Categories: Physics, Quantum mechanics.

The **Heisenberg picture** is an alternative formulation of quantum mechanics, and is equivalent to the traditionally-taught Schrödinger equation.

In the Schrödinger picture, the operators (observables) are fixed (as long as they do not depend on time), while the state \(\ket{\psi_S(t)}\) changes according to the Schrödinger equation, which can be written using the generator of translations \(\hat{U}(t)\) like so, for a time-independent \(\hat{H}_S\):

\[\begin{aligned} \ket{\psi_S(t)} = \hat{U}(t) \ket{\psi_S(0)} \qquad \quad \boxed{ \hat{U}(t) \equiv \exp\!\bigg(\!-\! i \frac{\hat{H}_S t}{\hbar} \bigg) } \end{aligned}\]

In contrast, the Heisenberg picture reverses the roles: the states \(\ket{\psi_H}\) are invariant, and instead the operators vary with time. An advantage of this is that the basis states remain the same.

Given a Schrödinger-picture state \(\ket{\psi_S(t)}\), and operator \(\hat{L}_S(t)\) which may or may not depend on time, they can be converted to the Heisenberg picture by the following change of basis:

\[\begin{aligned} \boxed{ \ket{\psi_H} \equiv \ket{\psi_S(0)} \qquad \hat{L}_H(t) \equiv \hat{U}^\dagger(t) \: \hat{L}_S(t) \: \hat{U}(t) } \end{aligned}\]

Since \(\hat{U}(t)\) is unitary, the expectation value of a given operator is unchanged:

\[\begin{aligned} \expval*{\hat{L}_H} &= \matrixel{\psi_H}{\hat{L}_H(t)}{\psi_H} = \matrixel{\psi_S(0)}{\hat{U}^\dagger(t) \: \hat{L}_S(t) \: \hat{U}(t)}{\psi_S(0)} \\ &= \matrixel*{\hat{U}(t) \psi_S(0)}{\hat{L}_S(t)}{\hat{U}(t) \psi_S(0)} = \matrixel{\psi_S(t)}{\hat{L}_S}{\psi_S(t)} = \expval*{\hat{L}_S} \end{aligned}\]

The Schrödinger and Heisenberg pictures therefore respectively correspond to active and passive transformations by \(\hat{U}(t)\) in Hilbert space. The two formulations are thus entirely equivalent, and can be derived from one another, as will be shown shortly.

In the Heisenberg picture, the states are constant, so the time-dependent Schrödinger equation is not directly useful. Instead, we will use it derive a new equation for \(\hat{L}_H(t)\). The key is that the generator \(\hat{U}(t)\) is defined from the Schrödinger equation:

\[\begin{aligned} \dv{t} \hat{U}(t) = - \frac{i}{\hbar} \hat{H}_S(t) \hat{U}(t) \end{aligned}\]

Where \(\hat{H}_S(t)\) may depend on time. We differentiate the definition of \(\hat{L}_H(t)\) and insert the other side of the Schrödinger equation when necessary:

\[\begin{aligned} \dv{\hat{L}_H}{t} &= \dv{\hat{U}^\dagger}{t} \hat{L}_S \hat{U} + \hat{U}^\dagger \hat{L}_S \dv{\hat{U}}{t} + \hat{U}^\dagger \dv{\hat{L}_S}{t} \hat{U} \\ &= \frac{i}{\hbar} \hat{U}^\dagger \hat{H}_S (\hat{U} \hat{U}^\dagger) \hat{L}_S \hat{U} - \frac{i}{\hbar} \hat{U}^\dagger \hat{L}_S (\hat{U} \hat{U}^\dagger) \hat{H}_S \hat{U} + \Big( \dv{\hat{L}_S}{t} \Big)_H \\ &= \frac{i}{\hbar} \hat{H}_H \hat{L}_H - \frac{i}{\hbar} \hat{L}_H \hat{H}_H + \Big( \dv{\hat{L}_S}{t} \Big)_H = \frac{i}{\hbar} [\hat{H}_H, \hat{L}_H] + \Big( \dv{\hat{L}_S}{t} \Big)_H \end{aligned}\]

We thus get the equation of motion for operators in the Heisenberg picture:

\[\begin{aligned} \boxed{ \dv{t} \hat{L}_H(t) = \frac{i}{\hbar} [\hat{H}_H(t), \hat{L}_H(t)] + \Big( \dv{t} \hat{L}_S(t) \Big)_H } \end{aligned}\]

This equation is closer to classical mechanics than the Schrödinger picture: inserting the position \(\hat{X}\) and momentum \(\hat{P} = - i \hbar \: \dv*{\hat{X}}\) gives the following Newton-style equations:

\[\begin{aligned} \dv{\hat{X}}{t} &= \frac{i}{\hbar} [\hat{H}, \hat{X}] = \frac{\hat{P}}{m} \\ \dv{\hat{P}}{t} &= \frac{i}{\hbar} [\hat{H}, \hat{P}] = - \dv{V(\hat{X})}{\hat{X}} \end{aligned}\]

For a proof, see Ehrenfest’s theorem, which is closely related to the Heisenberg picture.

© Marcus R.A. Newman, a.k.a. "Prefetch".
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