A canonical transformation is a Coordinate transformation that transforms a set of canonical coordinates that satisfy the Hamilton equations into another set that also satisfies them. It is said that canonical transformations preserve the form of the Hamilton equations.
Mathematically, the transformation , is canonical if for all Hamiltonians , there exists a function called the conjugate Hamiltonian such that, if the equations of motion in coordinates obey the Hamilton equations
then the equations in also obey the Hamilton equations, with as the Hamiltonian:
The solutions are related by the functions and as and .
Interpretation#
Being a special case of coordinate transformations, canonical transformations warrant a special interpretation. Say we have some transformation . What are and ? They are canonical coordinates and therefore points in phase space. What's the relation between them? There's a couple of ways we can see the pair :
- and are coordinates of two different points in the same coordinate system.
- and are coordinates of the same point in two different coordinate systems.
It's easier to see an example.
> We can see this in two ways. > > ![[Plot Rotation coordinate change interpretations.svg|100%]] > > 1. On the left, we see the transformation as moving the point from the blue point to the red point. > 2. On the right, we see the transformation rotating the axes themselves from blue to red. Note how the rotation goes in the *opposite* direction here. In this sense, we can interpret canonical transformations in one of two ways: 1. *Actively*, as moving the points from an old location to a new location. 2. *Passively*, as letting the points stay where they are and moving the axes instead, in the opposite direction of the points. In a way, both are intuitive, but the active interpretation is particularly convenient when working with single-parameter families of canonical transformations, such as the [[Hamiltonian flow]] or the aforementioned rotations. ### Canonicity criteria Determining if a transformation is canonical can be arduous: you have to prove that it is canonical for *all* possible Hamiltonians, and there's infinite of them. Sometimes this brute force way is a valid approach, such as in some of the examples below, but there is a better method to ensuring canonicity. For a transformation to be canonical, it must satisfy the so-called **canonicity criteria**. Assume we're working for some [[dynamical system]] $\dot{\mathbf{x}}=f(\mathbf{x},t)$. Consider a generic coordinate transformation $x_{i}=w_{i}(\tilde{\mathbf{x}},t)$. The inverse is $\tilde{x}_{i}=\tilde{w}_{i}(\mathbf{x},t)$. Motion in [[phase space]] is described by either $x_{i}(t)$ or $\tilde{x}_{i}(t)$, linked by the aforementioned equations as $x_{i}(t)=w(\tilde{\mathbf{x}}(t),t)$ and $\tilde{x}_{i}(t)=\tilde{w}(\mathbf{x}(t),t)$. The [[Differential|total time derivative]] is\dot{\tilde{x}}{i}=\sum{j=1}^{n} \underbrace{ \frac{ \partial \tilde{w}{i} }{ \partial x{j} } }{ \tilde{J}{ij} } (\mathbf{x}(t),t)\dot{x}_{j}(t)+\frac{ \partial \tilde{w} }{ \partial t } (\mathbf{x}(t),t)
where $\tilde{J}$ is the [[Jacobian]] of $\tilde{w}$. Since $\dot{x}_{j}=f_{j}(\mathbf{x},t)$, then\dot{\tilde{x}}{i}=\sum{j=1}^{2n} \tilde{J}{ij}(w(\tilde{\mathbf{x}}(t),t),t)f{j}(w(\tilde{\mathbf{x}}(t),t),t)+\frac{ \partial \tilde{w} }{ \partial t } (w(\tilde{\mathbf{x}}(t),t),t)
\dot{\tilde{x}}{i}=\sum{j=1}^{2n} \tilde{J}{ij}f{j}+\frac{ \partial \tilde{w} }{ \partial t }
Let's say that our coordinates solve the Hamilton equations $\dot{\mathbf{x}}=\mathrm{E}\nabla_{\mathbf{x}}H=f(\mathbf{x},t)$, with individual components $\dot{x}_{i}=\sum_{l}\mathrm{E}_{il}\frac{ \partial H }{ \partial x_{l} }=f_{i}(\mathbf{x},t)$. Then,\dot{\tilde{x}}{i}=\sum{j=1}^{2n} \tilde{J}{ij}\sum{l}\mathrm{E}{il}\frac{ \partial H }{ \partial x{l} } +\frac{ \partial \tilde{w}_{i} }{ \partial t }
\dot{\tilde{\mathbf{x}}}=\tilde{J}\mathrm{E}\nabla_{\mathbf{x}}H+\frac{ \partial \tilde{\mathbf{w}} }{ \partial t }
