The states aren't really important, and they aren't really physical. The fundamental thing is the operator algebra

When you're first learning quantum, you learn to think of states as "things", and operators/observables as "measurements we do to states". Given a state $\ket \psi$ and a Hermitian operator $\mathcal{O}$, we get the "expected value of measuring $\mathcal{O}$ on $\ket \psi$" by computing $\mathbb{E}_{\ket\psi}[\mathcal{O}] := \bra \psi \mathcal{O} \ket \psi$.

But it turns out we can also look at the problem from a different perspective. When you learn intro quantum, you don't actually spend that long learning about properties of states. Instead, you learn a lot about the properties of the observables. You study their commutation relations, and that sort of thing. Really, the basic objects that we study are these observables. And in fact, we can take observables to be our fundamental "things". Then, you can think of states as "ways of measuring observables".

And we can make this formal. We can say that a state is any positive, linear functional of norm 1 on the space of observables. By linear functional, I mean that it's a function that takes in operators and spits out real numbers. By positive, I mean that the functions are nonnegative on positive semidefinite operators. And by norm 1, I mean that these functions are 1 on the identity. It's pretty simple to check that the expected value function $\mathbb{E}_{\ket\psi}$ that I defined above has these properties.

Amazingly, the GNS construction tells us that ever positive linear function of norm 1 can be represented as a vector state (i.e. it is an expected value for some state vector). So the reasonable measurements that we can do to operators look like measuring state vectors! We can think of state vectors as being a convenient representation of the measurements we can do to our operators.

Functions on Groups

Given a finite group $G$, I want to consider the algebra of complex-valued functions on $G$ (i.e. $\{f:G \to \C\}$).

Why? There are several reasons to study this algebra of functions. From a purely group-theoretic perspective, it is interesting because it can help you understand groups better. Today, however, I'll mostly focus on applications to CS.

What algebra structure are we using?

There are multiple choices of algebra structure on the set of functions $\{f:G \to \C\}$. Addition and scalar multiplication are pretty straightforward; clearly they should be done pointwise. You might also want to multiply the functions pointwise. This defines a perfectly fine algebra, but then nothing in our algebra actually depends on the group structure. We want to take advantage of the group structure to study these functions. So we won't define multiplication this way. Instead, we take the slightly stranger definition that \[ f_1 * f_2 (g) := \sum_{g_1 g_2 = g} f_1(g_1)\cdot f_2(g_2) \] We call this product convolution.

Miscellaneous facts about $\C[G]$

For $h \in G$, we define the $\delta$-function \[ \delta_h(g) := \begin{cases} 1 & g = h\\0 & \text{otherwise}\end{cases} \] With the convolution product, the multiplicative identity is the function $\delta_e$.

You can check that \[ f \star \delta_e(g) = \sum_{g_1g_2 = g} f(g_1) \delta_e(g_2) = f(g)\delta_e(e) = f(g) \] One useful basis for $\C[G]$ is given by $\{\delta_h\;|\;h \in G\}$.

We can define a hermitian product on $\C[G]$ by \[ \inrp {f_1} {f_2} := \frac 1 {|G|} \sum_{g \in G} f_1(g) \cdot \overline{f_2(g)} \]

Example: $\C[\Z/n\Z]$

Suggestively, I'll denote $\delta_k$ by $x^k$. Then, elements of $\C[\Z/n\Z]$ are sums $a = \sum_{i=0}^{n-1} a_i x^i$. The product is given by \[ a\cdot b = \sum_{i = 0}^{n-1} \left[\sum_{j + k = i} a_j b_k \right]x^i \] So we see that $\C[\Z/n\Z] = \C[x]/(x^n-1)$. This is an important example for CS applications.

One useful tool for studying rings is studying modules over that ring. This trick will be very helpful to us.

Modules over $\C[G]$

Let $M$ be a module over $\C[G]$. First, we note that we can use our action of $\C[G]$ on $M$ to define an action of $\C$ on $M$. Simply define $\lambda \cdot m := (\lambda \delta_e) \cdot m$ for all $m \in M$. You can check that this makes $M$ into a complex vector space. Since $\delta_e$ is the multiplicative identity, it commutes with every element of $\C[G]$. So the action of every other element of $\C[G]$ on $M$ is linear with respect to this $\C$-action.

Furthermore, given a $\C$-vector space structure on $M$, the $\C[G]$ module structure is entirely determined by how each $\delta_g$ acts on $M$. So we can describe $M$ as a complex vector space along with a map $\rho:G \to \Aut(M)$.

