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Heisenberg picture

In quantum mechanics in the Heisenberg picture the state vector, |ψ> does not change with time, and an observable A satisfies

$\frac{d}{dt}A=(i\hbar)^{-1}[A,H]+\left(\frac{\partial A}{\partial t}\right)_{classical}.$

In some sense, the Heisenberg picture is more natural and fundamental than the Schrödinger picture, especially for relativistic theories. Lorentz invariance is manifest in the Heisenberg picture.

Moreover, the similarity to classical physics is easily seen: by replacing the commutator above by the Poisson bracket, the Heisenberg equation becomes an equation in Hamiltonian mechanics.

By the Stone-von Neumann theorem, the Heisenberg picture and the Schrödinger picture are unitarily equivalent.

Deriving Heisenberg's equation

Suppose we have an observable A (which is a Hermitian linear operator). The expectation value of A for a given state |ψ(t)> is given by:

$\lang A \rang _{t} = \lang \psi (t) | A | \psi(t) \rang$

or if we write following the Schrödinger equation

$| \psi (t) \rang = e^{-iHt / \hbar} | \psi (0) \rang$

(where H is the Hamiltonian and hbar is Plank's constant divided in 2*Pi) we get

$\lang A \rang _{t} = \lang \psi (0) | e^{iHt / \hbar} A e^{-iHt / \hbar} | \psi(0) \rang$

and so we define

$A(t) := e^{iHt / \hbar} A e^{-iHt / \hbar}$

Now,

${d \over dt} A(t) = {i \over \hbar} H e^{iHt / \hbar} A e^{-iHt / \hbar} + {i \over \hbar}e^{iHt / \hbar} A \cdot (-H) e^{-iHt / \hbar}$

(differentiating according to the product rule while assuming the A has no explicit depandance in time),

$= {i \over \hbar } e^{iHt / \hbar} \left( H A - A H \right) e^{-iHt / \hbar} = {i \over \hbar } \left( H A(t) - A(t) H \right)$

(the last passage is valid since exp(-iHt/hbar) commutes with H)

$= {i \over \hbar } [ H(t) , A ]$

(where [X,Y] is the commutator of two operators and defined as [X,Y]: = XY - YX)

So we got

${d \over dt} A(t) = {i \over \hbar } [ H(t) , A ]$

If we assume that A(t) does have explicit time dependeces it is easy to show that the Heisenberg's equation becomes

${d \over dt} A(t) = {i \over \hbar } [ H(t) , A ] + \left(\frac{\partial A}{\partial t}\right)_{classical}.$