If the transformation is canonical, then this must also be equal to $\dot{\tilde{\mathbf{x}}}=\mathrm{E}\nabla_{\tilde{\mathbf{x}}}K$. This provides an initial criterion for canonicity[^1]. > [!info] Canonicity lemma > A transformation is canonical if, for all $H(\mathbf{x},t)$, there exists a conjugate Hamiltonian $K(\tilde{\mathbf{x}},t)$ such that > $$\tilde{J}\mathrm{E}\nabla_{\mathbf{x}}H+\frac{ \partial \tilde{\mathbf{w}} }{ \partial t } =\mathrm{E}\nabla_{\tilde{\mathbf{x}}}KThe issue now is that we don't really know what $K$ *is*. We'll start from $\mathrm{E}\nabla_{\mathbf{x}}K$ and see what we can do.In particular, we call the conjugate Hamiltonian when , that is
\mathrm{E}\nabla_{\tilde{\mathbf{x}}}K=\tilde{J}\mathrm{E}\nabla_{\mathbf{x}}H+\frac{ \partial \tilde{\mathbf{w}} }{ \partial t } =\tilde{J}\mathrm{E}\tilde{J}^{T}(\tilde{J}^{T})^{-1}\nabla_{\mathbf{x}}H+\mathrm{E}\nabla_{\tilde{\mathbf{x}}}K_{0}=\ldots
In the first step, we used the first part of the canonicity lemma. In the second, we used the second part of the lemma and we arbitrarily added $\tilde{J}^{T}(\tilde{J}^{T})^{-1}$ since a [[matrix]] multiplied by its [[Invertible matrix|inverse]] is the [[Identity matrix]] and doesn't change anything. It's a weird step, but it allows us to use the following equality. First, notice that, since $\tilde{J}^{-1}=J$, $(\tilde{J}^{T})^{-1}=(\tilde{J}^{-1})^{T}=J^{T}$. Then\begin{align} ((\tilde{J}^{T})^{-1} \nabla_{\mathbf{x}}H){i}&=\sum{j}J_{ij}^{T}\frac{ \partial H }{ \partial x_{j} } \ (\text{symmetric: }J^{T}{ij}=J{ji})&=\sum_{j}J_{ji}\frac{ \partial H }{ \partial x_{j} } \ \left( \text{def: }J_{ji}=\frac{ \partial w_{j} }{ \partial \tilde{x}{i} } \right)&=\sum{j}\frac{ \partial w_{j} }{ \partial \tilde{x}{i} } \frac{ \partial H }{ \partial x{j} } \ (\text{def: }H(w(\tilde{x},t),t)\equiv\tilde{H}(\tilde{x},t))&=\frac{ \partial \tilde{H} }{ \partial \tilde{x}_{i} } \end{align}
(\tilde{J}^{T})^{-1}\nabla_{\mathbf{x}}H=\nabla_{\tilde{\mathbf{x}}}\tilde{H}
The meaning of this is that we can use the Jacobian of the transformation to transform from the [[Gradient]] in $\mathbf{x}$ of $H$ to the one in $\tilde{\mathbf{x}}$ of $\tilde{H}$ (basically, we use the derivatives of the transformation to transform the *gradient's* coordinates). Plugging this into our previous equation yields\begin{align} \ldots&=\tilde{J}\mathrm{E}\tilde{J}^{T}\nabla_{\tilde{\mathbf{x}}}\tilde{H}+\mathrm{E}\nabla_{\tilde{\mathbf{x}}}K_{0} \ \end{align}
We can see that we have a gradient of $K$ and one of $K_{0}$. We can move the one with $K_{0}$ on the other side to get\tilde{J}\mathrm{E}\tilde{J}^{T}\nabla_{\tilde{\mathbf{x}}}\tilde{H}=\mathrm{E}\nabla_{\tilde{\mathbf{x}}}(K-K_{0})
By applying $\mathrm{E}^{-1}=-\mathrm{E}$ on the left on both sides of the equation we get-\mathrm{E}\tilde{J}\mathrm{E}\tilde{J}^{T}\nabla_{\tilde{\mathbf{x}}}\tilde{H}=\nabla_{\tilde{\mathbf{x}}}(K-K_{0})
The [[Vector field]] $-\mathrm{E}\tilde{J}\mathrm{E}\tilde{J}^{T}\nabla_{\tilde{\mathbf{x}}}\tilde{H}$ is [[Vector field|irrotational]] for all $\tilde{H}$. This is because it is equal to a gradient, and the [[Curl]] of a gradient is always zero. This means that there exists some [[Potential]] whose gradient is the field itself. This won't be proven here, but knowing this, we can write-\mathrm{E}\tilde{J}\mathrm{E}\tilde{J}^{T}=c\mathrm{I}_{2n}\tag{1}
where $\mathrm{I}_{2n}$ is the $2n$-dimensional [[Identity matrix]] and $c$ is some real constant. With one last substitution we seec\nabla_{\tilde{\mathbf{x}}}\tilde{H}=\nabla_{\tilde{\mathbf{x}}}(K-K_{0})
\boxed{K=c \tilde{H}+K_{0}}\tag{2}
