7.3 The Heisenberg picture

7.3 The Heisenberg picture

Concept links · terms present in this machine draft; source roles are unverified: eigenvalue · adjoint representation

The treatment of time-dependence so far has used what physicists call the “Schr¨odinger picture” of quantum mechanics. States in $\mathcal{H}$ are functions of time, obeying the Schr¨odinger equation determined by a Hamiltonian observable $H,$ while observable self-adjoint operators $O$ are time-independent. Time evolution is given by a unitary transformation

$$U(t)=e^{-itH},|\psi (t)⟩=U(t)|\psi (0)⟩$$

$U(t)$ can instead be used to make a unitary transformation that puts the time-dependence in the observables, removing it from the states, giving something called the “Heisenberg picture.” This is done as follows:

$$|\psi (t)⟩\to |\psi (t){⟩}_H=U^{-1}(t)|\psi (t)⟩=|\psi (0)⟩,\mathcal{O}\to {\mathcal{O}}_H(t)=U^{-1}(t)\mathcal{O}U(t)$$

where the ${}^{66}{\mathrm{H}}^5$ subscripts indicate the Heisenberg picture choice for the treatment of time-dependence. It can easily be seen that the physically observable quantities given by eigenvalues and expectations values are identical in the two pictures:

$${}_H⟨\psi (t)|{\mathcal{O}}_H|\psi (t){⟩}_H=⟨\psi (t)|U(t)(U^{-1}(t)\mathcal{O}U(t))U^{-1}(t)|\psi (t)⟩=⟨\psi (t)|\mathcal{O}|\psi (t)⟩$$

In the Heisenberg picture the dynamics is given by a diferential equation not for the states but for the operators. Recall from our discussion of the adjoint representation (see equation 5.1) the formula

$$\begin{array}{rl}\frac{d}{dt}(e^{tX}Y{e}^{-tX})& =\left(\frac{d}{dt}(e^{tX}Y)\right)e^{-tX}+e^{tX}Y\left(\frac{d}{dt}{e}^{-tX}\right)\\ & =X{e}^{tX}Y{e}^{-tX}-e^{tX}Y{e}^{-tX}X\end{array}$$

Using this with

$$Y=\mathcal{O},X=iH$$

we find

$$\frac{d}{dt}{\mathcal{O}}_H(t)=[iH,{\mathcal{O}}_H(t)]=i[H,{\mathcal{O}}_H(t)]$$

and this equation determines the time evolution of the observables in the Heisenberg picture.

Applying this to the case of the spin $\begin{array}{l}\frac{1}{2}\end{array}$ system in a magnetic field, and taking for our observable S (the $S_j$ , taken together as a column vector) we find

$$\frac{d}{dt}{\mathbf{S}}_H(t)=i[H,{\mathbf{S}}_H(t)]=i\frac{eg}{2mc}[{\mathbf{S}}_H(t)\cdot \mathbf{B},{\mathbf{S}}_H(t)]\text{\hspace{1em}(7.3)}$$

We know from the discussion above that the solution will be

$${\mathbf{S}}_H(t)=U(t){\mathbf{S}}_H(0)U(t{)}^{-1}$$

for

$$U(t)=e^{-it\frac{ge|\mathbf{B}|}{2mc}\mathbf{S}\cdot \frac{\mathbf{B}}{|\mathbf{B}|}}$$

By equation 6.5 and the identification there of vectors and 2 by 2 matrices, the spin vector observable evolves in the Heisenberg picture by rotating about the magnetic field vector $B$ with angular velocity $\frac{ge|\tilde{\mathbf{B}}|}{2mc}$

Previous section: 7.2 The spin 1/2 particle in a magnetic field · Next section: 7.4 Complex projective space · 中文版本

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