# Relation between Schrödinger's equation and the path integral formulation of quantum mechanics

This article relates the Schrödinger equation with the path integral formulation of quantum mechanics using a simple nonrelativistic one-dimensional single-particle Hamiltonian composed of kinetic and potential energy.

## Background

### Schrödinger's equation

Schrödinger's equation, in bra–ket notation, is

${\displaystyle i\hbar {\frac {d}{dt}}|\psi \rangle ={\hat {H}}|\psi \rangle }$

where ${\displaystyle {\hat {H}}}$ is the Hamiltonian operator. We have assumed for simplicity that there is only one spatial dimension.

The Hamiltonian operator can be written

${\displaystyle {\hat {H}}={\frac {{\hat {p}}^{2}}{2m}}+V({\hat {q}})}$

where ${\displaystyle V({\hat {q}})}$ is the potential energy, m is the mass and we have assumed for simplicity that there is only one spatial dimension q.

The formal solution of the equation is

${\displaystyle |\psi (t)\rangle =\exp \left(-{\frac {i}{\hbar }}{\hat {H}}t\right)|q_{0}\rangle \equiv \exp \left(-{\frac {i}{\hbar }}{\hat {H}}t\right)|0\rangle }$

where we have assumed the initial state is a free-particle spatial state ${\displaystyle |q_{0}\rangle }$.

The transition probability amplitude for a transition from an initial state ${\displaystyle |0\rangle }$ to a final free-particle spatial state ${\displaystyle |F\rangle }$ at time T is

${\displaystyle \langle F|\psi (t)\rangle =\left\langle F{\bigg |}\exp \left(-{\frac {i}{\hbar }}{\hat {H}}T\right){\bigg |}0\right\rangle .}$

### Path integral formulation

The path integral formulation states that the transition amplitude is simply the integral of the quantity

${\displaystyle \exp \left({\frac {i}{\hbar }}S\right)}$

over all possible paths from the initial state to the final state. Here S is the classical action.

The reformulation of this transition amplitude, originally due to Dirac[1] and conceptualized by Feynman,[2] forms the basis of the path integral formulation.[3]

## From Schrödinger's equation to the path integral formulation

The following derivation[4] makes use of the Trotter product formula, which states that for self-adjoint operators A and B (satisfying certain technical conditions), we have

${\displaystyle e^{i(A+B)}\psi =\lim _{N\rightarrow \infty }(e^{iA/N}e^{iB/N})^{N}\psi }$,

even if A and B do not commute.

We can divide the time interval [0, T] into N segments of length

${\displaystyle \delta t={\frac {T}{N}}.}$

The transition amplitude can then be written

${\displaystyle \left\langle F{\bigg |}\exp \left(-{\frac {i}{\hbar }}{\hat {H}}T\right){\bigg |}0\right\rangle =\left\langle F{\bigg |}\exp \left(-{\frac {i}{\hbar }}{\hat {H}}\delta t\right)\exp \left(-{\frac {i}{\hbar }}{\hat {H}}\delta t\right)\cdots \exp \left(-{\frac {i}{\hbar }}{\hat {H}}\delta t\right){\bigg |}0\right\rangle .}$

Although the kinetic energy and potential energy operators do not commute, the Trotter product formula, cited above, says that over each small time-interval, we can ignore this noncommutativity and write

${\displaystyle \exp \left(-{\frac {i}{\hbar }}{\hat {H}}\delta t\right)\approx \exp \left({-{i \over \hbar }{{\hat {p}}^{2} \over 2m}\delta t}\right)\exp \left({-{i \over \hbar }V\left(q_{j}\right)\delta t}\right).}$

For notational simplicity, we delay making this substitution for the moment.

We can insert the identity matrix

${\displaystyle I=\int dq|q\rangle \langle q|}$

N − 1 times between the exponentials to yield

${\displaystyle \left\langle F{\bigg |}\exp \left(-{\frac {i}{\hbar }}{\hat {H}}T\right){\bigg |}0\right\rangle =\left(\prod _{j=1}^{N-1}\int dq_{j}\right)\left\langle F{\bigg |}\exp \left(-{\frac {i}{\hbar }}{\hat {H}}\delta t\right){\bigg |}q_{N-1}\right\rangle \left\langle q_{N-1}{\bigg |}\exp \left(-{\frac {i}{\hbar }}{\hat {H}}\delta t\right){\bigg |}q_{N-2}\right\rangle \cdots \left\langle q_{1}{\bigg |}\exp \left(-{\frac {i}{\hbar }}{\hat {H}}\delta t\right){\bigg |}0\right\rangle .}$

We now implement the substitution associated to the Trotter product formula, so that we have, effectively

${\displaystyle \left\langle q_{j+1}{\bigg |}\exp \left(-{\frac {i}{\hbar }}{\hat {H}}\delta t\right){\bigg |}q_{j}\right\rangle =\left\langle q_{j+1}{\Bigg |}\exp \left({-{i \over \hbar }{{\hat {p}}^{2} \over 2m}\delta t}\right)\exp \left({-{i \over \hbar }V\left(q_{j}\right)\delta t}\right){\Bigg |}q_{j}\right\rangle .}$

