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diff --git a/content/know/_index.md b/content/know/_index.md index 41aeef6..8993ee1 100644 --- a/content/know/_index.md +++ b/content/know/_index.md @@ -16,14 +16,11 @@ trying to make sense of obscure, poorly explained concepts. To help me remember what I learn, I write notes in LaTeX. This knowledge base is based on my notes, and -is freely available to anyone who might need it. +is freely available to anyone who needs it. I hope it helps you in your studies or work. -Currently there's isn't much here yet, -but I have over 200 pages of LaTeX waiting to be converted. -Keep in mind that I'm only human, -so there are probably mistakes in my work. -I take no responsibility for any injuries incurred as a consequence. +Keep in mind that I'm only human, so there are probably mistakes here. +I take no responsibility for any resulting injuries. If you're doing something important, you should check things yourself! diff --git a/content/know/concept/calculus-of-variations/index.pdc b/content/know/concept/calculus-of-variations/index.pdc new file mode 100644 index 0000000..fb043e0 --- /dev/null +++ b/content/know/concept/calculus-of-variations/index.pdc @@ -0,0 +1,236 @@ +--- +title: "Calculus of variations" +firstLetter: "C" +publishDate: 2021-02-24 +categories: +- Mathematics +- Physics + +date: 2021-02-24T18:50:06+01:00 +draft: false +markup: pandoc +--- + +# Calculus of variations + +The **calculus of variations** lays the mathematical groundwork +for Lagrangian mechanics. + +Consider a **functional** $J$, mapping a function $f(x)$ to a scalar value +by integrating over the so-called **Lagrangian** $L$, +which represents an expression involving $x$, $f$ and the derivative $f'$: + +$$\begin{aligned} + J[f] = \int_{x_0}^{x_1} L(f, f', x) \dd{x} +\end{aligned}$$ + +If $J$ in some way measures the physical "cost" (e.g. energy) of +the path $f(x)$ taken by a physical system, +the **principle of least action** states that $f$ will be a minimum of $J[f]$, +so for example the expended energy will be minimized. + +If $f(x, \alpha\!=\!0)$ is the optimal route, then a slightly +different (and therefore worse) path between the same two points can be expressed +using the parameter $\alpha$: + +$$\begin{aligned} + f(x, \alpha) = f(x, 0) + \alpha \eta(x) + \qquad \mathrm{or} \qquad + \delta f = \alpha \eta(x) +\end{aligned}$$ + +Where $\eta(x)$ is an arbitrary differentiable deviation. +Since $f(x, \alpha)$ must start and end in the same points as $f(x,0)$, +we have the boundary conditions: + +$$\begin{aligned} + \eta(x_0) = \eta(x_1) = 0 +\end{aligned}$$ + +Given $L$, the goal is to find an equation for the optimal path $f(x,0)$. +Just like when finding the minimum of a real function, +the minimum $f$ of a functional $J[f]$ is a stationary point +with respect to the deviation weight $\alpha$, +a condition often written as $\delta J = 0$. +In the following, the integration limits have been