Propagator for the Harmonic Oscillator

Quantum Mechanics A

Schrödinger Equation The most fundamental equation of quantum mechanics; given a Hamiltonian ${\mathcal {H}}$, it describes how a state $|\Psi \rangle$ evolves in time.

Physical Basis of Quantum Mechanics

Basic Concepts and Theory of Motion UV Catastrophe (Black-Body Radiation) Photoelectric Effect Stability of Matter Double Slit Experiment Stern-Gerlach Experiment The Principle of Complementarity The Correspondence Principle The Philosophy of Quantum Theory

Schrödinger Equation

Brief Derivation of Schrödinger Equation Relation Between the Wave Function and Probability Density Stationary States Heisenberg Uncertainty Principle Some Consequences of the Uncertainty Principle

Operators, Eigenfunctions, and Symmetry

Linear Vector Spaces and Operators Commutation Relations and Simultaneous Eigenvalues The Schrödinger Equation in Dirac Notation Transformations of Operators and Symmetry Time Evolution of Expectation Values and Ehrenfest's Theorem

Motion in One Dimension

One-Dimensional Bound States Oscillation Theorem The Dirac Delta Function Potential Scattering States, Transmission and Reflection Motion in a Periodic Potential Summary of One-Dimensional Systems

Discrete Eigenvalues and Bound States

Harmonic Oscillator Spectrum and Eigenstates Analytical Method for Solving the Simple Harmonic Oscillator Coherent States Charged Particles in an Electromagnetic Field WKB Approximation

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The Path Integral Formulation of Quantum Mechanics

Feynman Path Integrals The Free-Particle Propagator Propagator for the Harmonic Oscillator

Continuous Eigenvalues and Collision Theory

Differential Cross Section and the Green's Function Formulation of Scattering Central Potential Scattering and Phase Shifts Coulomb Potential Scattering

We will now evaluate the propagator for the harmonic oscillator. The Lagrangian for this system is

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

Before we begin, let us prove that the propagator will separate into two factors; one of these comes entirely from the classical motion of the system, and the other comes entirely from quantum fluctuations about said trajectory. To this end, let us write $x=x_{c}+y,\!$ where $x_{c}\!$ is the classical trajectory and $y\!$ is the fluctuation, which will be a new integration variable for the path integral. If we take $t_{i}\!$ and $t_{f}\!$ to be the initial and final times, respectively, then $y(t_{i})=y(t_{f})=0.\!$ Substituting this into the action, we get

$S=\int _{t_{i}}^{t_{f}}dt\,[{\tfrac {1}{2}}m({\dot {x}}_{c}+{\dot {y}})^{2}-{\tfrac {1}{2}}m\omega ^{2}(x_{c}+y)^{2}].$

We now expand out the squares, obtaining

$S=\int _{t_{i}}^{t_{f}}dt\,({\tfrac {1}{2}}m{\dot {x}}_{c}^{2}-{\tfrac {1}{2}}m\omega ^{2}x_{c}^{2})+\int _{t_{i}}^{t_{f}}dt\,({\tfrac {1}{2}}m{\dot {y}}^{2}-{\tfrac {1}{2}}m\omega ^{2}y^{2})+\int _{t_{i}}^{t_{f}}dt\,(m{\dot {x}}_{c}{\dot {y}}-m\omega ^{2}x_{c}y).$

If we integrate by parts in the third term, we get

$S=\int _{t_{i}}^{t_{f}}dt\,({\tfrac {1}{2}}m{\dot {x}}_{c}^{2}-{\tfrac {1}{2}}m\omega ^{2}x_{c}^{2})+\int _{t_{i}}^{t_{f}}dt\,({\tfrac {1}{2}}m{\dot {y}}^{2}-{\tfrac {1}{2}}m\omega ^{2}y^{2})-\int _{t_{i}}^{t_{f}}dt\,m({\ddot {x}}_{c}+\omega ^{2}x_{c})y.$

We know, however, that the classical motion obeys the equation, ${\ddot {x}}_{c}+\omega ^{2}x_{c}=0.\!$ As a result, the third term is zero, and the action separates into two contributions, one coming entirely from the classical motion, and the other coming entirely from quantum fluctuations. Denoting these two contributions as

$S_{c}(x_{c})=\int _{t_{i}}^{t_{f}}dt\,({\tfrac {1}{2}}m{\dot {x}}_{c}^{2}-{\tfrac {1}{2}}m\omega ^{2}x_{c}^{2})$

and

$S_{q}(y)=\int _{t_{i}}^{t_{f}}dt\,({\tfrac {1}{2}}m{\dot {y}}^{2}-{\tfrac {1}{2}}m\omega ^{2}y^{2}),\!$

the propagator may now be written as

$K(x_{f},t_{f};x_{i},t_{i})=e^{iS_{c}/\hbar }\int D[y(t)]\,e^{iS_{q}/\hbar }.$

We will now evaluate each of these contributions.