Representations

Characters

Now we've gone full circle and come back to $\C[G]$.

Note: $\chi_V$ is constant on conjugacy classes of $G$.

\[ \chi_V(hgh^{-1}) = \Tr[\rho(hgh^{-1})] = \Tr[\rho(h)\rho(g)\rho(h^{-1})] = \Tr[\rho(g)\rho(h^{-1})\rho(h)] = \Tr[\rho(g)] = \chi_V(g) \]

So we can also think the characters are complex-valued class functions.

There's a very nice theorem about the inner products of characters.

I don't have time to prove this at the moment. It's not too hard, though.

Therefore, the characters of irreps are an orthonormal subset of the group algebra.

Fact: the number of irreducible representations of $G$ is the same as the number of conjugacy classes of $G$. Therefore, characters of irreps form a basis for the space of class functions on $G$.

Example: $\C[\Z/n\Z]$

Recall that $\C[\Z/n\Z] = \C[x]/(x^n-1)$, the ring of polynomials modulo $(x^n-1)$. What do these look like in the irreducible character basis?

What are the irreps of $\Z/n\Z$?

Thus, to find the irreps of $\Z/n\Z$, we only have to consider homomorphisms $\phi : \Z/n\Z \to \Aut[\C] = \C \setminus \{0\}$. Since $\Z/n\Z$ is cyclic, these homomorphisms are determined by $\phi(1)$. Since 1 has order $n$ in $\Z/n\Z$, $\phi(1)$ must be an $n$th root of unity. Let $\omega$ be a primitive $n$th root of unity. Then for all $k = 0, \ldots, n-1$ we get a homomorphism $\phi_k : \Z/n\Z \to \C$ given by $\phi_k(1) = \omega^k$. Since we have found $n$ irreducible representations, we have them all. Since the representations are all 1-dimensional, these are also the irreducible characters.

We can express a function on $\Z/n\Z$ in the character basis by taking its inner products with the irreducible characters. \[ \inrp {\phi_k} f = \sum_{\ell=0}^{n-1} \phi_k(\ell) \overline{f(\ell)} = \sum_{\ell=0}^{n-1} \overline{f(\ell)} \omega^{\ell k}\] Recall, we expressed these functions $f$ as $f = \sum_{\ell = 0}^{n-1} f(\ell) x^{\ell}$. So this is essentially evaluating the polynomial on $\omega^k$. This also looks a lot like the Fourier transform. In fact, it is the Discrete Fourier Transform.

If we write the function as a column vector, we can express this operation by the matrix \[ \begin{pmatrix} 1 & 1 & 1 & 1 & \cdots & 1\\ 1 & \omega & \omega^2 & \omega^3 & \cdots & \omega^{n-1}\\ 1 & \omega^2 & \omega^4 & \omega^6 & \cdots & \omega^{2(n-1)}\\ 1 & \omega^3 & \omega^6 & \omega^9 & \cdots & \omega^{3(n-1)}\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & \omega^{n-1} & \omega^{2(n-1)} & \omega^{3(n-1)} & \cdots & \omega^{(n-1)(n-1)} \end{pmatrix} \begin{pmatrix} \overline{f(0)}\\ \overline{f(1)}\\ \overline{f(2)}\\ \overline{f(3)}\\ \vdots\\ \overline{f(n-1)} \end{pmatrix} \] It turns out that because this matrix is so regularly-structured, we can compute this matrix-vector product really quickly. The Fast Fourier Transform lets us compute this in $O(n \log n)$ time, which is must faster than the naive $O(n^2)$ time for regular matrix-vector products. The Fast Fourier Transform has applications all over CS. It's used in signal processing, data compression (e.g. image processing), solving PDEs, multiplying polynomials, multiplying integers, and convolution, among other things.

A Strange Tensor Product

I'm taking a class on quantum computation at the moment, so I've been thinking a lot about tensor products (which I've written about before here). The tensor product is a weird operation. On vector spaces, it has a fairly straightforward definition, although it has some very strange, unintuitive properties. But the tensor product of general modules is even weirder.

Here's an example. What is the tensor product of $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$ as $\mathbb{Z}$-modules? That is to say, what is $\mathbb{Z}/2\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/3\mathbb{Z}$? For convenience, I'll write $M$ for this tensor product from now on. Cyclic groups are not free $\mathbb{Z}$ modules, so we can't think about $M$ in the same way as we think about the tensor product of vector spaces. And the answer turns out to be very strange.

\[M = 0\]

This answer doesn't make very much sense to me. The tensor product of nonzero vector spaces $V \otimes W$ always contains copies of $V$ and $W$. How can two nonzero $\mathbb Z$-modules tensor to 0?