This is the *general* shape of the conjugate Hamiltonian. In the special case where $c=1$, the transformation is said to be **univalent**. This equation allows us to state another canonicity criterion. > [!info] First canonicity criterion > A transformation $\mathbf{x}=\mathbf{w}(\tilde{\mathbf{x}},t)$ is **canonical** if and only if there exists some $c\neq 0$ for which its Jacobian $J$ satisfies $cJ\mathrm{E}J^{T}=\mathrm{E}$. If $c=1$, the transformation is **univalent canonical** and its Jacobian is [[Symplectic matrix|symplectic]]. > [!quote]- Proof > $(\Rightarrow)$ If the transformation is canonical, the $c\neq 0$ must exist such that $cJ\mathrm{E}\tilde{J}=\mathrm{E}$. Starting from $(1)$ we can apply $\mathrm{E}$ to the right on both sides > $$-\tilde{J}\mathrm{E}\tilde{J}\mathrm{E}\mathrm{E}=c\mathrm{I}_{2n}\mathrm{E}\quad\Rightarrow \quad \tilde{J}\mathrm{E}\tilde{J}^{T}=c\mathrm{E}> since $J=\tilde{J}^{-1}$ and so $J\tilde{J}=\tilde{J}^{T}J^{T}=\mathrm{I}_{2n}$. > > $(\Leftarrow)$ If $c\neq 0$ exists, then the transformation must be canonical, that is, for all $H$ there exists some $K$ such that $E\nabla_{\tilde{\mathbf{x}}}K=\tilde{J}\mathrm{E}\nabla_{\mathbf{x}}H+\frac{ \partial \tilde{ \mathbf{w}} }{ \partial t }$ (part 1 of canonicity lemma). We already know that $\nabla_{\mathbf{x}}H=\tilde{J}^{T}\nabla_{\tilde{\mathbf{x}}}\tilde{H}$ so > $$\tilde{J}\mathrm{E}\nabla_{\mathbf{x}}H=\tilde{J}\mathrm{E}\tilde{J}^{T}\nabla_{\tilde{\mathbf{x}}}\tilde{H}=c\mathrm{E}\nabla_{\tilde{\mathbf{x}}}\tilde{H}We now apply to the left and to the right on both sides
With two more lemmas that won't be covered here, one can prove that if exists such that (part 2 of canonicity lemma), then there exists a such that the canonicity lemma is valid and thus the transformation is canonical.
We should stop for a moment to think about this result. What does mean? Well, if you set , you get . This is the definition of a Symplectic matrix, a set of matrices which form a group with useful properties that is very important in Hamiltonian mechanics1. Unfortunately, the makes things harder. In general, we need to analyze things further. Spoilers: while may not be symplectic when , everything else regarding Hamilton's equations stays the same (up to some scaling).
If we take the determinant of , remember that and that , we get and so
Now take a canonical transformation of Jacobian . We have for some . We compose it with the canonical transformation2 . Our original transformation now becomes . Its Jacobian is
So
So, if , the first canonicity criterion seems to hold. If , we can compose the coordinates with
which leads to
and so we go back to our canonicity criterion, up to a composition. In fact, this is fully general. We can always do this. What we just found is that we can always compose a generic canonical transformation with a specific canonical transformation such that the composition has . In other words, all canonical transformations can be reduced to univalent ones, provided we figure out the correct compositions.
Univalent transformations are such an important group of functions that a lot of study has gone into them. Most importantly, they are a part of a more general set of transformations called symplectic transformations, which are all the transformations whose Jacobian is symplectic. Since by the first canonicity criterion all univalent canonical transformations have a symplectic Jacobian, they are automatically all symplectic transformations.
As a bonus property, note . Since for all univalent transformations, then for them too. This is sometimes important, since appears when doing coordinate changes inside of integrals. For a relevant example, see for instance Liouville's theorem > Proof (analytical mechanics). It also means that, if we know that a transformation is canonical, we can check if it's also univalent by taking the determinant of its Jacobian and seeing if its equal to one.
Poisson bracket preservation#
A transformation is said to preserve the Poisson brackets if for all and we have
where and are the compositions of and with the transformation . Basically, the brackets are preserved if it doesn't matter if you apply the transformation before or after calculating the brackets.
