From 8b8caf2467a738c0b0ccac32163d426ffab2cbd8 Mon Sep 17 00:00:00 2001 From: Prefetch Date: Tue, 15 Oct 2024 18:08:29 +0200 Subject: Improve knowledge base --- source/know/concept/heisenberg-picture/index.md | 108 +++++++----- source/know/concept/interaction-picture/index.md | 182 ++++++++++---------- source/know/concept/self-steepening/index.md | 8 +- .../know/concept/time-evolution-operator/index.md | 184 +++++++++++++++++++++ 4 files changed, 339 insertions(+), 143 deletions(-) create mode 100644 source/know/concept/time-evolution-operator/index.md (limited to 'source/know') 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}$$ -For a proof, see [Ehrenfest's theorem](/know/concept/ehrenfests-theorem/), -which is closely related to the Heisenberg picture. +Where the commutators have been treated as known. +These equations would not be valid in the Schrödinger picture, +unless we took their expectation value +to get [Ehrenfest's theorem](/know/concept/ehrenfests-theorem/). diff --git a/source/know/concept/interaction-picture/index.md b/source/know/concept/interaction-picture/index.md index de469fa..8428bf3 100644 --- a/source/know/concept/interaction-picture/index.md +++ b/source/know/concept/interaction-picture/index.md @@ -13,17 +13,19 @@ is an alternative formulation of quantum mechanics, equivalent to both the Schrödinger picture and the [Heisenberg picture](/know/concept/heisenberg-picture/). -Recall that Schrödinger lets states $$\Ket{\psi_S(t)}$$ evolve in time, -but keeps operators $$\hat{L}_S$$ fixed (except for explicit time dependence). -Meanwhile, Heisenberg keeps states $$\Ket{\psi_H}$$ fixed, -and puts all time dependence on the operators $$\hat{L}_H(t)$$. - -However, in the interaction picture, +Recall that in the Schrödinger picture, +the states $$\Ket{\psi_S(t)}$$ evolve in time, +but time-independent operators $$\hat{L}_S$$ are fixed. +Meanwhile in the Heisenberg picture, +the states $$\Ket{\psi_H}$$ are constant, +and all time dependence is on the operators $$\hat{L}_H(t)$$ instead. + +In the interaction picture, both the states $$\Ket{\psi_I(t)}$$ and the operators $$\hat{L}_I(t)$$ evolve in $$t$$. -This might seem unnecessarily complicated, -but it turns out be convenient when considering -a time-dependent "perturbation" $$\hat{H}_{1,S}$$ +This may seem unnecessarily complicated, +but it turns out to be convenient when considering +a system with a time-dependent "perturbation" $$\hat{H}_{1,S}$$ to a time-independent Hamiltonian $$\hat{H}_{0,S}$$: $$\begin{aligned} @@ -31,29 +33,43 @@ $$\begin{aligned} = \hat{H}_{0,S} + \hat{H}_{1,S}(t) \end{aligned}$$ -With $$\hat{H}_S(t)$$ the full Schrödinger Hamiltonian. -We define the unitary conversion operator: +Despite being called a perturbation, +$$\hat{H}_{1, S}$$ need not be weak compared to $$\hat{H}_{0, S}$$. +Basically, any way of splitting $$\hat{H}_S$$ is valid +as long as $$\hat{H}_{0, S}$$ is time-independent, +but only a few ways are useful. + +We now define the unitary conversion operator $$\hat{U}(t)$$ as shown below. +Note its similarity to the [time-evolution operator](/know/concept/time-evolution-operator/) +$$\hat{K}_S(t)$$, but with the opposite sign in the exponent: $$\begin{aligned} \boxed{ \hat{U}(t) - \equiv \exp\!\bigg( i \frac{\hat{H}_{0,S} t}{\hbar} \bigg) + \equiv \exp\!\bigg( \frac{i}{\hbar} \hat{H}_{0,S} t \bigg) } \end{aligned}$$ -The interaction-picture states $$\Ket{\psi_I(t)}$$ and operators $$\hat{L}_I(t)$$ -are then defined to be: +The interaction-picture states $$\Ket{\psi_I(t)}$$ +and operators $$\hat{L}_I(t)$$ are then defined as follows: $$\begin{aligned} \boxed{ \Ket{\psi_I(t)} \equiv \hat{U}(t) \Ket{\psi_S(t)} - \qquad + } + \qquad\qquad + \boxed{ \hat{L}_I(t) \equiv \hat{U}(t) \: \hat{L}_S(t) \: \hat{U}{}^\dagger(t) } \end{aligned}$$ +Because $$\hat{H}_{0, S}$$ is time-independent, +it commutes with $$\hat{U}(t)$$, +so conveniently $$\hat{H}_{0, I} = \hat{H}_{0, S}$$. + + ## Equations of motion @@ -61,17 +77,17 @@ To find the equation of motion for $$\Ket{\psi_I(t)}$$, we differentiate it and multiply by $$i \hbar$$: $$\begin{aligned} - i \hbar \dv{}{t}\Ket{\psi_I} - &= i \hbar \Big( \dv{\hat{U}}{t} \Ket{\psi_S} + \hat{U} \dv{}{t}\Ket{\psi_S} \Big) - \\ - &= i \hbar \Big( i \frac{\hat{H}_{0,S}}{\hbar} \Big) \hat{U} \Ket{\psi_S} + \hat{U} \Big( i \hbar \dv{}{t}\Ket{\psi_S} \Big) + i \hbar \dv{}{t} \Ket{\psi_I} + &= i \hbar \dv{\hat{U}}{t} \Ket{\psi_S} + \hat{U} \bigg( i \hbar \dv{}{t}\Ket{\psi_S} \bigg) \end{aligned}$$ -We insert the Schrödinger equation into the second term, -and use $$\comm{\hat{U}}{\hat{H}_{0,S}} = 0$$: +We insert the definition of $$\hat{U}$$ in the first term +and the Schrödinger equation into the second, +and use the fact that $$\comm{\hat{H}_{0, S}}{\hat{U}} = 0$$ +thanks to the time-independence of $$\hat{H}_{0, S}$$: $$\begin{aligned} - i \hbar \dv{}{t}\Ket{\psi_I} + i \hbar \dv{}{t} \Ket{\psi_I} &= - \hat{H}_{0,S} \hat{U} \Ket{\psi_S} + \hat{U} \hat{H}_S \Ket{\psi_S} \\ &= \hat{U} \big( \!-\! \hat{H}_{0,S} + \hat{H}_S \big) \Ket{\psi_S} @@ -84,129 +100,103 @@ with $$\hat{H}_{1,I} = \hat{U} \hat{H}_{1,S} \hat{U}{}^\dagger$$: $$\begin{aligned} \boxed{ - i \hbar \dv{}{t}\Ket{\psi_I(t)} + i \hbar \dv{}{t} \Ket{\psi_I(t)} = \hat{H}_{1,I}(t) \Ket{\psi_I(t)} } \end{aligned}$$ Next, we do the same with an operator $$\hat{L}_I$$ -to find a description of its evolution in time: +in order to describe its evolution in time: $$\begin{aligned} - \dv{}{t}\hat{L}_I - &= \dv{\hat{U}}{t} \hat{L}_S \hat{U}{}^\dagger + \hat{U} \hat{L}_S \dv{\hat{U}{}^\dagger}{t} + \hat{U} \dv{\hat{L}_S}{t} \hat{U}{}^\dagger + \dv{\hat{L}_I}{t} + &= \dv{\hat{U}}{t} \hat{L}_S \hat{U}{}^\dagger + \hat{U} \hat{L}_S \dv{\hat{U}{}^\dagger}{t} + + \hat{U} \dv{\hat{L}_S}{t} \hat{U}{}^\dagger \\ &= \frac{i}{\hbar} \hat{U} \hat{H}_{0,S} \big( \hat{U}{}^\dagger \hat{U} \big) \hat{L}_S \hat{U}{}^\dagger - \frac{i}{\hbar} \hat{U} \hat{L}_S \big( \hat{U}{}^\dagger \hat{U} \big) \hat{H}_{0,S} \hat{U}{}^\dagger - + \Big( \dv{\hat{L}_S}{t} \Big)_I + + \bigg( \dv{\hat{L}_S}{t} \bigg)_I \\ &= \frac{i}{\hbar} \hat{H}_{0,I} \hat{L}_I - \frac{i}{\hbar} \hat{L}_I \hat{H}_{0,I} - + \Big( \dv{\hat{L}_S}{t} \Big)_I - = \frac{i}{\hbar} \comm{\hat{H}_{0,I}}{\hat{L}_I} + \Big( \dv{\hat{L}_S}{t} \Big)_I + + \bigg( \dv{\hat{L}_S}{t} \bigg)_I \end{aligned}$$ The result is analogous to the equation of motion in the Heisenberg picture: $$\begin{aligned} \boxed{ - \dv{}{t}\hat{L}_I(t) - = \frac{i}{\hbar} \comm{\hat{H}_{0,I}(t)}{\hat{L}_I(t)} + \Big( \dv{}{t}\hat{L}_S(t) \Big)_I + \dv{}{t} \hat{L}_I(t) + = \frac{i}{\hbar} \comm{\hat{H}_{0,I}(t)}{\hat{L}_I(t)} + \bigg( \dv{}{t}\hat{L}_S(t) \bigg)_I } \end{aligned}$$ +In other words, in the interaction picture, +the "simple" time-dependence (from $$\hat{H}_{0, S}$$) is given to the operators, +and the "complicated" dependence (from $$\hat{H}_{1, S}$$) to the states. +This means that the difficult part of a problem +can be solved in isolation in a kind of Schrödinger picture. -## Time evolution operator -Recall that an alternative form of the Schrödinger equation is as follows, -where a **time evolution operator** or -**generator of translations in time** $$K_S(t, t_0)$$ -brings $$\Ket{\psi_S}$$ from time $$t_0$$ to $$t$$: -$$\begin{aligned} - \Ket{\psi_S(t)} - = \hat{K}_S(t, t_0) \Ket{\psi_S(t_0)} - \qquad \quad - \hat{K}_S(t, t_0) - \equiv \exp\!\Big( \!-\! i \frac{\hat{H}_S (t - t_0)}{\hbar} \Big) -\end{aligned}$$ +## Time evolution operator -We want to find an analogous operator in the interaction picture, satisfying: +What about the time evolution operator $$\hat{K}_S(t)$$? +Its interaction version $$\hat{K}_I(t)$$ +is unsurprisingly obtained by the standard transform +$$\hat{K}_I = \hat{U} \hat{K}_S \hat{U}^\dagger$$: $$\begin{aligned} \Ket{\psi_I(t)} - \equiv \hat{K}_I(t, t_0) \Ket{\psi_I(t_0)} -\end{aligned}$$ - -Inserting this definition into the equation of motion for $$\Ket{\psi_I}$$ yields -an equation for $$\hat{K}_I$$, with the logical boundary condition $$\hat{K}_I(t_0, t_0) = 1$$: - -$$\begin{aligned} - i \hbar \dv{}{t}\Big( \hat{K}_I(t, t_0) \Ket{\psi_I(t_0)} \Big) - &= \hat{H}_{1,I}(t) \Big( \hat{K}_I(t, t_0) \Ket{\psi_I(t_0)} \Big) + &= \hat{U}(t) \Ket{\psi_S(t)} \\ - i \hbar \dv{}{t}\hat{K}_I(t, t_0) - &= \hat{H}_{1,I}(t) \hat{K}_I(t, t_0) -\end{aligned}$$ - -We turn this into an integral equation -by integrating both sides from $$t_0$$ to $$t$$: - -$$\begin{aligned} - i \hbar \int_{t_0}^t \dv{}{t'}K_I(t', t_0) \dd{t'} - = \int_{t_0}^t \hat{H}_{1,I}(t') \hat{K}_I(t', t_0) \dd{t'} -\end{aligned}$$ - -After evaluating the left integral, -we see an expression for $$\hat{K}_I$$ as a function of $$\hat{K}_I$$ itself: - -$$\begin{aligned} - K_I(t, t_0) - = 1 + \frac{1}{i \hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \hat{K}_I(t', t_0) \dd{t'} + &= \hat{U}(t) \: \hat{K}_S(t) \: \hat{U}^\dagger(t) \Ket{\psi_I(0)} + \\ + &\equiv \hat{K}_I(t) \Ket{\psi_I(0)} \end{aligned}$$ -By recursively inserting $$\hat{K}_I$$ once, we get a longer expression, -still with $$\hat{K}_I$$ on both sides: +But we can do better. By inserting this definition of $$\hat{K}_I$$ +into the interaction picture's analogue of Schrödinger's equation, +we get the following relation for $$\hat{K}_I$$: $$\begin{aligned} - K_I(t, t_0) - = 1 + \frac{1}{i \hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'} - + \frac{1}{(i \hbar)^2} \int_{t_0}^t \hat{H}_{1,I}(t') \int_{t_0}^{t'} \hat{H}_{1,I}(t'') \hat{K}_I(t'', t_0) \dd{t''} \dd{t'} + i \hbar \dv{}{t} \hat{K}_I(t) + &= \hat{H}_{1,I}(t) \: \hat{K}_I(t) \end{aligned}$$ -And so on. Note the ordering of the integrals and integrands: -upon closer inspection, we see that the $$n$$th term is -a [time-ordered product](/know/concept/time-ordered-product/) $$\mathcal{T}$$ -of $$n$$ factors $$\hat{H}_{1,I}$$: +In other words, $$\hat{K}_I$$ can be said to also obey +the standard equation of motion for states, despite being an operator. +We integrate both sides and use $$\hat{K}_I(0) = 1$$: $$\begin{aligned} - \hat{K}_I(t, t_0) - &= 1 + \int_{t_0}^t \hat{H}_{1,I}(t_1) \dd{t_1} - + \frac{1}{2} \int_{t_0}^{t} \int_{t_0}^{t_1} \mathcal{T} \Big\{ \hat{H}_{1,I}(t_1) \hat{H}_{1,I}(t_2) \Big\} \dd{t_1} \dd{t_2} - + \: ... - \\ - &= 1 + \sum_{n = 1}^\infty \frac{1}{n!} \frac{1}{(i \hbar)^n} - \int_{t_0}^{t} \cdots \int_{t_0}^{t_n} \mathcal{T} \Big\{ \hat{H}_{1,I}(t_1) \cdots \hat{H}_{1,I}(t_n) \Big\} \dd{t_1} \cdots \dd{t_n} - \\ - &= \sum_{n = 0}^\infty \frac{1}{n!} \frac{1}{(i \hbar)^n} - \mathcal{T} \bigg\{ \bigg( \int_{t_0}^{t} \hat{H}_{1,I}(t') \dd{t'} \bigg)^n \bigg\} + K_I(t) + = 1 + \frac{1}{i \hbar} \int_0^t \hat{H}_{1,I}(\tau) \: \hat{K}_I(\tau) \dd{\tau} \end{aligned}$$ -This construction is occasionally called the **Dyson series**. -We recognize the well-known Taylor expansion of $$\exp(x)$$, -leading us to a final expression for $$\hat{K}_I$$: +This equation can be recursively inserted into itself forever. +We recognize the resulting so called *Dyson series* +from the derivation of $$\hat{K}_S(t)$$ +for time-dependent Hamiltonians in the Schrödinger picture +([given here](/know/concept/time-evolution-operator/)), +so we know that the result is given by: $$\begin{aligned} \boxed{ - \hat{K}_I(t, t_0) - = \mathcal{T} \bigg\{ \exp\!