Although it is very counterintuitive, this fact is actually not very hard to prove. Suppose $a \otimes b \in M$. It's always true that $1 \cdot a \otimes b = a \otimes b$. But because $2$ and $3$ are relatively prime, we can express 1 as a linear combination of 2 and 3. In fact, $1 = 2 \cdot 2 - 3$. Then \[\begin{aligned} a \otimes b &= 1 \cdot (a \otimes b)\\ &= (2 \cdot 2 - 3) (a \otimes b)\\ &= 2 \cdot 2 (a \otimes b)- 3 (a \otimes b)\\ &= 2 [(2a) \otimes b] - a \otimes (3b)\\ &= 0 \end{aligned}\] Since we can use this trick to show that every element of $M$ is 0, $M$ must be the 0 module.

However, I don't find this proof very satisfying. To me, it doesn't feel like it explains why the tensor product should be 0. It just demonstrates that it is 0. Of course, the division between 'should' and 'is' is vague and subjective, but I would prefer a different perspective on the problem. Today, I came up with an alternative justification using tensor-hom adjunction. This justification isn't fully formal at the moment, but it gives me a sense of why I might think this tensor product should be 0.

Tensor-Hom Adjunction

To me, the tensor-hom adjunction is all about curried functions (which I've written about before here). The idea is essentially that if you have a bilinear function from $X \times Y$ to $Z$, you can express this as a linear function on $X$ that returns a linear function that takes in an element of $Y$ and returns an element of $Z$. This decomposition of a multi-argument function into single-variable functions that return other functions is precisely the idea of currying.

More formally, the tensor-hom adjunction gives us a natural isomorphism \[\text{Hom}(X \otimes Y, Z) \simeq \text{Hom}(X, \text{Hom}(Y, Z))\]

Using the Tensor-Hom Adjunction to Study Our Tensor Product

We can use this to study our module $M$. Let $Z$ be an arbitrary $\mathbb Z$-module. Then tensor-hom adjunction tells us that\[\text{Hom}(M, Z) = \text{Hom}(\mathbb Z /2 \mathbb Z \otimes \mathbb Z / 3 \mathbb Z, Z) \simeq \text{Hom}(\mathbb Z / 2 \mathbb Z, \text{Hom}(\mathbb Z / 3 \mathbb Z, Z))\]

The nonzero of $\mathbb Z / 3 \mathbb Z$ all have order 3. So their images under any $\mathbb Z$-linear map must have order 1 or order 3. Therefore, every element of $\text{Hom}(\mathbb Z / 3 \mathbb Z, Z)$ must have order 1 or 3 as an element of the group of $\mathbb Z$-linear maps. But the image of an element of $\mathbb Z / 2 \mathbb Z$ under a linear map must have order 1 or order 2. Since every element of the codomain has order 1 or order 3, we must map every element of $\mathbb Z / 2 \mathbb Z$ to the identity. So the only elements of $\text{Hom}(\mathbb Z / 2 \mathbb Z, \text{Hom}(\mathbb Z / 3 \mathbb Z, Z))$ are maps which send $\mathbb Z / 2 \mathbb Z$ to the zero map $\mathbb Z / 3 \mathbb Z \to Z$. This means that the only element of $\text{Hom}(M, Z)$ is the zero map, no matter what $Z$ is.

This is a pretty convincing argument to me that maybe $M$ should be 0. In particular, if $Z = M$, then we see that the only linear map from $M$ to itself is the zero map. So it looks like $M$ has to be 0. Furthermore, this argument generalizes naturally in the same way that the previous calculation does. The argument should work for $\mathbb Z / m \mathbb Z \otimes \mathbb Z / n \mathbb Z$ whenever $n$ and $m$ are relatively prime. So maybe it is a good perspective to keep in mind. We can always find nontrivial maps between nonzero vector spaces over a fixed field. But because modules can have restrictions on element orders, we can have modules which don't have nontrivial maps to each other. And this can result in their tensor products being 0.

I'm still not totally satisfied with my answer to this problem. The tensor product still seems to be a little spooky. If you have a different way of looking at the tensor product in general, or at this problem in particular, I'd love to hear it in the comments.

