diff options
Diffstat (limited to 'source/know/concept/heisenberg-picture')
| -rw-r--r-- | source/know/concept/heisenberg-picture/index.md | 108 |
1 files changed, 64 insertions, 44 deletions
diff --git a/source/know/concept/heisenberg-picture/index.md b/source/know/concept/heisenberg-picture/index.md index 359ecfe..54bf397 100644 --- a/source/know/concept/heisenberg-picture/index.md +++ b/source/know/concept/heisenberg-picture/index.md @@ -8,99 +8,117 @@ categories: layout: "concept" --- -The **Heisenberg picture** is an alternative formulation of quantum -mechanics, and is equivalent to the traditionally-taught Schrödinger equation. +The **Heisenberg picture** is an alternative formulation of quantum mechanics, +and is equivalent to the traditional 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 $$\hat{H}_S$$: +In the Schrödinger picture, +time-independent operators are constant by definition, +and the state $$\Ket{\psi_S(t)}$$ varies as follows, where $$\hat{U}(t)$$ +is the [time evolution operator](/know/concept/time-evolution-operator/): $$\begin{aligned} - \Ket{\psi_S(t)} = \hat{U}(t) \Ket{\psi_S(0)} - \qquad \quad - \boxed{ - \hat{U}(t) \equiv \exp\!\bigg(\!-\! i \frac{\hat{H}_S t}{\hbar} \bigg) - } + \Ket{\psi_S(t)} + = \hat{U}(t) \Ket{\psi_S(0)} \end{aligned}$$ -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. +In the Heisenberg picture, the roles are reversed: +the states $$\Ket{\psi_H}$$ are constants, +and instead the operators vary in time. +In some situations this approach can be more convenient, +and since we usually care about the evolution of observable quantities, +studying the corresponding operators directly +may make more sense than finding abstract quantum states. +Another advantage is that basis states remain fixed, +which can simplify calculations. -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: +Given a Schrödinger-picture state $$\Ket{\psi_S(t)}$$ +and an operator $$\hat{L}_S(t)$$ that may or may not depend on time, +they can be converted to the Heisenberg picture by the following transformation: $$\begin{aligned} \boxed{ - \Ket{\psi_H} \equiv \Ket{\psi_S(0)} - \qquad - \hat{L}_H(t) \equiv \hat{U}^\dagger(t) \: \hat{L}_S(t) \: \hat{U}(t) + \Ket{\psi_H} + \equiv \Ket{\psi_S(0)} + } + \qquad\qquad + \boxed{ + \hat{L}_H(t) + \equiv \hat{U}^\dagger(t) \: \hat{L}_S(t) \: \hat{U}(t) } \end{aligned}$$ -Since $$\hat{U}(t)$$ is unitary, the expectation value of a given operator is unchanged: +Note that if $$\hat{H}_S$$ is time-independent, +then it commutes with $$\hat{U}(t)$$, +meaning $$\hat{H}_H = \hat{H}_S$$, +so it can simply be labelled $$\hat{H}$$. +This is not true for time-dependent Hamiltonians. + +Since $$\hat{U}(t)$$ is unitary, +the expectation value of a given operator is unchanged: $$\begin{aligned} \expval{\hat{L}_H} &= \matrixel{\psi_H}{\hat{L}_H(t)}{\psi_H} - = \matrixel{\psi_S(0)}{\hat{U}^\dagger(t) \: \hat{L}_S(t) \: \hat{U}(t)}{\psi_S(0)} + \\ + &= \matrixel{\psi_S(0)}{\hat{U}^\dagger(t) \: \hat{L}_S(t) \: \hat{U}(t)}{\psi_S(0)} \\ &= \matrixel{\hat{U}(t) \psi_S(0)}{\hat{L}_S(t)}{\hat{U}(t) \psi_S(0)} - = \matrixel{\psi_S(t)}{\hat{L}_S}{\psi_S(t)} - = \expval{\hat{L}_S} + \\ + &= \matrixel{\psi_S(t)}{\hat{L}_S}{\psi_S(t)} + \\ + &= \expval{\hat{L}_S} \end{aligned}$$ The Schrödinger and Heisenberg pictures therefore respectively correspond to active and passive transformations by $$\hat{U}(t)$$ in [Hilbert space](/know/concept/hilbert-space/). -The two formulations are thus entirely equivalent, +The two formulations are entirely equivalent, and can be derived from one another, -as will be shown shortly. +as we will show 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: +Instead, we use it derive a new equation for $$\hat{L}_H(t)$$, +with the key being that $$\hat{U}(t)$$ itself +satisfies the Schrödinger equation by definition: $$\begin{aligned} - \dv{}{t}\hat{U}(t) = - \frac{i}{\hbar} \hat{H}_S(t) \: \hat{U}(t) + \dv{}{t} \hat{U}(t) + = - \frac{i}{\hbar} \hat{H}_S(t) \: \hat{U}(t) \end{aligned}$$ 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: +$$\hat{L}_H(t)$$ and insert the other side of the Schrödinger equation when necessary: $$\begin{aligned} - \dv{}{\hat{L}H}{t} + \dv{\hat{L}_H}{t} &= \dv{\hat{U}^\dagger}{t} \hat{L}_S \hat{U} + \hat{U}^\dagger \hat{L}_S \dv{\hat{U}}{t} + \hat{U}^\dagger \dv{\hat{L}_S}{t} \hat{U} \\ &= \frac{i}{\hbar} \hat{U}^\dagger \hat{H}_S (\hat{U} \hat{U}^\dagger) \hat{L}_S \hat{U} - \frac{i}{\hbar} \hat{U}^\dagger \hat{L}_S (\hat{U} \hat{U}^\dagger) \hat{H}_S \hat{U} - + \Big( \dv{\hat{L}_S}{t} \Big)_H + + \bigg( \dv{\hat{L}_S}{t} \bigg)_H \\ &= \frac{i}{\hbar} \hat{H}_H \hat{L}_H - \frac{i}{\hbar} \hat{L}_H \hat{H}_H - + \Big( \dv{\hat{L}_S}{t} \Big)_H - = \frac{i}{\hbar} \comm{\hat{H}_H}{\hat{L}_H} + \Big( \dv{\hat{L}_S}{t} \Big)_H + + \bigg( \dv{\hat{L}_S}{t} \bigg)_H \end{aligned}$$ -We thus get the equation of motion for operators in the Heisenberg picture: +We thus get the following equation of motion for operators in the Heisenberg picture: $$\begin{aligned} \boxed{ - \dv{}{t}\hat{L}_H(t) = \frac{i}{\hbar} \comm{\hat{H}_H(t)}{\hat{L}_H(t)} + \Big( \dv{}{t}\hat{L}_S(t) \Big)_H + \dv{}{t}\hat{L}_H(t) + = \frac{i}{\hbar} \comm{\hat{H}_H(t)}{\hat{L}_H(t)} + \bigg( \dv{}{t}\hat{L}_S(t) \bigg)_H } \end{aligned}$$ -This equation is closer to classical mechanics than the Schrödinger picture: -inserting the position $$\hat{X}$$ and momentum $$\hat{P} = - i \hbar \: \idv{}{\hat{X}}$$ -gives the following Newton-style equations: +This result is arguably more intuitive than the Schrödinger picture, +because it allows us to think about observables (i.e. operators) in a more classical way. +For example, inserting the position $$\hat{X}$$ +and momentum $$\hat{P} = - i \hbar \: \idv{}{\hat{X}}$$ +gives the following Newton-style relations: $$\begin{aligned} \dv{\hat{X}}{t} @@ -112,5 +130,7 @@ $$\begin{aligned} = - \dv{V(\hat{X})}{\hat{X}} \end{aligned}$$ -For a proof, see [Ehrenfest's theorem](/know/concept/ehrenfests-theorem/), -which is closely related to the Heisenberg picture. +Where the commutators have been treated as known. +These equations would not be valid in the Schrödinger picture, +unless we took their expectation value +to get [Ehrenfest's theorem](/know/concept/ehrenfests-theorem/). |