\bigg( \frac{1}{i \hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'} \bigg) \bigg\} + \hat{K}_I(t) + = \mathcal{T} \bigg\{ \exp\!\bigg( \frac{1}{i \hbar} \int_0^t \hat{H}_{1,I}(\tau) \dd{\tau} \bigg) \bigg\} } \end{aligned}$$ +Where $$\mathcal{T}$$ is the +[time-ordering meta-operator](/know/concept/time-ordered-product/), +which is conventionally written in this way +to say that it applies to the terms of a Taylor expansion of $$\exp(x)$$. +This means that the evolution of a quantum state in the interaction picture +is determined by the perturbation $$\hat{H}_{1, I}$$. + ## References 1. H. Bruus, K. Flensberg, *Many-body quantum theory in condensed matter physics*, 2016, Oxford. - diff --git a/source/know/concept/self-steepening/index.md b/source/know/concept/self-steepening/index.md index 409f6c9..015aa40 100644 --- a/source/know/concept/self-steepening/index.md +++ b/source/know/concept/self-steepening/index.md @@ -97,7 +97,9 @@ $$E \equiv \int_{-\infty}^\infty |A|^2 \dd{t}$$ anymore, which is often used to quantify simulation errors. Fortunately, another value can then be used instead: it can be shown that the "photon number" $$N$$ -is still conserved, defined as: +is still conserved, defined like so, +where $$\omega$$ is the absolute frequency +(as opposed to the relative frequency $$\Omega$$): $$\begin{aligned} \boxed{ @@ -193,9 +195,9 @@ pulls the pulse apart before a shock can occur. The early steepening is observable though. A simulation of self-steepening without dispersion is illustrated below -for the following initial power distribution, +for the following Gaussian power distribution, with $$T_0 = 25\:\mathrm{fs}$$, $$P_0 = 3\:\mathrm{kW}$$, -$$\beta_2 = 0$$, $$\gamma = 0.1/\mathrm{W}/\mathrm{m}$$, +$$\beta_2 = 0$$, $$\gamma_0 = 0.1/\mathrm{W}/\mathrm{m}$$, and a vacuum carrier wavelength $$\lambda_0 \approx 73\:\mathrm{nm}$$ (the latter determined by the simulation's resolution settings): diff --git a/source/know/concept/time-evolution-operator/index.md b/source/know/concept/time-evolution-operator/index.md new file mode 100644 index 0000000..f489ac6 --- /dev/null +++ b/source/know/concept/time-evolution-operator/index.md @@ -0,0 +1,184 @@ +--- +title: "Time evolution operator" +sort_title: "Time evolution operator" +date: 2024-10-15 +categories: +- Quantum mechanics +- Physics +layout: "concept" +--- + +In general, given a system whose governing equation is known, +the **time evolution operator** $$\hat{U}(t, t_0)$$ +transforms the state at time $$t_0$$ to the one at time $$t$$. +Although not specific to it, +this is most often used in quantum mechanics, +as governed by the Schrödinger equation: + +$$\begin{aligned} + i \hbar \dv{}{t} \ket{\psi(t)} + = \hat{H}(t) \ket{\psi(t)} +\end{aligned}$$ + +Such that the definition of $$\hat{U}(t)$$ is as follows, +where we have set $$t_0 = 0$$: + +$$\begin{aligned} + \ket{\psi(t)} + = \hat{U}(t) \ket{\psi(0)} +\end{aligned}$$ + +Clearly, $$\hat{U}(t)$$ must be unitary. +The goal is to find an expression that satisfies this relation. + + + +## Time-independent Hamiltonian + +We start by inserting the definition of $$\hat{U}(t)$$ +into the Schrödinger equation: + +$$\begin{aligned} + \dv{}{t} \hat{U}(t) \ket{\psi(0)} + = - \frac{i}{\hbar} \hat{H} \: \hat{U}(t) \ket{\psi(0)} +\end{aligned}$$ + +If we hide the state $$\ket{\psi(0)}$$, +then $$\hat{U}(t)$$ can be said to satisfy the equation in its own right: + +$$\begin{aligned} + \dv{}{t} \hat{U}(t) + = - \frac{i}{\hbar} \hat{H} \: \hat{U}(t) +\end{aligned}$$ + +If the Hamiltonian $$\hat{H}$$ is time-independent, +this is straightforward to integrate, yielding: + +$$\begin{aligned} + \boxed{ + \hat{U}(t) + = \exp\!\bigg( \!-\! \frac{i}{\hbar} t \hat{H} \bigg) + } +\end{aligned}$$ + +And the generalization to $$t_0 \neq 0$$ is trivial, +since we can just shift the time axis: + +$$\begin{aligned} + \hat{U}(t, t_0) + = \exp\!\bigg( \!-\! \frac{i}{\hbar} (t - t_0) \hat{H} \bigg) +\end{aligned}$$ + + + +## Time-dependent Hamiltonian + +Even when $$\hat{H}$$ is time-dependent, +$$\hat{U}(t)$$ can be said to satisfy the Schrödinger equation: + +$$\begin{aligned} + \dv{}{t} \hat{U}(t) + = - \frac{i}{\hbar} \hat{H}(t) \: \hat{U}(t) +\end{aligned}$$ + +Integrating from $$0$$ to $$t$$, +and using $$\hat{U}(0) = 1$$ (which should be clear from its definition): + +$$\begin{aligned} + \hat{U}(t) + = 1 + \frac{1}{i \hbar} \int_0^t \hat{H}(\tau_1) \: \hat{U}(\tau_1) \dd{\tau_1} +\end{aligned}$$ + +This is a self-consistent equation for $$\hat{U}(t)$$. +We can recursively insert it into itself, yielding: + +$$\begin{aligned} + \hat{U}(t) + &= 1 + \frac{1}{i \hbar} \int_0^t \hat{H}(\tau_1) + \bigg( 1 + \frac{1}{i \hbar} \int_0^{\tau_1} \hat{H}(\tau_2) \: \hat{U}(\tau_2) \dd{\tau_2} \bigg) \dd{\tau_1} + \\ + &= 1 + \frac{1}{i \hbar} \int_0^t \hat{H}(\tau_1) \dd{\tau_1} + + \frac{1}{(i \hbar)^2} \int_0^t \hat{H}(\tau_1) \int_0^{\tau_1} \hat{H}(\tau_2) \: \hat{U}(\tau_2) \dd{\tau_2} \dd{\tau_1} + \\ + &= 1 + \frac{1}{i \hbar} \int_0^t \hat{H}(\tau_1) \dd{\tau_1} + + \frac{1}{(i \hbar)^2} \int_0^t \hat{H}(\tau_1) \int_0^{\tau_1} \hat{H}(\tau_2) \dd{\tau_2} \dd{\tau_1} + + \frac{1}{(i \hbar)^3} \int_0^t \cdots \: \dd{\tau_1} +\end{aligned}$$ + +And so on. +Let us take a closer look at the third (i.e. second-order) term in this series, +noting that the integrals are ordered such that $$\tau_2 < \tau_1$$ always. +We can exploit this fact to introduce several +[Heaviside step functions](/know/concept/heaviside-step-function/) $$\Theta(t)$$: + +$$\begin{aligned} + &\quad \int_0^t \hat{H}(\tau_1) \int_0^{\tau_1} \hat{H}(\tau_2) \dd{\tau_2} \dd{\tau_1} + \\ + &= \frac{1}{2} \int_0^t \hat{H}(\tau_1) \int_0^{\tau_1} \hat{H}(\tau_2) \dd{\tau_2} \dd{\tau_1} + + \frac{1}{2} \int_0^t \hat{H}(\tau_2) \int_0^{\tau_2} \hat{H}(\tau_1) \dd{\tau_1} \dd{\tau_2} + \\ + &= \frac{1}{2} \int_0^t \! \hat{H}(\tau_1) + \int_0^{\tau_1} \! \Theta(\tau_1 \!