Today, I want to take Maxwell's equations and write them in the language of differential forms. The resulting equations are clearly covariant (i.e. they look the same after you apply a Lorentz transformation), and look a lot simpler than Maxwell's equations in vector notation. This is one of my favorite examples of how differential forms can make life easier.

Maxwell's Equations

In CGS units, Maxwell's equations are given by

1. $\nabla \cdot E = 4 \pi \rho$
2. $\nabla \cdot B = 0$
3. $\nabla \times E = -\frac 1 c \frac{\partial B}{\partial t}$
4. $\nabla \times B = \frac{4\pi}c J + \frac 1 c \frac{\partial E}{\partial t}$

$E$ is the electric field, $B$ is the magnetic field, $J$ is the electric current density, and $\rho$ is the electric charge density. $E, B$ and $J$ are vector fields and $\rho$ is a scalar field.

If we want to write Maxwell's equations with differential forms, we need to decide what type of forms will represent $E, B, J$ and $\rho$. Following Stern et al we will decide this based on how these fields are used in various equations.

First, we consider Faraday's law \[ \oint_C E \cdot d\ell = -\frac d {dt} \int_S B \cdot dA \]

$E$ is integrated over a curve, so $E$ naturally corresponds to a 1-form (Since one-forms are objects that you can integrate along curves). We will write the 1-form associated to $E$ as $\eta = E^\flat$. Meanwhile $B$ is integrated over a surface, so $B$ naturally corresponds to a 2-form. We will write this 2-form as $\beta = \star B^\flat$ as the 2-form associated to $B$. $\beta$ can be thought of as a function that measures the flux of $B$ through oriented parallelograms. $J$, the current density, is integrated over surfaces to find the current passing through the surface, so $J$ is naturally a 2-form. We will write this forms as $\mathscr J = \star J^\flat$. Finally, $\rho$ is integrated over volumes to find the enclosed charge, so $\rho$ is naturally a 3-form, and we will call the 3-form $\rho$.

We recall the following rules for translating from vector calculus to differential forms: \[\begin{aligned} (\nabla \cdot v)^\flat &= \star d \star v^\flat\\ (\nabla \times v)^\flat &= \star d v^\flat \end{aligned}\] We can use these rules to write Maxwell's equations in terms of $\eta, \beta, \mathscr J,$ and $\rho$.

We will start with the first equation. On the left hand side, $\nabla \cdot E$ becomes $\star d \star \eta$, which is a 0-form. We want to set it equal to the 3-form $4\pi \rho$. So we use the Hodge star to turn $\star d \star \eta$ into a 3 form and find that $\star \star d \star \eta = 4\pi \rho$. In 3D, $\star\star = 1$, so our equation is just

1'.  $d\star \eta = 4\pi \rho$

Now, we move on to the second equation. $\nabla \cdot B$ becomes $\star d \star (\star \beta)$. Since $\star\star = 1$, this just becomes $\star d \beta$. So our equation is $\star d \beta = 0$. Applying $\star$ to both sides gives $d\beta = 0$.

2'.  $d\beta = 0$

For the third equation, our substitutions gives us $\star d \eta = -\frac 1 c \frac{\partial\star \beta} {\partial t}$. Applying $\star$ to both sides yields $d\eta = -\frac 1 c \star \frac{\partial\star \beta} {\partial t}$. But we can pull the $\star$ inside the derivative to get

3'.  $d \eta = -\frac 1 c \frac{\partial \beta}{\partial t}$

Finally, we translate the last equation. Our substitution rules give us

4'.  $\star d \star \beta = \frac {4\pi} c \star \mathscr J + \frac 1 c \frac{\partial \eta}{\partial t}$


Putting all of the equations together, we have

1'.	$d \star \eta = 4 \pi \rho$
2'.	$d \beta = 0$
3'.	$d \eta = - \frac 1 c \frac{\partial \beta}{\partial t}$
4'.	$\star d \star \beta = \frac{4\pi}{c} \star \mathscr J + \frac 1 c \frac {\partial \eta}{\partial t}$

Now, we have written the equations using differential forms. But the equations still don't look very relativistic yet - we still have a big distinction between space and time derivatives. For our next step, we will stop thinking about forms on space that change over time, and instead think about forms on $3+1$-dimensional spacetime (i.e. spacetime with 3 spatial dimensions and 1 time dimension). In spacetime, we have both the spatial exterior derivative, the spatial Hodge star, the spacetime exterior derivative, and the spacetime Hodge star. We will denote the spatial operators $d_s, \star_s$ respectively, and we will denote the spacetime operators $d$ and $\star$.