> [!info] Characterization 2 > $\mathbf{x}=w(\tilde{\mathbf{x}},t)$ preserves the Poisson brackets if and only its Jacobian is symplectic. In other words, only symplectic transformations preserve the Poisson brackets. > [!quote]- Proof > Take the brackets of the components of $w$. > $$\{ w_{k},w_{l} \}=\sum_{i,j}\frac{ \partial w_{k} }{ \partial \tilde{x}_{i} } \mathrm{E}_{ij}\frac{ \partial w_{l} }{ \partial \tilde{x}_{j} } =\sum_{i,j}J_{ki}\mathrm{E}_{ij}J_{lj}=(J\mathrm{E}J^{T})_{kl}Then is symplectic.
Infinitesimal transformations#
An interesting case is that of infinitesimal canonical transformations, which generally have the form
where and are infinitesimal variations. Alternatively, in compact form, . The benefit of these is that, since we're working with infinitesimals, we can ignore any higher order infinitesimals like without loss of generality.
By "generally" I mean that not all variations lead to a canonical transformation: we need to prove that it does. We'll use the second canonicity criterion (must preserve Poisson brackets) to see which are valid. Preserving the Poisson brackets is equivalent to preserving the elementary functions (see characterization 1 above):
The cancels out from both sides so we get
Expanding the brackets
We define
so
This can use the following lemma:
> which is only true if $\mathrm{E}\Omega$ is antisymmetric. > > ($\Leftarrow$). Define $\mathbf{u}\equiv \mathrm{E}\mathbf{f}$. Then > $$\frac{ \partial u }{ \partial x } =\mathrm{E}\Omega> This is the same as asking that the [[Curl]] of $\mathbf{u}$ be zero. But if $\mathbf{u}$ is [[Vector field|irrotational]], there exists an $F$ such that $\mathbf{u}=-\nabla_{\mathbf{x}}F$ and so $\mathbf{f}=\mathrm{E}\nabla_{\mathbf{x}}F$. Our vector field here is $\delta \mathbf{x}$, and since the latter part of the lemma is true, then there exists some function $G(\mathbf{x},t)$ that allows us to writemeans that is symmetrical. Then
\boxed{\delta \mathbf{x}=\mathrm{E}\nabla_{\mathbf{x}}(\varepsilon G)=\varepsilon { \mathbf{x},G }}\tag{4}
where $\varepsilon\ll 1$, which is a parameter that we added to make sure $\varepsilon G$ is infinitesimal even while allowing $G$ to be any vector field. In non-compact form, this reads\boxed{\delta p_{k}=-\varepsilon \frac{ \partial G }{ \partial q_{k} },\qquad\delta q_{k}=\varepsilon \frac{ \partial G }{ \partial p_{k} } }
The function $G(x,t)$ is known as the **generator** of the infinitesimal canonical transformation, because it determines $\delta \mathbf{x}$ and therefore the variation of $\mathbf{x}$. As such, $G$ contains all of the information about the transformation. $(4)$ allows us to calculate the infinitesimal variation of any [[dynamical variable]] $f(\mathbf{x},t)$:\delta f(\mathbf{x},t)=f(\mathbf{x}+\delta \mathbf{x},t)-f(\mathbf{x},t)=df(\mathbf{x},t)[\delta \mathbf{x}]=\ldots
This is the [[Differential]] of $f(\mathbf{x},t)$. Applying the definition\ldots=\sum_{i=1}^{2n} \frac{ \partial f }{ \partial x_{i} }(\mathbf{x},t)\delta x_{i}=\ldots
and applying $(4)$\ldots=\sum_{i=1}^{2n} \frac{ \partial f }{ \partial x_{i} } \varepsilon \sum_{j=1}^{2n} E_{ij}\frac{ \partial G }{ \partial x_{j} } =\varepsilon \sum_{i,j=1}^{2n} \frac{ \partial f }{ \partial x_{i} } E_{ij}\frac{ \partial G }{ \partial x_{j} } =\varepsilon { f,G }
Basically, if we know the generator $G$, we also know the variation of $f$:\boxed{\delta f=\varepsilon { f,G }}\tag{5}
Among other things, this is used to prove the Hamiltonian variant of [[Nöther's theorem]]. #### Infinitesimal transformations of the Hamiltonian We can analyze what happens to the variation of $H$ during a time-independent coordinate change:\delta H=H(\mathbf{x}+\delta \mathbf{x})-H(\mathbf{x})=H(\tilde{\mathbf{x}})-\tilde{H}(\tilde{\mathbf{x}})=H(\tilde{\mathbf{x}})-K(\tilde{\mathbf{x}})