-\! \tau_2) \hat{H}(\tau_2) \dd{\tau_2} \dd{\tau_1} + + \frac{1}{2} \int_0^t \! \hat{H}(\tau_2) + \int_0^{\tau_2} \! \Theta(\tau_2 \!-\! \tau_1) \hat{H}(\tau_1) \dd{\tau_1} \dd{\tau_2} + \\ + &= \frac{1}{2} \int_0^t \int_0^t + \bigg( \Theta(\tau_1 \!-\! \tau_2) \: \hat{H}(\tau_1) \: \hat{H}(\tau_2) + + \Theta(\tau_2 \!-\! \tau_1) \: \hat{H}(\tau_1) \: \hat{H}(\tau_2) \bigg) \dd{\tau_1} \dd{\tau_2} + \\ + &= \frac{1}{2} \int_0^t \int_0^t \mathcal{T} \Big\{ \hat{H}(\tau_1) \: \hat{H}(\tau_2) \Big\} \dd{\tau_1} \dd{\tau_2} +\end{aligned}$$ + +Where we have recognized the +[time-ordering meta-operator](/know/concept/time-ordered-product/) $$\mathcal{T}$$. +The above procedure is easy to generalize to the higher-order terms, +so we arrive at the following expression for $$\hat{U}(t)$$: + +$$\begin{aligned} + \hat{U}(t) + &= 1 + \frac{1}{i \hbar} \int_0^t \hat{H}(\tau_1) \dd{\tau_1} + + \frac{1}{2} \frac{1}{(i \hbar)^2} \iint_0^t \mathcal{T} \Big\{ \hat{H}(\tau_1) \: \hat{H}(\tau_2) \Big\} \dd{\tau_2} \dd{\tau_1} + \\ + &\qquad+ \frac{1}{6} \frac{1}{(i \hbar)^3} \iiint_0^t + \mathcal{T} \Big\{ \hat{H}(\tau_1) \: \hat{H}(\tau_2) \: \hat{H}(\tau_3) \Big\} \dd{\tau_3} \dd{\tau_2} \dd{\tau_1} + + \: ... + \\ + &= 1 + \sum_{n = 1}^\infty \frac{1}{n!} \frac{1}{(i \hbar)^n} \int_0^t \!\cdots\! \int_0^t + \mathcal{T} \Big\{ \hat{H}(\tau_1) \cdots \hat{H}(\tau_n) \Big\} \dd{\tau_n} \cdots \dd{\tau_1} +\end{aligned}$$ + +This result is sometimes called a **Dyson series**. +Convention allows us to write it as follows, +despite such a use of $$\mathcal{T}$$ looking a bit strange: + +$$\begin{aligned} + \hat{U}(t) + &= 1 + \sum_{n = 1}^\infty \frac{1}{n!} \frac{1}{(i \hbar)^n} + \mathcal{T} \bigg\{ \bigg( \int_0^t \hat{H}(\tau) \dd{\tau} \bigg)^n \bigg\} +\end{aligned}$$ + +Here, we recognize the Taylor expansion of $$\exp(x)$$, +leading us to the desired result: + +$$\begin{aligned} + \boxed{ + \hat{U}(t) + = \mathcal{T} \bigg\{ \exp\!\bigg( \!-\! \frac{i}{\hbar} \int_0^t \hat{H}(\tau) \dd{\tau} \bigg) \bigg\} + } +\end{aligned}$$ + +Where once again $$\mathcal{T}$$ is being used according to convention. +Finally, the time axis can be shifted arbitrarily, +so many authors write the evolution operator from $$t_0$$ to $$t$$ as $$\hat{U}(t, t_0)$$: + +$$\begin{aligned} + \hat{U}(t, t_0) + = \mathcal{T} \bigg\{ \exp\!\bigg( \!-\! \frac{i}{\hbar} \int_{t_0}^t \hat{H}(\tau) \dd{\tau} \bigg) \bigg\} +\end{aligned}$$ + + + +## References +1. H. Bruus, K. Flensberg, + *Many-body quantum theory in condensed matter physics*, + 2016, Oxford. -- cgit v1.2.3