Contribution from Classical Path

We will begin by evaluating the "classical" contribution to the propagator. This is essentially just a problem of classical mechanics; we begin by solving for the classical motion of the particle. The equation of motion is, as stated earlier,

${\ddot {x}}_{c}+\omega ^{2}x_{c}=0.\!$

We impose the boundary conditions, $x(t_{i})=x_{i}\!$ and $x(t_{f})=x_{f}.\!$ The solution of the equation of motion that satisfies these boundary conditions is

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

and the corresponding velocity is

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

If we now substitute these expressions into the Lagrangian and simplify, we obtain

$L={\frac {m\omega ^{2}}{2\sin ^{2}{\omega (t_{f}-t_{i})}}}\{(x_{i}^{2}+x_{f}^{2})\cos[2\omega (t-t_{i})]-2x_{i}x_{f}\cos {\omega (t_{i}+t_{f}-2t)}\}.$

If we now substitute this into the action, we finally obtain

$S_{c}={\frac {m\omega }{2\sin {\omega (t_{f}-t_{i})}}}((x_{i}^{2}+x_{f}^{2})\cos {\omega (t_{f}-t_{i})}-2x_{i}x_{f}).$

Explicit evaluation of the path integral for the harmonic oscillator can be found here File:FeynmanHibbs H O Amplitude.pdf

Contribution From Fluctuations

Now, let's evaluate the path integral: $A=A(t)=\int _{y(0)=0}^{y(t)=0}D[y(t')]e^{{\frac {i}{\hbar }}\int _{0}^{t}({\frac {1}{2}}my'^{2}-{\frac {1}{2}}ky^{2})dt'}$

Note that the integrand is taken over all possible trajectory starting at point $x_{0}$ at time $t'=0$, ending at point $x$ at time $t'=t$.

Expanding this integral,

$A(t)=\left({\frac {m}{2\pi i\hbar }}\right)^{\frac {N}{2}}\int _{-\infty }^{\infty }dy_{1}\ldots dy_{N-1}e^{\left[{\frac {i}{\hbar }}\left({\frac {m}{2\Delta t}}y_{N-1}^{2}-{\frac {\Delta t}{2}}ky_{N-1}^{2}\right)\right]}e^{\left[{\frac {i}{\hbar }}\left({\frac {m}{2\Delta t}}(y_{N-1}-y_{N-2})^{2}-{\frac {\Delta t}{2}}ky_{N-2}^{2}\right)\right]}\ldots e^{\left[{\frac {i}{\hbar }}\left({\frac {m}{2\Delta t}}y_{1}^{2}-{\frac {\Delta t}{2}}ky_{1}^{2}\right)\right]}$

where $N\Delta t=t\!$.

Expanding the path trajectory in Fourier series, we have $y(t')=\sum _{n}a_{n}\sin \left({\frac {n\pi t'}{t}}\right)$

we may express $A(t)\!$ in the form

$A(t)=C\int _{-\infty }^{\infty }da_{1}\ldots da_{N-1}\exp {\left[\sum _{n=1}^{N-1}{\frac {im}{2\hbar }}\left(\left({\frac {n\pi }{t}}\right)^{2}-\omega ^{2}\right)a_{n}^{2}\right]}$

where C is a constant independent of the frequency which comes from the Jacobian of the transformation. The important point is that it does not depend on the frequency $\omega \!$. Thus, evaluating the integral of,

$A(t)=C'\prod _{n=1}^{N-1}\left[\left({\frac {n\pi }{t}}\right)^{2}-\omega ^{2}\right]^{-{\frac {1}{2}}}=C'\prod _{n=1}^{N-1}\left[\left({\frac {n\pi }{t}}\right)^{2}\right]^{-{\frac {1}{2}}}\prod _{n=1}^{N-1}\left[1-\left({\frac {\omega t}{n\pi }}\right)^{2}\right]^{-{\frac {1}{2}}}$

where C' is a constant directly related to C and still independent of the frequency of motion. Since the first product series in this final expression is also independent of the frequency of motion, we can absorb it into our constant C' to have a new constant, C. Simplifying further,

$A(t)=C''{\sqrt {\frac {\omega t}{\sin(\omega t)}}}$

In the limit $\omega \rightarrow 0$, we already know that

$C''={\sqrt {\frac {m}{2\pi i\hbar t}}}$

Thus,

$A(t)={\sqrt {\frac {m}{2\pi i\hbar t}}}{\sqrt {\frac {\omega t}{\sin(\omega t)}}}={\sqrt {\frac {m}{2\pi i\hbar \sin(\omega t)}}}$

and

$<x|{\hat {U}}(t,0)|x_{0}>={\sqrt {\frac {m}{2\pi i\hbar \sin(\omega t)}}}e^{{\frac {i}{\hbar }}\left({\frac {m\omega }{2sin(\omega t)}}((x^{2}+x_{0}^{2})cos(\omega t)-2xx_{0})\right)}$

Reference

For a more detailed evaluation of this problem, please see Barone, F. A.; Boschi-Filho, H.; Farina, C. 2002. "Three methods for calculating the Feynman propagator". American Association of Physics Teachers, 2003. Am. J. Phys. 71 (5), May 2003. pp 483-491.