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| -rw-r--r-- | content/know/concept/harmonic-oscillator/index.pdc | 3 | ||||
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diff --git a/content/know/category/classical-mechanics.md b/content/know/category/classical-mechanics.md new file mode 100644 index 0000000..2180143 --- /dev/null +++ b/content/know/category/classical-mechanics.md @@ -0,0 +1,9 @@ +--- +title: "Classical mechanics" +firstLetter: "C" +date: 2021-07-04T14:10:06+02:00 +draft: false +layout: "category" +--- + +This page will fill itself. diff --git a/content/know/concept/convolution-theorem/index.pdc b/content/know/concept/convolution-theorem/index.pdc index 1454cc0..2712c21 100644 --- a/content/know/concept/convolution-theorem/index.pdc +++ b/content/know/concept/convolution-theorem/index.pdc @@ -69,7 +69,8 @@ $$\begin{aligned} ## Laplace transform For functions $f(t)$ and $g(t)$ which are only defined for $t \ge 0$, -the convolution theorem can also be stated using the Laplace transform: +the convolution theorem can also be stated using +the [Laplace transform](/know/concept/laplace-transform/): $$\begin{aligned} \boxed{(f * g)(t) = \hat{\mathcal{L}}^{-1}\{\tilde{f}(s) \: \tilde{g}(s)\}} diff --git a/content/know/concept/hamiltonian-mechanics/index.pdc b/content/know/concept/hamiltonian-mechanics/index.pdc new file mode 100644 index 0000000..0a7306e --- /dev/null +++ b/content/know/concept/hamiltonian-mechanics/index.pdc @@ -0,0 +1,312 @@ +--- +title: "Hamiltonian mechanics" +firstLetter: "H" +publishDate: 2021-07-03 +categories: +- Physics +- Classical mechanics + +date: 2021-07-03T14:39:14+02:00 +draft: false +markup: pandoc +--- + +# Hamiltonian mechanics + +**Hamiltonian mechanics** is an alternative formulation of classical mechanics, +which equivalent to Newton's laws, +but often mathematically advantageous. +It is built on the shoulders of [Lagrangian mechanics](/know/concept/lagrangian-mechanics/), +which is in turn built on [variational calculus](/know/concept/calculus-of-variations/). + + +## Definitions + +In Lagrangian mechanics, use a Lagrangian $L$, +which depends on position $q(t)$ and velocity $\dot{q}(t)$, +to define the momentum $p(t)$ as a derived quantity. +Hamiltonian mechanics switches the roles of $\dot{q}$ and $p$: +the **Hamiltonian** $H$ is a function of $q$ and $p$, +and the velocity $\dot{q}$ is derived from it: + +$$\begin{aligned} + \pdv{L(q, \dot{q})}{\dot{q}} = p + \qquad \quad + \pdv{H(q, p)}{p} \equiv \dot{q} +\end{aligned}$$ + +Conveniently, this switch turns out to be +[Legendre transformation](/know/concept/legendre-transform/): +$H$ is the Legendre transform of $L$, +with $p = \partial L / \partial \dot{q}$ taken as +the coordinate to replace $\dot{q}$. +Therefore: + +$$\begin{aligned} + \boxed{ + H(q, p) \equiv \dot{q} \: p - L(q, \dot{q}) + } +\end{aligned}$$ + +This almost always works, +because $L$ is usually a second-order polynomial of $\dot{q}$, +and thus convex as required for Legendre transformation. +In the above expression, +$\dot{q}$ must be rewritten in terms of $p$ and $q$, +which is trivial, since $p$ is proportional to $\dot{q}$ by definition. + +The Hamiltonian $H$ also has a direct physical meaning: +for a mass $m$, and for $L = T - V$, +it is straightforward to show that $H$ represents the total energy $T + V$: + +$$\begin{aligned} + H + = \dot{q} \: p - L + = m \dot{q}^2 - L + = 2 T - (T - V) + = T + V +\end{aligned}$$ + +Just as Lagrangian mechanics, +Hamiltonian mechanics scales well for large systems. +Its definition is generalized as follows to $N$ objects, +where $p$ is shorthand for $p_1, ..., p_N$: + +$$\begin{aligned} + \boxed{ + H(q, p) + \equiv \bigg( \sum_{n = 1}^N \dot{q}_n \: p_n \bigg) - L(q, \dot{q}) + } +\end{aligned}$$ + +The positions and momenta $(q, p)$ form a phase space, +i.e. they fully describe the state. + +An extremely useful concept in Hamiltonian mechanics +is the **Poisson bracket** (PB), +which is a binary operation on two quantities $A(q, p)$ and $B(q, p)$, +denoted by $\{A, B\}$: + +$$\begin{aligned} + \boxed{ + \{ A, B \} + \equiv \sum_{n = 1}^N \Big( \pdv{A}{q_n} \pdv{B}{p_n} - \pdv{A}{p_n} \pdv{B}{q_n} \Big) + } +\end{aligned}$$ + + +## Canonical equations + +Lagrangian mechanics has a single Euler-Lagrange equation per object, +yielding $N$ second-order equations of motion in total. +In contrast, Hamiltonian mechanics has $2 N$ first-order equations of motion, +known as **Hamilton's canonical equations**: + +$$\begin{aligned} + \boxed{ + - \pdv{H}{q_n} = \dot{p}_n + \qquad + \pdv{H}{p_n} = \dot{q}_n + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-canoneq"/> +<label for="proof-canoneq">Proof</label> +<div class="hidden"> +<label for="proof-canoneq">Proof.</label> +For the first equation, +we differentiate $H$ with respect to $q_n$, +and use the chain rule: +$$\begin{aligned} + \pdv{H}{q_n} + &= \pdv{q_n} \Big( \sum_{j} \dot{q}_j \: p_j - L \Big) + \\ + &= \sum_{j} \bigg( \Big( \dot{q}_j \pdv{p_j}{q_n} + p_j \pdv{\dot{q}_j}{q_n} \Big) + - \Big( \pdv{L}{q_n} + \pdv{L}{\dot{q}_j} \pdv{\dot{q}_j}{q_n} \Big) \bigg) + \\ + &= \sum_{j} \Big( p_j \pdv{\dot{q}_j}{q_n} - \pdv{L}{q_n} - p_j \pdv{\dot{q}_j}{q_n} \Big) + = - \pdv{L}{q_n} +\end{aligned}$$ + +We use the Euler-Lagrange equation here, +leading to the desired equation: + +$$\begin{aligned} + - \pdv{L}{q_n} = - \dv{t} \Big( \pdv{L}{\dot{q}_n} \Big) = - \dv{p_n}{t} = - \dot{p}_n +\end{aligned}$$ + +The second equation is somewhat trivial, +since $H$ is defined to satisfy it in the first place. +Nevertheless, we can prove it by brute force, +using the same approach as above: +$$\begin{aligned} + \pdv{H}{p_n} + &= \pdv{p_n} \Big( \sum_{j} \dot{q}_j \: p_j - L \Big) + \\ + &= \sum_{j} \bigg( \Big( \dot{q}_j \pdv{p_j}{p_n} + p_j \pdv{\dot{q}_j}{p_n} \Big) + - \Big( \pdv{L}{q_j} \pdv{q_j}{p_n} + \pdv{L}{\dot{q}_j} \pdv{\dot{q}_j}{p_n} \Big) \bigg) + \\ + &= \dot{q}_n + \sum_{j} \Big( p_j \pdv{\dot{q}_j}{p_n} + - 0 \pdv{L}{q_j} - p_j \pdv{\dot{q}_j}{p_n} \Big) + = \dot{q}_n +\end{aligned}$$ +</div> +</div> + +Just like in Lagrangian mechanics, if $H$ does not explicitly contain $q_n$, +then $q_n$ is called a **cyclic coordinate**, and leads to the conservation of $p_n$: + +$$\begin{aligned} + \dot{p}_n = - \pdv{H}{q_n} = 0 + \quad \implies \quad + p_n = \mathrm{conserved} +\end{aligned}$$ + +Of course, there may be other conserved quantities. +Generally speaking, the $t$-derivative of an arbitrary quantity $A(q, p, t)$ is as follows, +where $\pdv*{t}$ is a "soft" derivative +(only affects explicit occurrences of $t$), +and $\dv*{t}$ is a "hard" derivative +(also affects implicit $t$ inside $q$ and $p$): + +$$\begin{aligned} + \boxed{ + \dv{A}{t} + = \{ A, H \} + \pdv{A}{t} + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-diff-t"/> +<label for="proof-diff-t">Proof</label> +<div class="hidden"> +<label for="proof-diff-t">Proof.