where $x+\delta x=\tilde{x}$ and the last step is true for univalent canonical, time-independent transformations. If the transformation is dependent on $t$, we have\delta H=H(\tilde{\mathbf{x}})-K(\tilde{\mathbf{x}})=H(\tilde{\mathbf{x}})-\tilde{H}(\tilde{\mathbf{x}})-K_{0}(\tilde{\mathbf{x}})=\underbrace{ H(\mathbf{x}+\delta \mathbf{x})-H(\mathbf{x}) }{ \varepsilon { H,G } }-K{0}(\tilde{\mathbf{x}})\tag{5}
\mathrm{E}\nabla_{\tilde{\mathbf{x}}}K_{0}=\frac{ \partial \tilde{ w} }{ \partial t }(w(\tilde{\mathbf{x}},t),t)=\frac{ \partial }{ \partial t } (\mathbf{x}+\delta \mathbf{x}(\mathbf{x},t))=\frac{ \partial \delta \mathbf{x} }{ \partial t } =\frac{ \partial }{ \partial t } \varepsilon \mathrm{E}\nabla_{\mathbf{x}}G(\mathbf{x},t)=\mathrm{E}\nabla_{\mathbf{x}}\left( \varepsilon \frac{ \partial G }{ \partial t } \right)
The last term in the brackets is a function of $(\mathbf{x},t)$. Define\Lambda(\tilde{\mathbf{x}},t)\equiv \varepsilon \frac{ \partial G }{ \partial t } (\tilde{\mathbf{x}}-\delta \mathbf{x},t)
\frac{ \partial \Lambda }{ \partial \tilde{x}{j} }=\sum{i}\varepsilon \frac{ \partial ^{2}G }{ \partial x_{i}\partial t }(\tilde{\mathbf{x}}-\delta \mathbf{x},t) \frac{ \partial \tilde{\mathbf{x}}{i}+\delta x{i} }{ \partial \tilde{x}{j} } =\frac{ \partial }{ \partial x{j} } \left( \varepsilon \frac{ \partial G }{ \partial t } \right)(\tilde{\mathbf{x}}-\delta \mathbf{x},t)
\nabla_{\mathbf{x}}\left( \varepsilon \frac{ \partial G }{ \partial t } \right)=\nabla_{\tilde{\mathbf{x}}}\Lambda
\mathrm{E}\nabla_{\tilde{\mathbf{x}}}K_{0}=\mathrm{E}\nabla_{\tilde{\mathbf{x}}}\Lambda
K_{0}(\tilde{\mathbf{x}})=\Lambda(\tilde{\mathbf{x}})=\varepsilon \frac{ \partial G }{ \partial t }
Substituting this back in $(5)$ yields\boxed{\delta H=\varepsilon \left( { H,G }- \frac{ \partial G }{ \partial t } \right)}
#### Finite transformations We now ask the question: since we have a theory on infinitesimal transformations, how do we go back to finite transformations from these? The idea here is to interpret infinitesimal transformations as an *approximation* of finite, continuous one-parameter transformations. The approximation is best when the parameter is small and the two are exactly equal when the parameter is zero. For a practical example, rotations are one-parameter transformations, where the parameter is the angle of rotation. A small angle leads to an infinitesimal rotation, which is (in theory) an infinitesimal transformation. We'll use the active interpretation here, so we think of the transformation as moving the actual point in phase space. We'll call the parameter $\varepsilon$. As $\varepsilon$ varies, the transformation moves the point, tracing a [[Curve]] in phase space given by\tilde{\mathbf{x}}(\varepsilon)=\tilde{\mathbf{w}}(x;\varepsilon)\equiv \Phi^{\varepsilon}(\mathbf{x})
(The similar choice of notation to the [[Hamiltonian flow]] is no coincidence: the flow is also a one-parameter transformation, with the parameter being time). For simplicity, we ignore time dependence in the transformation ($\mathbf{w}(\mathbf{x},t)\equiv \mathbf{w}(\mathbf{x})$). If $\varepsilon\ll1$, the transformation becomes infinitesimal and there exists some generator $G$ such that\delta \mathbf{x}=\varepsilon { \mathbf{x},G }=\varepsilon \mathrm{E}\nabla_{\mathbf{x}}G(\mathbf{x},t)
In the limit $\varepsilon\to 0$ we get\frac{d\tilde{\mathbf{x}}}{d\varepsilon}(0)={ \mathbf{x},G }(\mathbf{x},t)=\mathrm{E}\nabla_{\mathbf{x}}G(\tilde{\mathbf{x}}(0),t)
This limit can be repeated for every point on the curve $\tilde{\mathbf{x}}(\varepsilon)$, so it must be true for any $\varepsilon$:\frac{d\tilde{\mathbf{x}}}{d\varepsilon}(\varepsilon)=\mathrm{E}\nabla_{\mathbf{x}}G(\tilde{\mathbf{x}}(\varepsilon),t)