The Exterior Derivative and Hodge Star in Spacetime

Before we can write Maxwell's equations in spacetime, we have to learn about how our spatial operators are related to our spacetime operators. We will use the convention that coordinates in spacetime are written like \[ (x^0, x^1, x^2, x^3) = (ct, x, y, z) \]

The Exterior Derivative

Now, we will look at the relationship between $d$ and $d_s$. Let $\omega = \sum_I \omega_I dx^I$ be a spatial differential form (i.e. no component of $\omega$ involves a $dx^0$). We can compute $d\omega$ as follows \[\begin{aligned} d\omega &= \sum_I d \omega_i \wedge dx^I\\ &= (dx^0 \wedge \partial_0 + d_s)\omega \end{aligned}\]

So we see that $d = dx^0 \wedge \partial_0 + d_s$. The spacetime exterior derivative is just the spatial exterior derivative with an extra term related to the time derivative.

The Hodge Star of Spatial Forms

Now, we will relate $\star$ and $\star_s$. Let $\omega$ be a spatial $k$-form. The spacetime Hodge star is defined by the property that \[ \omega \wedge \star \omega = \left\langle \omega, \omega\right\rangle \mu \] Here $\mu = dx^0 \wedge dx^1 \wedge dx^2 \wedge dx^3$ is the spacetime volume form. Let $\mu_s = dx^1 \wedge dx^2 \wedge dx^3$ be the spatial volume form. Clearly $\mu = dx^0 \wedge \mu_s$. Furthermore, we know that $\omega \wedge \star_s \omega = \left\langle \omega, \omega\right\rangle \mu_s$. Therefore, \[\begin{aligned} \omega \wedge \star \omega &= \langle \omega, \omega \rangle \mu\\ &= dx^0 \wedge \langle \omega, \omega \rangle \mu_s\\ &= dx^0 \wedge \omega \wedge \star_s\omega\\ &= \omega \wedge (-1)^k (dx^0 \wedge \star_s \omega) \end{aligned}\] So $\star \omega = (-1)^{k} \; dx^0 \wedge \star_s \omega$ when $\omega$ is purely a spatial $k$-form.

The Hodge Star of Forms with a Time Component

Now, suppose that $\omega = dx^0 \wedge \omega_s$ where $\omega_s$ is a spatial form. Then $\left\langle \omega, \omega\right\rangle = -\left\langle \omega_s,\omega_s\right\rangle$, so we need $\omega \wedge \star \omega = -\left\langle\omega_s,\omega_s\right\rangle\mu$. We know that $\omega_s \wedge \star_s \omega_s = \left\langle\omega_s,\omega_s\right\rangle \mu_s$. Thus, \[\omega \wedge \star_s \omega_s = dx^0 \wedge \omega_s \wedge \star_s \omega_s = dx^0 \wedge \mu_s = \mu \] So, $\star \omega = -\star_s \omega$ when $\omega$ is the wedge product of $dx^0$ and a spatial form.

Finally, we note for completeness that $\star dx^0 = -\mu_s$.

Covariant formulation of Maxwell's equations

Finally, we've developed all of the tools we need to write Maxwell's equations in spacetime. We will begin with the homogeneous equations (equations two and three). Maxwell's second equation tells us that $d\beta = 0$ and Maxwell's third equation tells us that $d_s \eta + \frac 1 c \frac{\partial \beta}{\partial t} = 0$. Because, we write coordinates in spacetime as \[ (x^0, x^1, x^2, x^3) = (ct, x, y, z) \] it turns out that $\frac 1 c \frac{\partial \beta}{\partial t} = \frac{\partial \beta}{\partial x^0} =: \partial_0 \beta$. So we can write Maxwell's third equation as $d_s \eta + \partial_0 \beta = 0$

The second equation is an equation of 3-forms, and the third equation is an equation of 2-forms. We can make them both equations of 3-forms by wedging the third equation with $dx^0$. Then we have $d_s \beta = 0$ and $dx^0 \wedge(d_s \eta + \partial_0\beta) = 0$. We can add these together to get $d_s \beta + dx^0 \wedge \partial_0 \beta + dx^0 \wedge d_s\eta = 0$. We note that because we are adding together forms of different types, their sum is 0 if and only if the individual terms in the sum are 0. So this equation expresses both Maxwell's second law and his third law. Inspecting the sum, we see that the first two terms are our expression for $d\beta$! Furthermore, the last term is $d(\eta \wedge dx^0)$. \[\begin{aligned} d(\eta\wedge dx^0) &= d\eta \wedge dx^0\\ &= (d_s \eta + dx^0 \wedge \partial_0 c\eta) \wedge dx^0 \\ &= d_s \eta \wedge dx^0\\ &= dx^0 \wedge d_s \eta \end{aligned}\] Therefore, we can write Maxwell's second and third equations as $d\beta + d(\eta \wedge dx^0) = 0$, or $d(\beta + \eta \wedge dx^0) = 0$. To simplify even more, we call $F := \beta + \eta \wedge dx^0$ the Faraday tensor, and simply write $dF = 0$.