</label> +We differentiate via the multivariate chain rule, +insert the canonical equations, +and eventually recognize the PB definition: + +$$\begin{aligned} + \dv{A}{t} + &= \sum_{n} \Big( \pdv{A}{q_n} \pdv{q_n}{t} + \pdv{A}{p_n} \pdv{p_n}{t} \Big) + \pdv{A}{t} + \\ + &= \sum_{n} \Big( \pdv{A}{q_n} \dot{q}_n + \pdv{A}{p_n} \dot{p}_n \Big) + \pdv{A}{t} + \\ + &= \sum_{n} \Big( \pdv{A}{q_n} \pdv{H}{p_n} - \pdv{A}{p_n} \pdv{H}{q_n} \Big) + \pdv{A}{t} +\end{aligned}$$ +</div> +</div> + +Assuming that $H$ does not explicitly depend on $t$, +the above property naturally leads us to an alternative +way of writing Hamilton's canonical equations: + +$$\begin{aligned} + \dot{q}_n = \{ q_n, H \} + \qquad \quad + \dot{p}_n = \{ p_n, H \} +\end{aligned}$$ + + + +## Canonical coordinates + +So far, we have assumed that the phase space coordinates $(q, p)$ +are the *positions* and *canonical momenta*, respectively, +and that led us to Hamilton's canonical equations. + +In theory, we could make a transformation of the following general form: + +$$\begin{aligned} + q \to Q(q, p) + \qquad \quad + p \to P(q, p) +\end{aligned}$$ + +However, most choices of $(Q, P)$ would not preserve Hamilton's equations. +Any $(Q, P)$ that do keep this form +are known as **canonical coordinates**, +and the corresponding transformation is a **canonical transformation**. +That is, any $(Q, P)$ that satisfy: + +$$\begin{aligned} + - \pdv{H}{Q_n} = \dot{P}_n + \qquad \quad + \pdv{H}{P_n} = \dot{Q}_n +\end{aligned}$$ + +Then we might as well write $H(q, p)$ as $H(Q, P)$. +So, which $(Q, P)$ fulfill this? +It turns out that the following must be satisfied for all $n, j$, +where $\delta_{nj}$ is the Kronecker delta: + +$$\begin{aligned} + \boxed{ + \{ Q_n, Q_j \} = \{ P_n, P_j \} = 0 + \qquad + \{ Q_n, P_j \} = \delta_{nj} + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-cantrans"/> +<label for="proof-cantrans">Proof</label> +<div class="hidden"> +<label for="proof-cantrans">Proof.</label> +Assuming that $Q_n$, $P_n$ and $H$ do not explicitly depend on $t$, +we use our expression for the $t$-derivative of an arbitrary quantity, +and apply the multivariate chain rule to it: + +$$\begin{aligned} + \dot{Q}_n + &= \{Q_n, H\} + = \sum_{n} \bigg( \pdv{Q_n}{q_n} \pdv{H}{p_n} - \pdv{Q_n}{p_n} \pdv{H}{q_n} \bigg) + \\ + &= \sum_{n, j} \bigg( \pdv{Q_n}{q_n} \Big( \pdv{H}{Q_j} \pdv{Q_j}{p_n} + \pdv{H}{P_j} \pdv{P_j}{p_n} \Big) + - \pdv{Q_n}{p_n} \Big( \pdv{H}{Q_j} \pdv{Q_j}{q_n} + \pdv{H}{P_j} \pdv{P_j}{q_n} \Big) \bigg) + \\ + &= \sum_{n, j} \bigg( \pdv{H}{Q_j} \Big( \pdv{Q_n}{q_n} \pdv{Q_j}{p_n} - \pdv{Q_n}{p_n} \pdv{Q_j}{q_n} \Big) + + \pdv{H}{P_j} \Big( \pdv{Q_n}{q_n} \pdv{P_j}{p_n} - \pdv{Q_n}{p_n} \pdv{P_j}{q_n} \Big) \bigg) + \\ + &= \sum_{j} \bigg( \pdv{H}{Q_j} \{Q_n, Q_j\} + \pdv{H}{P_j} \{Q_n, P_j\} \bigg) +\end{aligned}$$ + +This is equivalent to Hamilton's equation $\dot{Q}_n = \pdv*{H}{P_n}$ +if and only if $\{Q_n, Q_j\} = 0$ for all $n$ and $j$, +and if $\{Q_n, P_j\} = \delta_{nj}$. + +Next, we do the exact same thing with $P_n$ instead of $Q_n$, +giving an analogous result: + +$$\begin{aligned} + \dot{P}_n + &= \{P_n, H\} + = \sum_{n} \bigg( \pdv{P_n}{q_n} \pdv{H}{p_n} - \pdv{P_n}{p_n} \pdv{H}{q_n} \bigg) + \\ + &= \sum_{n, j} \bigg( \pdv{P_n}{q_n} \Big( \pdv{H}{Q_j} \pdv{Q_j}{p_n} + \pdv{H}{P_j} \pdv{P_j}{p_n} \Big) + - \pdv{P_n}{p_n} \Big( \pdv{H}{Q_j} \pdv{Q_j}{q_n} + \pdv{H}{P_j} \pdv{P_j}{q_n} \Big) \bigg) + \\ + &= \sum_{n, j} \bigg( \pdv{H}{Q_j} \Big( \pdv{P_n}{q_n} \pdv{Q_j}{p_n} - \pdv{P_n}{p_n} \pdv{Q_j}{q_n} \Big) + + \pdv{H}{P_j} \Big( \pdv{P_n}{q_n} \pdv{P_j}{p_n} - \pdv{P_n}{p_n} \pdv{P_j}{q_n} \Big) \bigg) + \\ + &= \sum_{j} \bigg( \pdv{H}{Q_j} \{P_n, Q_j\} + \pdv{H}{P_j} \{P_n, P_j\} \bigg) +\end{aligned}$$ + +Which is equivalent to Hamilton's equation $\dot{P}_n = -\pdv*{H}{Q_n}$ +if and only if $\{P_n, P_j\} = 0$, +and $\{Q_n, P_j\} = - \delta_{nj}$. +The PB is anticommutative, +i.e. $\{A, B\} = - \{B, A\}$. +</div> +</div> + +If you have experience with quantum mechanics, +the latter equation should look suspiciously similar +to the *canonical commutation relation* $[\hat{Q}, \hat{P}] = i \hbar$. + + + +## References +1. R. Shankar, + *Principles of quantum mechanics*, 2nd edition, + Springer. diff --git a/content/know/concept/harmonic-oscillator/index.pdc b/content/know/concept/harmonic-oscillator/index.pdc index 5d97a24..4219602 100644 --- a/content/know/concept/harmonic-oscillator/index.pdc +++ b/content/know/concept/harmonic-oscillator/index.pdc @@ -68,7 +68,8 @@ $$\begin{aligned} x(t) = \cos\!(\omega_0 t) \end{aligned}$$ -When using Lagrangian mechanics or Hamiltonian mechanics, +When using [Lagrangian](/know/concept/lagrangian-mechanics/) +or Hamiltonian mechanics, we need to know the potential energy $V(x)$ added to the system by a displacement to $x$. This equals the work done by the displacement, diff --git a/content/know/concept/lagrangian-mechanics/index.pdc b/content/know/concept/lagrangian-mechanics/index.pdc index 24e1f93..dcf555b 100644 --- a/content/know/concept/lagrangian-mechanics/index.pdc +++ b/content/know/concept/lagrangian-mechanics/index.pdc @@ -4,6 +4,7 @@ firstLetter: "L" publishDate: 2021-07-01 categories: - Physics +- Classical mechanics date: 2021-07-01T18:44:43+02:00 draft: false @@ -16,7 +17,10 @@ markup: pandoc which is equivalent to Newton's laws, but offers some advantages. Its mathematical backbone is the -[calculus of variations](/know/concept/calculus-of-variations/). +[calculus of variations](/know/concept/calculus-of-variations/), +and hence it is built on the **principle of least action**, +which states that the path taken by a system +will be a minimum of the **action** (i.e. energy cost) of that path. For a moving object with position $x(t)$ and velocity $\dot{x}(t)$, we define the Lagrangian $L$ as the difference diff --git a/content/know/concept/laplace-transform/index.pdc b/content/know/concept/laplace-transform/index.pdc new file mode 100644 index 0000000..bd7673b --- /dev/null +++ b/content/know/concept/laplace-transform/index.pdc @@ -0,0 +1,125 @@ +--- +title: "Laplace transform" +firstLetter: "L" +publishDate: 2021-07-02 +categories: +- Mathematics +- Physics + +date: 2021-07-02T15:48:30+02:00 +draft: false +markup: pandoc +--- + +# Laplace transform + +The **Laplace transform** is an integral transform +that losslessly converts a function $f(t)$ of a real variable $t$, +into a function $\tilde{f}(s)$ of a complex variable $s$, +where $s$ is sometimes called the **complex frequency**, +analogously to the [Fourier transform](/know/concept/fourier-transform/). +The transform is defined as follows: + +$$\begin{aligned} + \boxed{ + \tilde{f}(s) + \equiv \hat{\mathcal{L}}\{f(t)\} + \equiv \int_0^\infty f(t) \exp\!