Integration of this equation yields the curve $\tilde{\mathbf{x}}(\varepsilon)$, which means that we have a *finite* transformation starting from an infinitesimal one. Basically, we just made a set of Hamilton equations where time is replaced by $\varepsilon$ and the Hamiltonian is replaced by $G$. In this sense, the curve $\tilde{\mathbf{x}}(\varepsilon)=\Phi^{\varepsilon}(\mathbf{x})$ is the Hamiltonian flow of $G$. > [!success] Moment maps > Every generator $G(\mathbf{p},\mathbf{q},t)$ can be interpreted as a Hamiltonian whose flow is the finite canonical transformations it generates. In this sense, $G$ is also called a **[[moment map]]** and denoted $\mu$. Knowing this, we can look at it in reverse. We now that the usual Hamiltonian has a Hamiltonian flow whose purpose is to translate points in time and we know that this time translation is a canonical transformation. But then $H$ *must be the generator*! > [!success] Time translations > The Hamiltonian is the generator of time translations. #### Infinitesimal rotations [[rotation|Rotations]] are a particularly important and ubiquitous kind of transformation. Infinitesimal rotations specifically appear quite a few times in analytical mechanics. It's fair to ask: are they canonical? What is the generator of infinitesimal rotations? Let's define some rotationR(\alpha)=\mathrm{I}+\Omega(\alpha)
where $R(\alpha)\in SO(3)$ is a rotation on the $\hat{\omega}$ axis of angle $\alpha$ (which must be infinitesimal for this approximation to hold), $\mathrm{I}$ is the [[Identity matrix]] and $\Omega(\alpha)$ is the infinitesimal rotation defined by its components as\Omega(\alpha){ij}=-\alpha \sum{k}\epsilon_{ijk}\hat{\omega}_{k}
where $\epsilon_{ijk}$ is the [[Tensore di Levi-Civita|Levi-Civita tensor]] and $\boldsymbol{\omega}=\alpha\hat{\omega}$ is the angular speed. We'll prove canonicity by showing that the Poisson brackets are preserved (second canonicity criterion) and we'll do so by showing that the fundamental Poisson brackets of the transformed coordinates are the same, which is a sufficient proof of canonicity. Speaking of, the transformed coordinates are\tilde{p}{j}=\sum{k}R_{jk}p_{k},\quad \tilde{q}{i}=\sum{m}R_{im}q_{m}
{ \tilde{q}{j},\tilde{q}{l} }=0,\qquad{ \tilde{p}{j},\tilde{p}{l} }=\sum_{k,s}R_{jk}R_{ls}\underbrace{ { p_{k},p_{s} } }_{ 0 }=0
{ \tilde{q}{i},\tilde{p}{j} }=\sum_{k,m}R_{im}R_{jk}\underbrace{ { q_{m},p_{k} } }{ \delta{mk} }=\sum_{k}R_{ik}R_{jk}=\sum_{k}R_{ik}R_{kj}^{T}=(\underbrace{ RR^{T} }{ \mathrm{I} }){ij}=\delta_{ij}
So all the fundamental brackets are [[Transformation invariance|invariant]] under rotation. This proves our point: *all rotations are canonical*. Now that that's done, we want to answer the second question. To find the generator, we'll use $(5)$ for some generator and see if that leads to the infinitesimal rotation. Before that, we need to check what the infinitesimal variation is due to the rotation itself. We know that $\tilde{q}_{i}=q_{i}+\delta q_{i}$ and $\tilde{p}_{j}=p_{j}+\delta p_{j}$, so comparing with the abovep_{j}+\delta p_{j}=\tilde{p}{j}=\sum{k}R_{jk}p_{k}=\sum_{k}(\delta_{jk}+\Omega_{jk})p_{k}=p_{j}+\sum_{k}\Omega_{jk}p_{k}
\delta p_{j}=\sum_{k}\Omega_{jk}p_{k}
The same can be done for $\tilde{q}_{i}$. This is our ground truth: we want to find the generator $G$ that leads to the same variation $\delta p_{j}$. We already have a very clear candidate for rotation generation: the [[angular momentum]] $\mathbf{L}=\mathbf{q}\times \mathbf{p}$. The variation of $\mathbf{p}$ as per $(5)$ due to $\mathbf{L}$ projected over the axis of rotation $\hat{\omega}$ then is:\delta \mathbf{p}=\alpha{ \mathbf{p},\hat{\omega}\cdot \mathbf{L} }