Now, we consider $d \star F$. We'll start by computing $\star F$. \[ \star F = \star(\beta + \eta \wedge dx^0) = \star \beta + \star(\eta \wedge dx^0) \] Since $\beta$ is a spatial 2-form, $\star \beta = dx^0 \wedge \star_s \beta$. Since $\eta$ is a spatial 1-form, \[ \star(\eta \wedge dx^0) = -\star(dx^0 \wedge \eta) = -(- \star_s \eta) = \star_s \eta\] Putting this together shows us that $\star F = dx^0 \wedge \star_s \beta + \star_s \eta$. Now, we can take the exterior derivative. \[\begin{aligned} d \star F &= d(dx^0 \wedge \star_s \beta + \star_s \eta)\\ &= d_s \star_s \eta + dx^0 \wedge (\partial_0 \star_s \eta - d_s \star_s d_s \beta) \end{aligned}\]

Maxwell's first equation tells us that $d_s \star_s \eta = 4\pi \rho$. Maxwell's fourth equation tells us that $\partial_0\eta - \star_s d_s \star_s \beta = -\frac{4\pi} c \star_s \mathscr J$. Therefore, \[\begin{aligned} d\star F &= 4\pi \star \rho\;dx^0 -\frac {4\pi} c dx^0 \wedge \mathscr J\\ \end{aligned}\] We define $c\rho - dx^0 \wedge \mathscr J = \mathfrak J$ to be the four-current. Now, our equation reads $d \star F = \frac {4\pi} c \mathfrak J$. This lets us finally express Maxwell's equations (in cgs units) as \[\begin{aligned} dF = 0 \quad\text{and}\quad d \star F = \frac{4\pi} c \mathfrak J \end{aligned}\]

Final Thoughts

I think this form of Maxwell's equations is very pretty. Because $F$ and $\mathfrak J$ are coordinate-independent objects, you can tell these equations must also be Lorentz covariant. And the equations don't distinguish between space and time coordinates.

Noether's theorem tells us that conserved quantities come from symmetries of physical systems. For example, momentum is conserved because the laws of physics are translation invariant.
This deep insight is helpful for understanding when quantities should be conserved. A mass falling off of a building is allowed to gain momentum because the system is not translation invariant - as you move vertically, the gravitational potential changes. However, a train moving along its tracks should conserve momentum because no relevant physical quantity changes as you move around the surface of the earth.

Proving Noether's Theorem

In the system of Hamiltonian Mechanics, the proof of Noether's theorem is surprisingly simple and elegant. First, we need to set up some machinery. Recall Hamilton's equations of motion \[\begin{aligned} \dot p_i &= \frac{\partial H} {\partial q_i}\\ -\dot q_i &= \frac{\partial H} {\partial p_i} \end{aligned}\] Hamilton's equations define a vector field $X_H = (\dot q, \dot p)$ on phase space that describes how a particle evolves over time. The trajectory of a particle starting at position $q$ with momentum $p$ is the integral curve of $X_H$ passing through point $(q,p)$.
We can express Hamilton's equations more simply using a symplectic form. A symplectic form is a closed, nondegenerate differential 2-form. Using $\Omega$, Hamilton's equations become \[ dH = \iota_{X_H} \Omega \] Where $d$ is the exterior derivative and $\iota_{X_H} \Omega$ is the interior product, a one-form defined by $(\iota_{X_H} \Omega)(X_1) = \Omega(X_H, X_1)$.
Now, let $X_G$ be an infinitesimal symmetry transformation. Then $\mathcal{L}_{X_G}H = 0$. That is to say, if we move space a small amount in the $X_G$ direction, the Hamiltonian stays the same. This is exactly what we mean by a symmetry. Furthermore, let $X_G$ be the symplectic gradient of some potential function $U_G$. i.e. $dU_G = \iota_{X_G} \Omega$. Then \[\begin{aligned} 0 &= \mathcal{L}_{X_G} H\\ &= \iota_{X_G} dH + d \iota_{X_G} H &&\text{Cartan's magic formula}\\ &= \iota_{X_G} dH + 0 &&H\;\text{doesn't take arguments, so}\; \iota_{X_G}H = 0\\ &= \iota_{X_G} \iota_{X_H} \Omega &&\text{definition of}\;X_H \\ &= \Omega(X_H, X_G) && \text{definition of the}\; \iota \; \text{operation}\\ &= -\Omega(X_G, X_H) && \Omega\;\text{is antisymmetric}\\ &= -\iota_{X_H} \iota_{X_G} \Omega && \text{definition of the}\;\iota\;\text{operation}\\ &= -\iota_{X_H} dU_G && \text{definition of}\; X_G\\ &= -\iota_{X_H} dU_G + d\iota_{X_H} U_G &&U_G\;\text{doesn't take arguments, so}\; \iota_{X_H}U_G = 0\\ &= -\mathcal{L}_{X_H} U_G && \text{Cartan's magic formula}\\ \end{aligned}\] Therefore, the quantity $U_G$ does not change when we flow along the vector field $X_H$. But flow along $X_H$ is time evolution! So $U_G$ is a conserved quantity over time!