(- s t) \dd{t} + } +\end{aligned}$$ + +Depending on $f(t)$, this integral may diverge. +This is solved by restricting the domain of $\tilde{f}(s)$ +to $s$ where $\mathrm{Re}\{s\} > s_0$, +for an $s_0$ large enough to compensate for the growth of $f(t)$. + + +## Derivatives + +The derivative of a transformed function is the transform +of the original mutliplied by its variable. +This is especially useful for transforming ODEs with variable coefficients: + +$$\begin{aligned} + \boxed{ + \tilde{f}'(s) = - \hat{\mathcal{L}}\{t f(t)\} + } +\end{aligned}$$ + +This property generalizes nicely to higher-order derivatives of $s$, so: + +$$\begin{aligned} + \boxed{ + \dv[n]{\tilde{f}}{s} = (-1)^n \hat{\mathcal{L}}\{t^n f(t)\} + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-dv-s"/> +<label for="proof-dv-s">Proof</label> +<div class="hidden"> +<label for="proof-dv-s">Proof.</label> +The exponential $\exp\!(- s t)$ is the only thing that depends on $s$ here: + +$$\begin{aligned} + \dv[n]{\tilde{f}}{s} + &= \dv[n]{s} \int_0^\infty f(t) \exp\!(- s t) \dd{t} + \\ + &= \int_0^\infty (-t)^n f(t) \exp\!(- s t) \dd{t} + = (-1)^n \hat{\mathcal{L}}\{t^n f(t)\} +\end{aligned}$$ +</div> +</div> + +The Laplace transform of a derivative introduces the initial conditions into the result. +Notice that $f(0)$ is the initial value in the original $t$-domain: + +$$\begin{aligned} + \boxed{ + \hat{\mathcal{L}}\{ f'(t) \} = - f(0) + s \tilde{f}(s) + } +\end{aligned}$$ + +This property generalizes to higher-order derivatives, +although it gets messy quickly. +Once again, the initial values of the lower derivatives appear: + +$$\begin{aligned} + \boxed{ + \hat{\mathcal{L}} \big\{ f^{(n)}(t) \big\} + = - \sum_{j = 0}^{n - 1} s^j f^{(n - 1 - j)}(0) + s^n \tilde{f}(s) + } +\end{aligned}$$ + +Where $f^{(n)}(t)$ is shorthand for the $n$th derivative of $f(t)$, +and $f^{(0)}(t) = f(t)$. +As an example, $\hat{\mathcal{L}}\{f'''(t)\}$ becomes +$- f''(0) - s f'(0) - s^2 f(0) + s^3 \tilde{f}(s)$. + +<div class="accordion"> +<input type="checkbox" id="proof-dv-t"/> +<label for="proof-dv-t">Proof</label> +<div class="hidden"> +<label for="proof-dv-t">Proof.</label> +We integrate by parts and use the fact that $\lim_{x \to \infty} \exp\!(-x) = 0$: + +$$\begin{aligned} + \hat{\mathcal{L}} \big\{ f^{(n)}(t) \big\} + &= \int_0^\infty f^{(n)}(t) \exp\!(- s t) \dd{t} + \\ + &= \big[ f^{(n - 1)}(t) \exp\!(- s t) \big]_0^\infty + s \int_0^\infty f^{(n-1)}(t) \exp\!(- s t) \dd{t} + \\ + &= - f^{(n - 1)}(0) + s \big[ f^{(n - 2)}(t) \exp\!(- s t) \big]_0^\infty + s^2 \int_0^\infty f^{(n-2)}(t) \exp\!(- s t) \dd{t} +\end{aligned}$$ + +And so on. +By partially integrating $n$ times in total we arrive at the conclusion. +</div> +</div> + + + +## References +1. O. Bang, + *Applied mathematics for physicists: lecture notes*, 2019, + unpublished. diff --git a/content/know/concept/legendre-transform/index.pdc b/content/know/concept/legendre-transform/index.pdc index 290a89a..9cb6824 100644 --- a/content/know/concept/legendre-transform/index.pdc +++ b/content/know/concept/legendre-transform/index.pdc @@ -27,9 +27,9 @@ $$\begin{aligned} y(x) = f'(x_0) (x - x_0) + f(x_0) = f'(x_0) x - C \end{aligned}$$ -The Legendre transform $L(f')$ is defined such that $L(f'(x_0)) = C$ (or -sometimes $-C$ instead) for all $x_0 \in [a, b]$, where $C$ is the -constant corresponding to the tangent line at $x = x_0$. This yields: +The Legendre transform $L(f')$ is defined such that $L(f'(x_0)) = C$ +(or sometimes $-C$) for all $x_0 \in [a, b]$, +where $C$ corresponds to the tangent line at $x = x_0$. This yields: $$\begin{aligned} L(f'(x)) = f'(x) \: x - f(x) @@ -85,7 +85,8 @@ $$\begin{aligned} \end{aligned}$$ Legendre transformation is important in physics, -since it connects Lagrangian and Hamiltonian mechanics to each other. +since it connects [Lagrangian](/know/concept/lagrangian-mechanics/) +and Hamiltonian mechanics to each other. It is also used to convert between thermodynamic potentials. diff --git a/content/know/concept/path-integral-formulation/index.pdc b/content/know/concept/path-integral-formulation/index.pdc new file mode 100644 index 0000000..c66aed8 --- /dev/null +++ b/content/know/concept/path-integral-formulation/index.pdc @@ -0,0 +1,188 @@ +--- +title: "Path integral formulation" +firstLetter: "P" +publishDate: 2021-07-03 +categories: +- Physics +- Quantum mechanics + +date: 2021-07-03T14:39:50+02:00 +draft: false +markup: pandoc +--- + +# Path integral formulation + +In quantum mechanics, the **path integral formulation** +is an alternative description of quantum mechanics, +which is equivalent to the "traditional" Schrödinger equation. +Whereas the latter is based on [Hamiltonian mechanics](/know/concept/hamiltonian-mechanics/), +the former comes from [Lagrangian mechanics](/know/concept/lagrangian-mechanics/). + +It expresses the [propagator](/know/concept/propagator/) $K$ +using the following sum over all possible paths $x(t)$, +which all go from the initial position $x_0$ at time $t_0$ +to the destination $x_N$ at time $t_N$: + +$$\begin{aligned} + \boxed{ + K(x_N, t_N; x_0, t_0) + = A \sum_{\mathrm{all}\:x(t)} \exp(i S[x] / \hbar) + } +\end{aligned}$$ + +Where $A$ normalizes. +$S[x]$ is the classical action of the path $x$, whose minimization yields +the Euler-Lagrange equation from Lagrangian mechanics. +Note that each path is given an equal weight, +even unrealistic paths that make big detours. + +This apparent problem solves itself, +thanks to the fact that paths close to the classical optimum $x_c(t)$ +have an action close to $S_c = S[x_c]$, +while the paths far away have very different actions. +Since $S[x]$ is inside a complex exponential, +this means that paths close to $x_c$ add contructively, +and the others add destructively and cancel out. + +An interesting way too look at it is by varying $\hbar$: +as its value decreases, minor action differences yield big phase differences, +which make the quantum wave function stay closer to $x_c$. +In the limit $\hbar \to 0$, quantum mechanics thus turns into classical mechanics. + +## Time-slicing derivation + +The most popular way to derive the path integral formulation proceeds as follows: +starting from the definition of the propagator $K$, +we divide the time interval $t_N - t_0$ into $N$ "slices" +of equal width $\Delta t = (t_N - t_0) / N$, +where $N$ is large: + +$$\begin{aligned} + K(x_N, t_N; x_0, t_0) + &= \matrixel{x_N}{e^{- i \hat{H} (t_N - t_0) / \hbar}}{x_0} + = \matrixel{x_N}{e^{- i \hat{H} \Delta t / \hbar} \cdots e^{- i \hat{H} \Delta t / \hbar}}{x_0} +\end{aligned}$$ + +Between the exponentials we insert $N\!