On the $i$-th axis we get\begin{align} \delta p_{i}=\alpha{ p_{i},\hat{\omega}\cdot \mathbf{L} }&=\alpha\left{ p_{i},\sum_{j}\hat{\omega}{j}L{j} \right}=\alpha \sum_{j}\hat{\omega}{j}{ p{i},L_{j} }=\alpha \sum_{j}\hat{\omega}{j}\sum{k}\epsilon_{ijk}p_{k} \ &=\sum_{k}\left( -\alpha \sum_{j}\epsilon_{ikj}\hat{\omega}{j} \right)p{j}=\sum_{k}\Omega_{ij}p_{k} \end{align}
where we used that $L_{j}=(\mathbf{q}\times \mathbf{p})_{j}$ which can be expressed with the Levi-Civita definition of the [[Vector product]] $L_{j}=\sum_{jkm}\epsilon_{jkm}q_{k}p_{m}$ and then explicitly calculated the brackets. This matches our previous variation, so it confirms our suspicion: *the projection of the angular momentum over the rotation axis generates the rotation*. The same can be proven for $\delta q_{i}$ to complete the proof. > [!success] Rotations > 1. Infinitesimal rotations are canonical transformations. > 2. The angular momentum is the generator of $SO(3)$ rotations. The projection of the angular momentum over an axis generates an infinitesimal rotation over that axis. Here's a useful fact about rotation invariance. Consider a function $f(\mathbf{p},\mathbf{q})$ that is invariant under rotation, that is, $f(R\mathbf{p},R\mathbf{q})=f(\mathbf{p},\mathbf{q})$. In this case, the function can be written as\boxed{f(\mathbf{p},\mathbf{q})=g(\lvert \mathbf{p} \rvert ^{2},\lvert \mathbf{q} \rvert ^{2},\mathbf{p}\cdot \mathbf{q})}
This is because $\lvert \mathbf{p} \rvert^{2}$, $\lvert \mathbf{q} \rvert^{2}$ and $\mathbf{p}\cdot \mathbf{q}$ are all invariant under rotation, therefore for $f$ to be invariant, it must exclusively depend on these. > [!quote]- Proof > Just check for Poisson brackets on all three quantities. > $$\{ L_{k},\lvert \mathbf{p} \rvert ^{2} \}=\sum_{i}\{ L_{i},p_{i}^{2} \}=\sum_{i}2p_{i}\{ L_{i},p_{i} \}=\sum_{i,j}2\epsilon_{kij}p_{j}p_{i}=0> because again, $e_{kij}$ is antisymmetric and the term in brackets is symmetric. > > Since $\{ L_{k},\lvert \mathbf{p} \rvert^{2} \}=\{ L_{k},\lvert \mathbf{q} \rvert^{2} \}=\{ L_{k},\mathbf{p}\cdot \mathbf{q} \}=0$, all three of these are invariant under rotation and completes our proof. > > As further confirmation, we can also prove that $f$ is invariant by calculating the Poisson brackets directly using the chain rule: > $$\begin{align} > \{ L_{k},f \}&=\sum_{i}\left( \frac{ \partial L_{k} }{ \partial q_{i} } \frac{ \partial f }{ \partial p_{i} } -\frac{ \partial L_{k} }{ \partial p_{i} } \frac{ \partial f }{ \partial q_{i} } \right) \\ > &=\sum_{i}\left( \frac{ \partial L_{k} }{ \partial q_{i} } \sum_{j=1}^{3}\frac{ \partial g }{ \partial \alpha_{j} } \frac{ \partial \alpha_{j} }{ \partial p_{i} } -\frac{ \partial L_{k} }{ \partial p_{i} } \sum_{j=1}^{3} \frac{ \partial g }{ \partial \alpha_{j} } \frac{ \partial \alpha_{j} }{ \partial q_{i} } \right) \\ > &=\sum_{j=1}^{3} \frac{ \partial g }{ \partial \alpha_{j} } \underbrace{ \sum_{i}\left( \frac{ \partial L_{k} }{ \partial q_{i} } \frac{ \partial \alpha_{j} }{ \partial p_{i} } -\frac{ \partial L_{k} }{ \partial p_{i} } \frac{ \partial \alpha_{j} }{ \partial q_{i} } \right) }_{ \{ L_{k},\alpha_{j} \}=0 } \\ > &=0 > \end{align}since is antisymmetric and is symmetric, and the contraction of the product of a antisymmetric and symmetric tensor is always zero. Similarly for . Then
\{ L_{k},\mathbf{p}\cdot \mathbf{q} \}&=\sum_{i}\{ L_{k},p_{i}q_{i} \}=\sum_{i}(p_{i}\{ L_{k},q_{i} \}+\{ L_{k},p_{i} \}q_{i}) \\ &=\sum_{i,j}(\epsilon_{kij}q_{j}p_{i}+\epsilon_{kij}p_{i}q_{j})=\sum_{i,j}\epsilon_{kij}(q_{j}p_{i}+p_{i}q_{j})=0 \end{align}
Examples#
