x ) 2 ] K 0 . Also,
∂ 2 K 0 ∂ x f 2 = [ i m ℏ T − m 2 ( Δ x ) 2 ℏ 2 T 2 ] K 0 . \frac{\partial^2K_0}{\partial x_f^2}
=
\left[
\frac{im}{\hbar T}
-
\frac{m^2(\Delta x)^2}{\hbar^2T^2}
\right]K_0. ∂ x f 2 ∂ 2 K 0 = [ ℏ T im − ℏ 2 T 2 m 2 ( Δ x ) 2 ] K 0 . Multiplying by − ℏ 2 / ( 2 m ) -\hbar^2/(2m) − ℏ 2 / ( 2 m ) i ℏ ∂ T K 0 i\hbar\,\partial_TK_0 i ℏ ∂ T K 0
Use the momentum representation to verify the composition law for the free-particle propagator.
Solution Write each kernel as a momentum integral:
K 0 ( x f , t f ; x , t ) = ∫ d p 2 π ℏ e i p ( x f − x ) / ℏ e − i p 2 ( t f − t ) / ( 2 m ℏ ) K_0(x_f,t_f;x,t)
=
\int\frac{dp}{2\pi\hbar}
e^{ip(x_f-x)/\hbar}
e^{-ip^2(t_f-t)/(2m\hbar)} K 0 ( x f , t f ; x , t ) = ∫ 2 π ℏ d p e i p ( x f − x ) /ℏ e − i p 2 ( t f − t ) / ( 2 m ℏ ) and similarly for K 0 ( x , t ; x i , t i ) K_0(x,t;x_i,t_i) K 0 ( x , t ; x i , t i ) x x x
∫ d x e i ( q − p ) x / ℏ = 2 π ℏ δ ( q − p ) . \int dx\,e^{i(q-p)x/\hbar}
=
2\pi\hbar\,\delta(q-p). ∫ d x e i ( q − p ) x /ℏ = 2 π ℏ δ ( q − p ) . The delta distribution sets q = p q=p q = p
∫ d p 2 π ℏ e i p ( x f − x i ) / ℏ e − i p 2 ( t f − t i ) / ( 2 m ℏ ) , \int\frac{dp}{2\pi\hbar}
e^{ip(x_f-x_i)/\hbar}
e^{-ip^2(t_f-t_i)/(2m\hbar)}, ∫ 2 π ℏ d p e i p ( x f − x i ) /ℏ e − i p 2 ( t f − t i ) / ( 2 m ℏ ) , which is the single free kernel for the total time.
Why does the free-particle kernel not decay at large ∣ x f − x i ∣ |x_f-x_i| ∣ x f − x i ∣ T T T
Solution The magnitude of the real-time free kernel is set by the prefactor:
∣ K 0 ∣ = ( m 2 π ℏ ∣ T ∣ ) 1 / 2 . \left|K_0\right|
=
\left(
\frac{m}{2\pi\hbar |T|}
\right)^{1/2}. ∣ K 0 ∣ = ( 2 π ℏ∣ T ∣ m ) 1/2 . The dependence on x f − x i x_f-x_i x f − x i
