Heisenberg picture

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.

Mathematical details

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

:$frac\{d\}\{dt\}A=(ihbar)^\{-1\}\; [A,H]\; +left(frac\{partial\; A\}\{partial\; t\}\; ight)\_mathrm\{classical\},$

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.

See also Schrödinger picture, and Ehrenfest theorem

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 $|psi(t)\; ang$ is given by:

:$lang\; A\; ang\; \_\{t\}\; =\; lang\; psi\; (t)\; |\; A\; |\; psi(t)\; ang$

or if we write following the Schrödinger equation

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

(where "H" is the Hamiltonian and ħ is Planck's constant divided by "2·π")we get

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

and so we define

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

Now,

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

(differentiating according to the product rule),

:$=\; \{i\; over\; hbar\; \}\; e^\{iHt\; /\; hbar\}\; left(\; H\; A\; -\; A\; H\; ight)\; e^\{-iHt\; /\; hbar\}\; +\; left(frac\{partial\; A\}\{partial\; t\}\; ight)\_mathrm\{classical\}\; =\; \{i\; over\; hbar\; \}\; left(\; H\; A(t)\; -\; A(t)\; H\; ight)\; +\; left(frac\{partial\; A\}\{partial\; t\}\; ight)\_mathrm\{classical\}$

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

:$=\; \{i\; over\; hbar\; \}\; [\; H\; ,\; A(t)\; ]\; +\; left(frac\{partial\; A\}\{partial\; t\}\; ight)\_mathrm\{classical\}$(where ["X", "Y"] is the commutator of two operators and defined as ["X", "Y"] := "XY" − "YX")

So we get:$\{d\; over\; dt\}\; A(t)\; =\; \{i\; over\; hbar\; \}\; [\; H\; ,\; A(t)\; ]\; +\; left(frac\{partial\; A\}\{partial\; t\}\; ight)\_mathrm\{classical\}.$

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.

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

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

:$dot\{p\}(0)=-momega^\{2\}\; x(0)$ :$dot\{x\}(0)=frac\{p(0)\}\{m\}$

leads to:

:$x(t)=x\_\{0\}cos(omega\; t)+frac\{p\_\{0\{omega\; m\}sin(omega\; t)$:$p(t)=p\_\{0\}cos(omega\; t)-momega!x\_\{0\}sin(omega\; t)$

Now, we are ready to directly compute the commutator relations::$[x(t\_\{1\}),\; x(t\_\{2\})]\; =frac\{ihbar\}\{momega\}sin(omega\; t\_\{2\}-omega\; t\_\{1\})$ :$[p(t\_\{1\}),\; p(t\_\{2\})]\; =ihbar\; momegasin(omega\; t\_\{2\}-omega\; t\_\{1\})$:$[x(t\_\{1\}),\; p(t\_\{2\})]\; =ihbar\; cos(omega\; t\_\{2\}-omega\; t\_\{1\})$

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

Further reading

* cite book
last = Cohen-Tannoudji
first = Claude
authorlink = Claude Cohen-Tannoudji
coauthors = Bernard Diu, Frank Laloe
title = Quantum Mechanics (Volume One)
publisher = Wiley
date = 1977
location = Paris
pages = 312-314
isbn = 047116433X

See also

* Schrödinger picture
* Interaction picture
* Quantum mechanics
* Schrödinger equation
* Bra-ket notation
* Matrix mechanics