> with Hamiltonian $H=p^{2}/2m$, that yields the equations of motion > $$\begin{cases} > \dot{p}=0 \\ > \dot{q}=p/m > \end{cases}> If we plug these in the equations of motion and invert we get > $$\begin{cases} > \dot{\tilde{p}}=\omega \tilde{p}\cos \tilde{q}\sin \tilde{q} \\ > \dot{\tilde{q}}=\omega \cos ^{2}\tilde{q} > \end{cases}The derivatives of the transformation are
> We need some property to verify that such a $K$ can exist. Since we're working with derivatives, we may as well use the [[Schwarz theorem]] as a check. The following must be true: > $$\frac{ \partial }{ \partial \tilde{p} } \frac{ \partial K }{ \partial \tilde{q} } =\frac{ \partial }{ \partial \tilde{q} } \frac{ \partial K }{ \partial \tilde{p} }This is correct, but we have not proven if this is a canonical transformation or not. To do so, we need to check if these new equations of motions take the form of Hamilton equations for some Hamiltonian :
\dot{\tilde{p}}=\omega \tilde{p}\cos \tilde{q}\sin \tilde{q}=-\frac{ \partial K }{ \partial \tilde{q} } \\ \dot{\tilde{q}}=\omega \cos ^{2}\tilde{q}=\frac{ \partial K }{ \partial \tilde{p} } \end{cases}\quad(?)\tag{1}
> This is impossibile, so $K$ cannot exists. As such, this transformation is not canonical. > [!example]- Scalings > Here's a scaling transformation > $$p_{i}=\alpha \tilde{p}_{i},\quad q_{i}=\beta \tilde{q}_{i}but doing the math with the values we are hoping for in leads to
> In abstract, the equations of motion $\dot{p}_{i}=-\frac{ \partial H }{ \partial q_{i} }(q,p,t)$ and $\dot{q}_{i}=\frac{ \partial H }{ \partial p_{i} }(q,p,t)$ become > $$\dot{\tilde{p}}_{i}=- \frac{1}{\alpha}\frac{ \partial H }{ \partial q_{i} } (\alpha \tilde{p},\beta \tilde{q},t),\quad \dot{\tilde{q}}_{i}=\frac{1}{\beta}\frac{ \partial H }{ \partial p_{i} } (\alpha \tilde{p},\beta \tilde{q},t)with derivatives
> Thus, this is a canonical transformation of conjugate Hamiltonian $K$. > [!example]- > Consider the transformation $p_{i}=\tilde{p}_{i}$ and $q_{i}=\tilde{q}_{i}+\alpha t \tilde{p}_{i}$. > $$K(\tilde{p},\tilde{q},t)=H(\tilde{p},\tilde{q}+\alpha t \tilde{p})- \frac{\alpha}{2}\tilde{\mathbf{p}}^{2}We can pretty readily see that these are equivalent to
> If $\alpha=1/m$, we have > $$\tilde{K}=\frac{\tilde{\mathbf{p}}^{2}}{2m}- \frac{1}{2m}\tilde{\mathbf{p}}^{2}=0\quad\to \quad \dot{\tilde{p}}=0,\quad \dot{\tilde{q}}=0
> By the criterion we calculate > $$J\mathrm{E}J^{T}=\begin{pmatrix}1 & 0 \\ \alpha t & 1\end{pmatrix}\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}1 & \alpha t \\ 0 & 1\end{pmatrix}=\begin{pmatrix}0 & -1 \\ 1 & -\alpha t\end{pmatrix}\begin{pmatrix}1 & \alpha t \\ 0 & 1\end{pmatrix}=\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}=\mathrm{E}So, in the modified coordinate systems, the equations of motion become trivial: and are just constants. Of course, the issue was getting here, i.e. knowing what transformation to do in the first place. (TODO: Fix this, 08/05/2025)
> $$\{ \tilde{q}+\alpha t\tilde{p},\tilde{q}+\alpha t\tilde{p} \}=\cancel{ \{ \tilde{q},\tilde{q} \} }+\{ \tilde{q},\alpha t\tilde{p} \}+\{ \alpha t\tilde{p},\tilde{q} \}+\{ \alpha t\tilde{p},\alpha t\tilde{q} \}=0So the transformation is not only canonical, but also univalent (since ). Thus, is a symplectic transformation.
We can also find this out by using the second canonicity criterion. We just need to prove that the Poisson brackets are preserved. The fundamental Poisson brackets are
> The gradient of $K_{0}$ must therefore be > $$\nabla_{\tilde{x}}K_{0}=\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}???(TODO: Finish this, 13/05/2025 end of lesson)
According the first criterion, there exists some for which . We now that . The condition is that , which means . We can calculate , so let's do that first:
> The modified Hamiltonian is > $$K(\tilde{w},t)=H(w(\tilde{x},t),t)- \frac{\alpha \tilde{x}_{1}^{2}}{2}(TODO: Finish this, 13/05/2025 towards the end)