-\!1$ identity operators +$\hat{I} = \int \ket{x} \bra{x} \dd{x}$, +and define $x_j = x(t_j)$ for an arbitrary path $x(t)$: + +$$\begin{aligned} + K + &= \idotsint \matrixel{x_N}{e^{- i \hat{H} \Delta t / \hbar}}{x_{N-1}} \cdots \matrixel{x_1}{e^{- i \hat{H} \Delta t / \hbar}}{x_0} + \dd{x_1} \cdots \dd{x_{N - 1}} +\end{aligned}$$ + +For sufficiently small time steps $\Delta t$ (i.e. large $N$ +we make the following approximation +(which would be exact, were it not for the fact that +$\hat{T}$ and $\hat{V}$ are operators): + +$$\begin{aligned} + e^{- i \hat{H} \Delta t / \hbar} + = e^{- i (\hat{T} + \hat{V}) \Delta t / \hbar} + \approx e^{- i \hat{T} \Delta t / \hbar} e^{- i \hat{V} \Delta t / \hbar} +\end{aligned}$$ + +Since $\hat{V} = V(x_j)$, +we can take it out of the inner product as a constant factor: + +$$\begin{aligned} + \matrixel{x_{j+1}}{e^{- i \hat{T} \Delta t / \hbar} e^{- i \hat{V} \Delta t / \hbar}}{x_j} + = e^{- i V(x_j) \Delta t / \hbar} \matrixel{x_{j+1}}{e^{- i \hat{T} \Delta t / \hbar}}{x_j} +\end{aligned}$$ + +Here we insert the identity operator +expanded in the momentum basis $\hat{I} = \int \ket{p} \bra{p} \dd{p}$, +and commute it with the kinetic energy $\hat{T} = \hat{p}^2 / (2m)$ to get: + +$$\begin{aligned} + \matrixel{x_{j+1}}{e^{- i \hat{T} \Delta t / \hbar}}{x_j} + = \int_{-\infty}^\infty \braket{x_{j+1}}{p} \exp\!\Big(\!-\! i \frac{p^2 \Delta t}{2 m \hbar}\Big) \braket{p}{x_j} \dd{p} +\end{aligned}$$ + +In the momentum basis $\ket{p}$, +the position basis vectors +are represented by plane waves: + +$$\begin{aligned} + \braket{p}{x_j} + = \frac{1}{\sqrt{2 \pi \hbar}} \exp\!\Big( \!-\! i \frac{x_j p}{\hbar} \Big) + \qquad + \braket{x_{j+1}}{p} + = \frac{1}{\sqrt{2 \pi \hbar}} \exp\!\Big( i \frac{x_{j+1} p}{\hbar} \Big) +\end{aligned}$$ + +With this, we return to the inner product and further evaluate the integral: + +$$\begin{aligned} + \matrixel{x_{j+1}}{e^{- i \hat{T} \Delta t / \hbar}}{x_j} + &= \frac{1}{2 \pi \hbar} \int_{-\infty}^\infty + \exp\!\Big(\!-\! i \frac{p^2 \Delta t}{2 m \hbar}\Big) \exp\!\Big(i \frac{(x_{j+1} - x_j) p}{\hbar}\Big) \:dp + \\ + &= \frac{1}{2 \pi \hbar} \sqrt{\frac{2 \pi m \hbar}{i \Delta t}} \exp\!\Big( i \frac{m (x_{j+1} - x_j)^2}{2 \hbar \Delta t} \Big) +\end{aligned}$$ + +Inserting this back into the definition of the propagator $K(x_N, t_N; x_0, t_0)$ yields: + +$$\begin{aligned} + K + = \Big( \frac{- i m}{2 \pi \hbar \Delta t} \Big)^{\!N / 2} + \idotsint + \exp\!\bigg(\! \sum_{j = 0}^{N - 1} i \Big( \frac{m (x_{j+1} \!-\! x_j)^2}{2 \hbar \Delta t} - \frac{V(x_j) \Delta t}{\hbar} \Big) \!\bigg) + \dd{x_1} \cdots \dd{x_{N-1}} +\end{aligned}$$ + +For large $N$ and small $\Delta t$, the sum in the exponent becomes an integral: + +$$\begin{aligned} + \frac{i}{\hbar} \sum_{j = 0}^{N - 1} \Big( \frac{m (x_{j+1} \!-\! x_j)^2}{2 \Delta t^2} - V(x_j) \Big) \Delta t + \quad \to \quad + \frac{i}{\hbar} \int_{t_0}^{t_N} \Big( \frac{1}{2} m \dot{x}^2 - V(x) \Big) \dd{\tau} +\end{aligned}$$ + +Upon closer inspection, this integral turns out to be the classical action $S[x]$, +with the integrand being the Lagrangian $L$: + +$$\begin{aligned} + S[x(t)] + = \int_{t_0}^{t_N} L(x, \dot{x}, \tau) \dd{\tau} + = \int_{t_0}^{t_N} \Big( \frac{1}{2} m \dot{x}^2 - V(x) \Big) \dd{\tau} +\end{aligned}$$ + +The definition of the propagator $K$ is then further reduced to the following: + +$$\begin{aligned} + K + = \Big( \frac{- i m}{2 \pi \hbar \Delta t} \Big)^{\!N / 2} + \idotsint \exp(i S[x] / \hbar) \dd{x_1} \cdots \dd{x_{N-1}} +\end{aligned}$$ + +Finally, for the purpose of normalization, +we define the integral over all paths $x(t)$ as follows, +where we write $D[x]$ instead of $\dd{x}$: + +$$\begin{aligned} + \int D[x] + \equiv \lim_{N \to \infty} \Big( \frac{- i m}{2 \pi \hbar \Delta t} \Big)^{\!N / 2} \idotsint \dd{x_1} \cdots \dd{x_{N-1}} +\end{aligned}$$ + +We thus arrive at **Feynman's path integral**, +which sums over all possible paths $x(t)$: + +$$\begin{aligned} + K + = \int \exp(i S[x] / \hbar) \:D[x] + = A \sum_{\mathrm{all}\:x(t)} \exp(i S[x] / \hbar) +\end{aligned}$$ + + + +## References +1. R. Shankar, + *Principles of quantum mechanics*, 2nd edition, + Springer. +2. L.E. Ballentine, + *Quantum mechanics: a modern development*, 2nd edition, + World Scientific. diff --git a/content/know/concept/propagator/index.pdc b/content/know/concept/propagator/index.pdc new file mode 100644 index 0000000..2f18c4d --- /dev/null +++ b/content/know/concept/propagator/index.pdc @@ -0,0 +1,81 @@ +--- +title: "Propagator" +firstLetter: "P" +publishDate: 2021-07-04 +categories: +- Physics +- Quantum mechanics + +date: 2021-07-04T10:46:47+02:00 +draft: false +markup: pandoc +--- + +# Propagator + +In quantum mechanics, the **propagator** $K(x_f, t_f; x_i, t_i)$ +gives the probability amplitude that a particle +starting at $x_i$ at $t_i$ ends up at position $x_f$ at $t_f$. +It is defined as follows: + +$$\begin{aligned} + \boxed{ + K(x_f, t_f; x_i, t_i) + \equiv \matrixel{x_f}{\hat{U}(t_f, t_i)}{x_i} + } +\end{aligned}$$ + +Where $\hat{U} \equiv \exp\!(- i t \hat{H} / \hbar)$ is the time-evolution operator. +The probability that a particle travels +from $(x_i, t_i)$ to $(x_f, t_f)$ is then given by: + +$$\begin{aligned} + P + &= \big| K(x_f, t_f; x_i, t_i) \big|^2 +\end{aligned}$$ + +Given a general (i.e. non-collapsed) initial state $\psi_i(x) \equiv \psi(x, t_i)$, +we must integrate over $x_i$: + +$$\begin{aligned} + P + &= \bigg| \int_{-\infty}^\infty K(x_f, t_f; x_i, t_i) \: \psi_i(x_i) \dd{x_i} \bigg|^2 +\end{aligned}$$ + +And if the final state $\psi_f(x) \equiv \psi(x, t_f)$ +is not a basis vector either, then we integrate twice: + +$$\begin{aligned} + P + &= \bigg| \iint_{-\infty}^\infty \psi_f^*(x_f) \: K(x_f, t_f; x_i, t_i) \: \psi_i(x_i) \dd{x_i} \dd{x_f} \bigg|^2 +\end{aligned}$$ + +Given a $\psi_i(x)$, the propagator can also be used +to find the full final wave function: + +$$\begin{aligned} + \boxed{ + \psi(x_f, t_f) + = \int_{-\infty}^\infty \psi_i(x_i) K(x_f, t_f; x_i, t_i) \:dx_i + } +\end{aligned}$$ + +Sometimes the name "propagator" is also used to refer to +the so-called *fundamental solution* or *Green's function* $G$ +of the time-dependent Schrödinger equation, +which is related to $K$ by: + +$$\begin{aligned} + \boxed{ + G(x_f, t_f; x_i, t_i) + = - \frac{i}{\hbar} \: \Theta(t_f - t_i) \: K(x_f, t_f; x_i, t_i) + } +\end{aligned}$$ + +Where $\Theta(t)$ is the [Heaviside step function](/know/concept/heaviside-step-function/). +The definition of $G$ is that it satisfies the following equation, +where $\delta$ is the [Dirac delta function](/know/concept/dirac-delta-function/): + +$$\begin{aligned} + \Big( i \hbar \pdv{t_f} - \hat{H} \Big) G = \delta(x_f - x_i) \: \delta(t_f - t_i) +\end{aligned}$$ |