omitted: + +$$\begin{aligned} + 0 + &= \delta J + = \pdv{J}{\alpha} \Big|_{\alpha = 0} + = \int \pdv{L}{\alpha} \dd{x} + = \int \pdv{L}{f} \pdv{f}{\alpha} + \pdv{L}{f'} \pdv{f'}{\alpha} \dd{x} + \\ + &= \int \pdv{L}{f} \eta + \pdv{L}{f'} \eta' \dd{x} + = \Big[ \pdv{L}{f'} \eta \Big]_{x_0}^{x_1} + \int \pdv{L}{f} \eta - \frac{d}{dx} \Big( \pdv{L}{f'} \Big) \eta \dd{x} +\end{aligned}$$ + +The boundary term from partial integration vanishes due to the boundary +conditions for $\eta(x)$. We are thus left with: + +$$\begin{aligned} + 0 + = \int \eta \bigg( \pdv{L}{f} - \dv{x} \Big( \pdv{L}{f'} \Big) \bigg) \dd{x} +\end{aligned}$$ + +This holds for all $\eta$, but $\eta$ is arbitrary, so in fact +only the parenthesized expression matters: + +$$\begin{aligned} + \boxed{ + 0 = \pdv{L}{f} - \dv{x} \Big( \pdv{L}{f'} \Big) + } +\end{aligned}$$ + +This is known as the **Euler-Lagrange equation** of the Lagrangian $L$, +and its solutions represent the optimal paths $f(x, 0)$. + + +## Multiple functions + +Suppose that the Lagrangian $L$ depends on multiple independent functions +$f_1, f_2, ..., f_N$: + +$$\begin{aligned} + J[f_1, ..., f_N] = \int_{x_0}^{x_1} L(f_1, ..., f_N, f_1', ..., f_N', x) \dd{x} +\end{aligned}$$ + +In this case, every $f_n(x)$ has its own deviation $\eta_n(x)$, +satisfying $\eta_n(x_0) = \eta_n(x_1) = 0$: + +$$\begin{aligned} + f_n(x, \alpha) = f_n(x, 0) + \alpha \eta_n(x) +\end{aligned}$$ + +The derivation procedure is identical to the case $N = 1$ from earlier: + +$$\begin{aligned} + 0 + &= \pdv{J}{\alpha} \Big|_{\alpha = 0} + = \int \pdv{L}{\alpha} \dd{x} + = \int \sum_{n} \Big( \pdv{L}{f_n} \pdv{f_n}{\alpha} + \pdv{L}{f_n'} \pdv{f_n'}{\alpha} \Big) \dd{x} + \\ + &= \int \sum_{n} \Big( \pdv{L}{f_n} \eta_n + \pdv{L}{f_n'} \eta_n' \Big) \dd{x} + \\ + &= \Big[ \sum_{n} \pdv{L}{f_n'} \eta_n \Big]_{x_0}^{x_1} + + \int \sum_{n} \eta_n \bigg( \pdv{L}{f_n} - \frac{d}{dx} \Big( \pdv{L}{f_n'} \Big) \bigg) \dd{x} +\end{aligned}$$ + +Once again, $\eta_n(x)$ is arbitrary and disappears at the boundaries, +so we end up with $N$ equations of the same form as for a single function: + +$$\begin{aligned} + \boxed{ + 0 = \pdv{L}{f_1} - \dv{x} \Big( \pdv{L}{f_1'} \Big) + \quad \cdots \quad + 0 = \pdv{L}{f_N} - \dv{x} \Big( \pdv{L}{f_N'} \Big) + } +\end{aligned}$$ + + +## Higher-order derivatives + +Suppose that the Lagrangian $L$ depends on multiple derivatives of $f(x)$: + +$$\begin{aligned} + J[f] = \int_{x_0}^{x_1} L(f, f', f'', ..., f^{(N)}, x) \dd{x} +\end{aligned}$$ + +Once again, the derivation procedure is the same as before: + +$$\begin{aligned} + 0 + &= \pdv{J}{\alpha} \Big|_{\alpha = 0} + = \int \pdv{L}{\alpha} \dd{x} + = \int \pdv{L}{f} \pdv{f}{\alpha} + \sum_{n} \pdv{L}{f^{(n)}} \pdv{f^{(n)}}{\alpha} \dd{x} + \\ + &= \int \pdv{L}{f} \eta + \sum_{n} \pdv{L}{f^{(n)}} \eta^{(n)} \dd{x} +\end{aligned}$$ + +The goal is to turn each $\eta^{(n)}(x)$ into $\eta(x)$, so we need to +partially integrate the $n$th term of the sum $n$ times. In this case, +we will need some additional boundary conditions for $\eta(x)$: + +$$\begin{aligned} + \eta'(x_0) = \eta'(x_1) = 0 + \qquad \cdots \qquad + \eta^{(N-1)}(x_0) = \eta^{(N-1)}(x_1) = 0 +\end{aligned}$$ + +This eliminates the boundary terms from partial integration, leaving: + +$$\begin{aligned} + 0 + &= \int \eta \bigg( \pdv{L}{f} + \sum_{n} (-1)^n \dv[n]{x} \Big( \pdv{L}{f^{(n)}} \Big) \bigg) \dd{x} +\end{aligned}$$ + +Once again, because $\eta(x)$ is arbitrary, the Euler-Lagrange equation becomes: + +$$\begin{aligned} + \boxed{ + 0 = \pdv{L}{f} + \sum_{n} (-1)^n \dv[n]{x} \Big( \pdv{L}{f^{(n)}} \Big) + } +\end{aligned}$$ + + +## Multiple coordinates + +Suppose now that $f$ is a function of multiple variables. +For brevity, we only consider two variables $x$ and $y$, +but the results generalize effortlessly to larger amounts. +The Lagrangian now depends on all the partial derivatives of $f(x, y)$: + +$$\begin{aligned} + J[f] = \iint_{(x_0, y_0)}^{(x_1, y_1)} L(f, f_x, f_y, x, y) \dd{x} \dd{y} +\end{aligned}$$ + +The arbitrary deviation $\eta$ is then also a function of multiple variables: + +$$\begin{aligned} + f(x, y; \alpha) = f(x, y; 0) + \alpha \eta(x, y) +\end{aligned}$$ + +The derivation procedure starts in the exact same way as before: + +$$\begin{aligned} + 0 + &= \pdv{J}{\alpha} \Big|_{\alpha = 0} + = \iint \pdv{L}{\alpha} \dd{x} \dd{y} + \\ + &= \iint \pdv{L}{f} \pdv{f}{\alpha} + \pdv{L}{f_x} \pdv{f_x}{\alpha} + \pdv{L}{f_y} \pdv{f_y}{\alpha} \dd{x} \dd{y} + \\ + &= \iint \pdv{L}{f} \eta + \pdv{L}{f_x} \eta_x + \pdv{L}{f_y} \eta_y \dd{x} \dd{y} +\end{aligned}$$ + +We partially integrate for both $\eta_x$ and $\eta_y$, yielding: + +$$\begin{aligned} + 0 + &= \int \Big[ \pdv{L}{f_x} \eta \Big]_{x_0}^{x_1} \dd{y} + \int \Big[ \pdv{L}{f_y} \eta \Big]_{y_0}^{y_1} \dd{x} + \\ + &\quad + \iint \eta \bigg( \pdv{L}{f} - \dv{x} \Big( \pdv{L}{f_x} \Big) - \dv{y} \Big( \pdv{L}{f_y} \Big) \bigg) \dd{x} \dd{y} +\end{aligned}$$ + +But now, to eliminate these boundary terms, we need extra conditions for $\eta$: + +$$\begin{aligned} + \forall y: \eta(x_0, y) = \eta(x_1, y) = 0 + \qquad + \forall x: \eta(x, y_0) = \eta(x, y_1) = 0 +\end{aligned}$$ + +In other words, the deviation $\eta$ must be zero on the whole "box". +Again relying on the fact that $\eta$ is arbitrary, the Euler-Lagrange +equation is: + +$$\begin{aligned} + 0 = \pdv{L}{f} - \dv{x} \Big( \pdv{L}{f_x} \Big) - \dv{y} \Big( \pdv{L}{f_y} \Big) +\end{aligned}$$ + +This generalizes nicely to functions of even more variables $x_1, x_2, ..., x_N$: + +$$\begin{aligned} + \boxed{ + 0 = \pdv{L}{f} - \sum_{n} \dv{x_n} \Big( \pdv{L}{f_{x_n}} \Big) + } +\end{aligned}$$ diff --git a/content/know/concept/ehrenfests-theorem/index.pdc b/content/know/concept/ehrenfests-theorem/index.pdc new file mode 100644 index 0000000..bdcb908 --- /dev/null +++ b/content/know/concept/ehrenfests-theorem/index.pdc @@ -0,0 +1,137 @@ +--- +title: "Ehrenfest's theorem" +firstLetter: "E" +publishDate: 2021-02-24 +categories: +- Quantum mechanics +- Physics + +date: 2021-02-24T14:53:13+01:00 +draft: false +markup: pandoc +--- + +# Ehrenfest's theorem + +In quantum mechanics, **Ehrenfest's theorem** gives a general expression for the +time evolution of an observable's expectation value $\expval*{\hat{L}}$. + +The time-dependent Schrödinger equation is as follows, +where prime denotes differentiation with respect to time $t$: + +$$\begin{aligned} + \ket{\psi'} = \frac{1}{i \hbar} \hat{H} \ket{\psi} + \qquad + \bra{\psi'} = - \frac{1}{i \hbar} \bra{\psi} \hat{H} +\end{aligned}$$ + +Given an observable operator $\hat{L}$ and a state $\ket{\psi}$, +the time-derivative of the expectation value $\expval*{\hat{L}}$ is as follows +(due to the product rule of differentiation): + +$$\begin{aligned} + \dv{\expval*{\hat{L}}}{t} + &= \matrixel{\psi}{\hat{L}}{\psi'} + \matrixel{\psi'}{\hat{L}}{\psi} + \matrixel{\psi}{\hat{L}'}{\psi} + \\ + &= \frac{1}{i \hbar} \matrixel{\psi}{\hat{L}\hat{H}}{\psi} + - \frac{1}{i \hbar} \matrixel{\psi}{\hat{H}\hat{L}}{\psi} + + \expval{\dv{\hat{L}}{t}} +\end{aligned}$$ + +The first two terms on the right can be rewritten using a commutator, +yielding the general form of Ehrenfest's theorem: + +$$\begin{aligned} + \boxed{ + \dv{\expval*{\hat{L}}}{t} + = \frac{1}{i \hbar} \expval{[\hat{L}, \hat{H}]} + \expval{\dv{\hat{L}}{t}} + } +\end{aligned}$$ + +In practice, since most operators are time-independent, +the last term often vanishes. + +As a interesting side note, in the [Heisenberg picture](/know/concept/heisenberg-picture/), +this relation proves itself, +when one simply wraps all terms in $\bra{\psi}$ and $\ket{\psi}$. + +Two observables of particular interest are the position $\hat{X}$ and momentum $\hat{P}$. +Applying the above theorem to $\hat{X}$ yields the following, +which we reduce using the fact that $\hat{X}$ commutes +with the potential $V(\hat{X})$, +because one is a function of the other: + +$$\begin{aligned} + \dv{\expval*{\hat{X}}}{t} + &= \frac{1}{i \hbar} \expval{[\hat{X}, \hat{H}]} + = \frac{1}{2 i \hbar m} \expval{[\hat{X}, \hat{P}^2] + 2 m [\hat{X}, V(\hat{X})]} + = \frac{1}{2 i \hbar m} \expval{[\hat{X}, \hat{P}^2]} + \\ + &= \frac{1}{2 i \hbar m} \expval{\hat{P} [\hat{X}, \hat{P}] + [\hat{X}, \hat{P}] \hat{P}} + = \frac{2 i \hbar}{2 i \hbar m} \expval*{\hat{P}} + = \frac{\expval*{\hat{P}}}{m} +\end{aligned}$$ + +This is the first part of the "original" form of Ehrenfest's theorem, +which is reminiscent of classical Newtonian mechanics: + +$$\begin{gathered} + \boxed{ + \dv{\expval*{\hat{X}}}{t} = \frac{\expval*{\hat{P}}}{m} + } +\end{gathered}$$ + +Next, applying the general formula to the expected momentum $\expval*{\hat{P}}$ +gives us: + +$$\begin{aligned} + \dv{\expval*{\hat{P}}}{t} + &= \frac{1}{i \hbar} \expval{[\hat{P}, \hat{H}]} + = \frac{1}{2 i \hbar m} \expval{[\hat{P}, \hat{P}^2] + 2 m [\hat{P}, V(\hat{X})]} + = \frac{1}{i \hbar} \expval{[\hat{P}, V(\hat{X})]} +\end{aligned}$$ + +To find the commutator, we go to the $\hat{X}$-basis and use a test +function $f(x)$: + +$$\begin{aligned} + \comm{- i \hbar \dv{x}}{V(x)} \: f(x) + &= - i \hbar \frac{dV}{dx} f(x) - i \hbar V(x) \frac{df}{dx} + i \hbar V(x) \frac{df}{dx} + = - i \hbar \frac{dV}{dx} f(x) +\end{aligned}$$ + +By inserting this result back into the previous equation, we find the following: + +$$\begin{aligned} + \dv{\expval*{\hat{P}}}{t} + &= - \frac{i \hbar}{i \hbar} \expval{\frac{d V}{d \hat{X}}} + = - \expval{\frac{d V}{d \hat{X}}} +\end{aligned}$$ + +This is the second part of Ehrenfest's theorem, +which is also similar to Newtonian mechanics: + +$$\begin{gathered} + \boxed{ + \dv{\expval*{\hat{P}}}{t} = - \expval{\pdv{V}{\hat{X}}} + } +\end{gathered}$$ + +There is an important consequence of Ehrenfest's original theorems +for the symbolic derivatives of the Hamiltonian $\hat{H}$ +with respect to $\hat{X}$ and $\hat{P}$: + +$$\begin{gathered} + \boxed{ + \expval{\pdv{\hat{H}}{\hat{P}}} + = \dv{\expval*{\hat{X}}}{t} + } + \qquad \quad + \boxed{ + - \expval{\pdv{\hat{H}}{\hat{X}}} + = \dv{\expval*{\hat{P}}}{t} + } +\end{gathered}$$ + +These are easy to prove yourself, +and are analogous to Hamilton's canonical equations. diff --git a/content/know/concept/heisenberg-picture/index.pdc b/content/know/concept/heisenberg-picture/index.pdc new file mode 100644 index 0000000..2dd4118 --- /dev/null +++ b/content/know/concept/heisenberg-picture/index.pdc @@ -0,0 +1,117 @@ +--- +title: "Heisenberg picture" +firstLetter: "H" +publishDate: 2021-02-24 +categories: +- Quantum mechanics +- Physics + +date: 2021-02-24T16:46:26+01:00 +draft: false +markup: pandoc +--- + +# Heisenberg picture + +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 +$\hat{U}(t) = \exp{} (- i t \hat{H} / \hbar)$ like so: + +$$\begin{aligned} + \ket{\psi_S(t)} = \hat{U}(t) \ket{\psi_S(0)} +\end{aligned}$$ + +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. + +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: + +$$\begin{aligned} + \boxed{ + \ket{\psi_H} = \ket{\psi_S(0)} + \qquad + \hat{L}_H(t) = \hat{U}^\dagger(t) \hat{L}_S(t) \hat{U}(t) + } +\end{aligned}$$ + +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](/know/concept/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 $\hat{P} = - i \hbar \: d/d\hat{X}$ +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](/know/concept/ehrenfests-theorem/), +which is closely related to the Heisenberg picture. diff --git a/content/know/concept/hilbert-space/index.pdc b/content/know/concept/hilbert-space/index.pdc index 1faf08a..c557fb7 100644 --- a/content/know/concept/hilbert-space/index.pdc +++ b/content/know/concept/hilbert-space/index.pdc @@ -13,7 +13,7 @@ markup: pandoc # Hilbert space -A **Hilbert space**, also known as an **inner product space**, is an +A **Hilbert space**, also called an **inner product space**, is an abstract **vector space** with a notion of length and angle. |