We can insert the identity

${\displaystyle I=\int {dp \over 2\pi }|p\rangle \langle p|}$

into the amplitude to yield

{\displaystyle {\begin{aligned}\left\langle q_{j+1}{\bigg |}\exp \left(-{\frac {i}{\hbar }}{\hat {H}}\delta t\right){\bigg |}q_{j}\right\rangle &=\exp \left(-{\frac {i}{\hbar }}V\left(q_{j}\right)\delta t\right)\int {\frac {dp}{2\pi }}\left\langle q_{j+1}{\bigg |}\exp \left(-{\frac {i}{\hbar }}{\frac {p^{2}}{2m}}\delta t\right){\bigg |}p\right\rangle \langle p|q_{j}\rangle \\&=\exp \left(-{\frac {i}{\hbar }}V\left(q_{j}\right)\delta t\right)\int {\frac {dp}{2\pi }}\exp \left(-{\frac {i}{\hbar }}{\frac {p^{2}}{2m}}\delta t\right)\left\langle q_{j+1}|p\right\rangle \left\langle p|q_{j}\right\rangle \\&=\exp \left(-{\frac {i}{\hbar }}V\left(q_{j}\right)\delta t\right)\int {\frac {dp}{2\pi \hbar }}\exp \left(-{\frac {i}{\hbar }}{\frac {p^{2}}{2m}}\delta t-{\frac {i}{\hbar }}p\left(q_{j+1}-q_{j}\right)\right)\end{aligned}}}

where we have used the fact that the free particle wave function is

${\displaystyle \langle p|q_{j}\rangle ={\frac {\exp \left({\frac {i}{\hbar }}pq_{j}\right)}{\sqrt {\hbar }}}}$.

The integral over p can be performed (see Common integrals in quantum field theory) to obtain

${\displaystyle \left\langle q_{j+1}{\bigg |}\exp \left(-{\frac {i}{\hbar }}{\hat {H}}\delta t\right){\bigg |}q_{j}\right\rangle =\left({-im \over 2\pi \delta t\hbar }\right)^{1 \over 2}\exp \left[{i \over \hbar }\delta t\left({1 \over 2}m\left({q_{j+1}-q_{j} \over \delta t}\right)^{2}-V\left(q_{j}\right)\right)\right]}$

The transition amplitude for the entire time period is

${\displaystyle \left\langle F{\bigg |}\exp \left(-{\frac {i}{\hbar }}{\hat {H}}T\right){\bigg |}0\right\rangle =\left({-im \over 2\pi \delta t\hbar }\right)^{N \over 2}\left(\prod _{j=1}^{N-1}\int dq_{j}\right)\exp \left[{i \over \hbar }\sum _{j=0}^{N-1}\delta t\left({1 \over 2}m\left({q_{j+1}-q_{j} \over \delta t}\right)^{2}-V\left(q_{j}\right)\right)\right].}$

If we take the limit of large N the transition amplitude reduces to

${\displaystyle \left\langle F{\bigg |}\exp \left({-{i \over \hbar }{\hat {H}}T}\right){\bigg |}0\right\rangle =\int Dq(t)\exp \left[{i \over \hbar }S\right]}$

where S is the classical action given by

${\displaystyle S=\int _{0}^{T}dtL\left(q(t),{\dot {q}}(t)\right)}$

and L is the classical Lagrangian given by

${\displaystyle L\left(q,{\dot {q}}\right)={1 \over 2}m{\dot {q}}^{2}-V(q)}$

Any possible path of the particle, going from the initial state to the final state, is approximated as a broken line and included in the measure of the integral

${\displaystyle \int Dq(t)=\lim _{N\to \infty }\left({\frac {-im}{2\pi \delta t\hbar }}\right)^{\frac {N}{2}}\left(\prod _{j=1}^{N-1}\int dq_{j}\right)}$

This expression actually defines the manner in which the path integrals are to be taken. The coefficient in front is needed to ensure that the expression has the correct dimensions, but it has no actual relevance in any physical application.

This recovers the path integral formulation from Schrödinger's equation.

## References

1. ^ Dirac, P. A. M. (1958). The Principles of Quantum Mechanics, Fourth Edition. Oxford. ISBN 0-19-851208-2.
2. ^ Brown, Laurie M. (1958). Feynman's Thesis: A New Approach to Quantum Theory. World Scientific. ISBN 981-256-366-0.
3. ^ A. Zee (2010). Quantum Field Theory in a Nutshell, Second Edition. Princeton University. ISBN 978-0-691-14034-6.
4. ^ See Hall 2015 Section 20.2
• Hall, Brian C. (2013), Quantum Theory for Mathematicians, Graduate Texts in Mathematics, 267, Springer, ISBN 978-1461471158