Examples

Translation in One Dimension

Suppose we have a one-dimensional physical system. Furthermore, suppose our Hamiltonian is invariant under translation. The vector field that infinitesimally moves things in the $x$ direction is the vector field that points in the $x$ direction. We need to express this vector field as a symplectic gradient. So we want a function $U(x, p)$ that satisfies \[\begin{aligned} \frac{\partial U(x, p)} {\partial x} &= 0\\ \frac{\partial U(x, p)} {\partial p} &= -1 \end{aligned}\] Clearly, $U(x, p) = -p$. Satisfies this condition. So the quantity $-p$ (and therefore $p$ as well), is conserved in this physical system! Just like that, we have shown the conservation of momentum!

Rotation in Two Dimensions

Suppose we have a two-dimensional physical system that is invariant under rotation. A rotation by an angle $\theta$ is given by the matrix \[\begin{pmatrix} \cos \theta & -\sin \theta\\\sin \theta & \cos \theta\end{pmatrix}\] For very small $\theta$, this matrix is approximately \[\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\] Rotation affects both position and momentum the same way. So an infinitesimal rotation is given by the transformation $\dot x = -y, \dot y = x, \dot p_x = -p_y, \dot p_y = p_x$. Now, to express this vector field as a symplectic gradient, we need a function $U(x, y, p_x, p_y)$ satisfying \[\begin{aligned} \frac {\partial U} {\partial x} &= \dot p_x = -p_y & \frac {\partial U} {\partial y} &= \dot p_y = p_x\\ \frac {\partial U} {\partial p_x} &= -\dot x = y & \frac {\partial U} {\partial p_y} &= -\dot y = -x \end{aligned}\] To satisfy these conditions, we pick $U(x, y, p_x, p_y) = yp_x - xp_y$, which is the angular momentum!

In quantum mechanics, we represent a particle as a vector in a 'state space' $V$. If we have two particles, we represent the pair as a vector in a product vector space $V_1 \otimes V_2$. This product space is called the 'tensor product'. But what is this tensor product? And why is does it represent pairs of particles?

Warm Up: The Direct Product

Before we talk about the tensor product of vector spaces, we'll go over a more intuitive way of taking the product of vector spaces. Given vector spaces $V$ and $W$, the direct product, $V \times W$, is defined as the set of all pairs of vectors $(v, w)$ for $v \in V$ and $w \in W$. We add together vectors in this new space component by component. \[ (v_1, w_1) + (v_2, w_2) = (v_1 + v_2, w_1 + w_2) \] And we scale up vectors by scaling up both components \[ \lambda(v, w) = (\lambda v, \lambda w)\]

Tensor Products of Vector Spaces

To get the tensor product $V \otimes W$, we can modify the direct product. We still want to look at pairs $(v, w)$. But we'll change the definitions of multiplication and addition a bit. In our new definition of scalar multiplication, multiplying our vector by a scalar only scales one of the components. \[ \lambda(v, w) = (\lambda v, w) = (v, \lambda w)\] For addition, we now only define addition if one of the components matches. \[ (v_1, w) + (v_2, w) = (v_1 + v_2, w)\] The sum only works because the second component in each term is $w$. We get a similar sum if the first components are equal and the second components are different. For all other sums, we just define them as themselves. $(v_1, w_1) + (v_2, w_2)$ is just defined to be itself. It cannot be simplified.

