Heisenberg picture

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In physics, the Heisenberg picture is that formulation of quantum mechanics where the operators (observables and others) are time-dependent and the state vectors are time-independent. It stands in contrast to the Schrödinger picture in which operators are constant and the states evolve in time. The two pictures only differ by a time-dependent basis change.

The Heisenberg Picture is the formulation of matrix mechanics in an arbitrary basis, where the Hamiltonian is not necessarily diagonal.

1 Mathematical details
2 Deriving Heisenberg's equation
3 Commutator relations
4 Further reading
5 See also

[edit] Mathematical details

In quantum mechanics in the Heisenberg picture the state vector, $|\psi \rang$ , does not change with time, and an observable A satisfies

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

where H is the Hamiltonian and [·,·] is the commutator of A and H. 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.

[edit] 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 $|\psi(t)\rang$ 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 ħ is Planck's constant divided by 2·π) 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,

${d \over dt} A(t) = {i \over \hbar} H e^{iHt / \hbar} A e^{-iHt / \hbar} + \left(\frac{\partial A}{\partial t}\right)_\mathrm{classical} + {i \over \hbar}e^{iHt / \hbar} A \cdot (-H) e^{-iHt / \hbar}$

(differentiating according to the product rule),

$= {i \over \hbar } e^{iHt / \hbar} \left( H A - A H \right) e^{-iHt / \hbar} + \left(\frac{\partial A}{\partial t}\right)_\mathrm{classical} = {i \over \hbar } \left( H A(t) - A(t) H \right) + \left(\frac{\partial A}{\partial t}\right)_\mathrm{classical}$

(the last passage is valid since : $e^{-iHt/ \hbar}$ commutes with H)

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

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

So we get

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

Making use of the operator identity

${e^B A e^{-B}} = A + [B,A] + \frac{1}{2!} [B,[B,A]] + \frac{1}{3!}[B,[B,[B,A]]]+\cdots$

we see that for a time independent observable A, we get:

$A(t)=A+\frac{it}{\hbar}[H,A]-\frac{t^{2}}{2!\hbar^{2}}[H,[H,A]] - \frac{it^3}{3!\hbar^3}[H,[H,[H,A]]] + \cdots.$

Due to the relationship between Poisson Bracket and Commutators this relation also holds for classical mechanics.

[edit] Commutator relations

Obviously, commutator relations are quite different than in the Schrodinger picture because of the time dependency of operators. For example, consider the operators