Heisenberg picture: Definition from Answers.com

Quantum mechanics

$\Delta x\, \Delta p \ge \frac{\hbar}{2}$

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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 See also
5 Further reading

Mathematical details

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

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

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.

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 time-independent 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,

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

(differentiating according to the product rule),

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

(the last passage is valid since : $e^{-iHt/ \hbar}$ commutes with H.) We are now left with the Heisenberg equation of motion:

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

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

Now, if we make 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:

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

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

Commutator relations

Obviously, commutator relations are quite different than in the Schrödinger picture because of the time dependency of operators. For example, consider the operators $x (t 1), x (t 2), p (t 1)$ and $p (t 2)$ . The time evolution of those operators depends on the Hamiltonian of the system. For the one-dimensional harmonic oscillator

$H=\frac{p^{2}}{2m}+\frac{m\omega^{2}x^{2}}{2}$

The evolution of the position and momentum operators is given by:

${d \over dt} x(t) = {i \over \hbar } [ H , x(t) ]=\frac {p}{m}$

${d \over dt} p(t) = {i \over \hbar } [ H , p(t) ]= -m \omega^{2} x$

By differentiating both equations one more time and solving them with proper initial conditions

$\dot{p}(0)=-m\omega^{2} x_0$

$\dot{x}(0)=\frac{p_0}{m}$

leads to:

$x(t)=x_{0}\cos(\omega t)+\frac{p_{0}}{\omega m}\sin(\omega t)$

$p(t)=p_{0}\cos(\omega t)-m\omega\!x_{0}\sin(\omega t)$

Now, we are ready to directly compute the commutator relations:

$[x(t_{1}), x(t_{2})]=\frac{i\hbar}{m\omega}\sin(\omega t_{2}-\omega t_{1})$

$[p(t_{1}), p(t_{2})]=i\hbar m\omega\sin(\omega t_{2}-\omega t_{1})$

$[x(t_{1}), p(t_{2})]=i\hbar \cos(\omega t_{2}-\omega t_{1})$

Note that for $t 1 = t 2$ , one simply gets the well-known canonical commutation relations.

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Heisenberg picture

Contents

Mathematical details

Deriving Heisenberg's equation

Commutator relations

See also

Further reading