# What are the mathematical formulations of quantum mechanics?

Oct 15, 2017

They are:

• Schrodinger formulation (wave mechanics)
• Heisenberg formulation (matrix mechanics)
• Feynman Path Integral formulation

The major differences are:

• Schrodinger formulated time-dependent wave functions and time-independent operators.
• Heisenberg formulated time-independent ket vectors and time-dependent operators.
• Feynman formulated an integration over all quantum paths, which could be described as the convolution of the Green's function G(x,x'; t) response to an impulse with a weight given by the stationary state $\psi \left(x , {t}_{0}\right)$.

For simplicity we do this in one dimension.

SCHRODINGER FORMULATION

The time-dependent Schrodinger equation is:

iℏ(delPsi)/(delt) = hatHPsi

where Psi(x,t) = e^(-iEt//ℏ)psi(x) is the time-dependent wave function and $\hat{H}$ is the time-independent Hamiltonian of the system.

As you can see, $\Psi = \Psi \left(x , t\right)$, but $\hat{H}$ is not time-dependent. Schrodinger postulated that:

1. The wave function $\Psi$ was time-dependent, but the operators associated with it are time-independent.
2. The wave function $\psi$ was an electric charge density spread out over allspace.
3. With the input of Max Born, $\psi$ is the probability amplitude, while ${\int}_{\text{allspace" psi^"*}} \psi d \tau$ is the probability of finding the electron somewhere.

In essence, Schroedinger was a guy who believed in wave mechanics and particle-wave duality, i.e. that quantum particles could be described by the de Broglie relation:

$\lambda = \frac{h}{m v}$

This is probably the easiest to understand out of all the formulations.

HEISENBERG FORMULATION

Heisenberg, for the life of him, saw it this way:

1. A system is described by some arbitrary ket vector | n >>, known as a state vector, independent of time.
2. The inner product of | n >> with << x | describing all the possible position bra vectors gives a complete set of eigenstates $\left\langlex | n\right\rangle = {\psi}_{n} \left(x\right)$.
3. The operators could be functions of time, and any time translation operator is unitary.

Under this formulation, the time-dependent Schrodinger equation becomes:

iℏ (d << x | n >>)/(dt) = hatH << x | n >>

From this, one could formulate the following relationships:

Overlap of definite position and momentum

<< x | p_x >> = 1/sqrt(2piℏ) e^(ip_x x//ℏ)

Matrix product of position and momentum with themselves

${\int}_{- \infty}^{\infty}$ | x >> << x |$\mathrm{dx} = {\int}_{- \infty}^{\infty}$ | p_x >> << p_x |${\mathrm{dp}}_{x}$ $\equiv 1$

Dirac Delta function

${\int}_{- \infty}^{\infty} \left\langlex | {p}_{x}\right\rangle \left\langle{p}_{x} | x '\right\rangle {\mathrm{dp}}_{x} = \left\langlex | x '\right\rangle$

= 1/(2piℏ) int_(-oo)^(oo) e^(ip_x(x-x')//ℏ)dp_x = delta(x-x')

${\int}_{- \infty}^{\infty} \left\langle{p}_{x} | x\right\rangle \left\langlex | {p}_{x} '\right\rangle \mathrm{dx} = \left\langle{p}_{x} | {p}_{x} '\right\rangle$

= 1/(2piℏ) int_(-oo)^(oo) e^(ix(p_x-p_x')//ℏ)dx = delta(p_x-p_x')

Relation between wave functions in position/momentum representations

${\phi}_{n} \left({p}_{x}\right) = \left\langle{p}_{x} | n\right\rangle$

$= {\int}_{- \infty}^{\infty} \left\langle{p}_{x} | x\right\rangle \left\langlex | n\right\rangle \mathrm{dx}$

= 1/sqrt(2piℏ) int_(-oo)^(oo) e^(-ip_x x//ℏ) psi_n(x) dx

${\psi}_{n} \left(x\right) = \left\langlex | n\right\rangle$

$= {\int}_{- \infty}^{\infty} \left\langlex | {p}_{x}\right\rangle \left\langle{p}_{x} | n\right\rangle {\mathrm{dp}}_{x}$

= 1/sqrt(2piℏ) int_(-oo)^(oo) e^(ip_x x//ℏ) phi_n(p_x) dp_x

FEYNMAN PATH INTEGRAL FORMULATION

Probably the hardest one to understand, I think... Feynman wanted to describe the transition probability amplitude of the quantum particle by summing over all possible quantum paths:

| psi (x,t') >> $= {\int}_{- \infty}^{\infty} \left\langle\psi \left(x ' , t '\right) | \psi \left({x}_{0} , {t}_{0}\right)\right\rangle \mathrm{dx} ' \cdot$ | psi(x',t') >>

Under this formulation, define the unitary time translator

hatU_t = e^(-ihatH(t-t_0)//ℏ)Theta(t-t_0),

where $\Theta \left(t - {t}_{0}\right) = {\int}_{- \infty}^{t} \delta \left(t ' - {t}_{0}\right) \mathrm{dt} '$ is the Heaviside step function and $\hat{H}$ is the Hamiltonian of the system.

Then a Green function can be defined in terms of ${\hat{U}}_{t}$ in position space:

G(x,x'; t) = << x | hatU_t | x' >>

where $x '$ is the position at which an initial impulse to the system was imparted at time ${t}_{0}$, and $x$ is the propagated position at time $t$.

The Feynman Path Integral is then the Green's function formulation of a system. For instance, if

iℏ(delPsi)/(del t) = hatH Psi,

then since $\frac{\partial}{\partial t} \left(\Theta \left(t - {t}_{0}\right)\right) = \delta \left(t - {t}_{0}\right)$,

(hatH - iℏ(del)/(delt)) G(x,x'; t)

= (hatH - iℏ(del)/(delt)) << x | e^(-ihatH(t-t_0)//ℏ)Theta(t-t_0) | x' >>

= [ . . . ] = -iℏ delta(x-x')delta(t-t_0)

As it turns out, if the Hamiltonian of a system is given by, for example,

$\hat{H} = {p}_{x}^{2} / \left(2 m\right) + V \left(\hat{x}\right)$,

then for a small enough time step $\frac{t}{N} \to \mathrm{dt}$, for a single time step ($N = 1$):

G(x,x'; t) = [ . . . ]

= 1/(2piℏ) int_(-oo)^(oo) e^(-ip_x^2(t-t_0)//2mNℏ)e^(-iV(hatx)(t-t_0)//Nℏ) e^(ip_x(x-x')//ℏ) dp_x

= sqrt((Nm)/(2pi iℏ(t-t_0))) "exp"[(i(t-t_0))/(Nℏ) (N^2/2 m((x - x')/(t-t_0))^2 - V(x'))],
$N = 1$

This does indeed relate back to the wave function:

Psi_n(x,t) = int_(-oo)^(oo) underbrace(G(x,x'; t))_(<< x | hatU_t | x' >>) underbrace(psi_n(x',t_0))_(<< x' | n >>) dx'

In a way (and this made the most sense to me):

The wave function can be given by the integral over the path described by the convolution of the impulse on the system with a weight given by the initial state.