Finally, instead of writing $(v, w)$, we instead write $v \otimes w$. This way, it looks different from the elements of $V \times W$. We call this new space the tensor product of $V$ and $W$.

Simple Example

Let's look at the tensor product of $\mathbb{R}$ with $\mathbb{R}$. The simplest elements of $\mathbb{R} \otimes \mathbb{R}$ look like $1 \otimes 2 + 3 \otimes 4$. Using our rules we defined earlier, we can do things like

\[\begin{aligned} 2 \otimes 3 + 4 \otimes 6 &= 2 \otimes 3 + 4 \otimes 2\cdot 3\\ &= 2 \otimes 3 + 8 \otimes 3\\ &= 10 \otimes 3 \end{align*}

But Why?

The definition of a tensor product looks fairly arbitrary. But it winds up having some nice properties that turn out to be interesting, and surprisingly natural, to study. In order to describe these properties and why they're useful, we will have to make an expedition into the wonderful land of algebra.

Over the last 100 years, mathematicians have realized that when studying mathematical objects, it is incredibly useful to study functions between these objects. When you're looking at vector space, the natural functions to study are linear functions. A linear function is a function that commutes with addition and scalar multiplication. That is to say, it is a function $f:V \to W$ such that $f(v_1 + v_2) = f(v_1) + f(v_2)$, and $f(\lambda v) = \lambda f(v)$.

A lot of functions that take in multiple arguments also have a similar property. For example, let's look at multiplication of real numbers. Mathematically, we can write this as a function that takes in two numbers and spits one number back out. i.e., $m : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$. If we fix the first number and vary the second number, this is a linear function!

\[\begin{aligned} m(a, b_1 + b_2) &= a(b_1 + b_2)\\ &= ab_1 + ab_2\\ &= m(a, b_1) + m(a, b_2)\\ m(a, \lambda b) &= a\cdot \lambda b\\ &= \lambda(a b)\\ &= \lambda \cdot m(a, b) \end{aligned}\]

But it actually has a stronger property. If we fix the second argument and vary the first argument, we also get a linear function out. So this function $m$ is a linear function in either argument. We call such a function bilinear (or multilinear if it takes more than 2 arguments). Multilinear functions pop up all over the place. The cross product, dot product and determinant are all multilinear.

From the definition of multilinearity, we get some identities that multilinear functions have to satisfy. Suppose $f$ is multilinear. Then

\[\begin{aligned} f(a, \lambda b) &= \lambda f(a, b) = f(\lambda a, b)\\ f(a, b_1 + b_2) &= f(a, b_1) + f(a, b_2)\\ f(a_1 + a_2, b) &= f(a_1, b) + f(a_2, b) \end{aligned}\] Do these equations look familiar? They're exactly the rules that we made the tensor product follow.

Tensor Products: An Algebraic Perspective

To an algebraist, tensor products are fundamentally about linear maps. The tensor product of two vector spaces $V$ and $W$ is defined by a universal property. Suppose $h$ is a function that takes in an element of $V$ and an element of $W$ and returns an element of a third vector space $Z$. We can write this as a function $h:V \times W \to Z$. Furthermore, let $h$ be bilinear. Then we can extend $h$ to a unique linear map $\bar h$ from $V \otimes W$ to $Z$. If we want to be fancy, we can draw this requirement as the following commutative diagram

The map $\varphi:V \times W \to V \otimes W$ is in a sense, the 'most general' bilinear map out of $V \times W$. We can write any bilinear map $V \times W \to Z$ as the composition of $\varphi$ with some linear map $V \otimes W \to Z$.

This is what tensor products are really about. The confusing definition given above is made specifically so that tensor products play nicely with bilinear maps. Whenever you see a tensor product, you should look around for bilinear maps.

Back to Quantum Mechanics

In quantum mechanics, linear functions play a central role. You look at quantum states as vectors in some large vector space, and special linear functions on this vector space correspond to quantities you can observe. These are called "observables".

Now, what if we have two particles? Each one independently is a vector in some vector space, but how do we describe the pair? Suppose we have some observable that we can measure on the pair. When restricted to one particle, it should be a linear operator like an ordinary observable. So really, our observables for the pair should be bilinear functions from the direct product of the vector spaces. This means that they are linear operators on the tensor product space!

The commutative diagram is from wikipedia