What is energy? What is physics?

In your high-school physics classes, you may have studied the dynamics of various physical systems, but you may have also heard that physics is the "general science", that is universally applicable. One may wonder if physics can be formulated as some general framework that can handle any type of system without more specific assumptions about such system, and what interesting conclusions can be drawn about physics in such a general setting.

Fundamentally, we seek to make predictions about some observables. You may imagine that all the observable information about a system is given by some abstract "phase space" of "state vectors" such that each observable is some function of the state vector. For example if the phase space is parameterized by position and momentum, then the position and momentum observables would be projection functions and various other observables can be written as compositions thereof of functions with these projections e.g. $E=\frac{p^2}{2m}+U(x,p)$.

The question of why position and momentum are sufficient to specify our phase space in so many real-life situations is a rather advanced one -- here's a paper that explains that this general "first-derivative sufficiency" in physics arises as a special case of something to do with Fisher information [1] -- but I don't understand Fisher information.

The Lie theory of dynamics

One may say that we are particularly interested in the time-evolution of such observables, $dA/dt$. If we can find a differential equation for this, then we can find the value of $A$ given some initial condition. But more generally we may not have an initial condition as such and maybe the problem we're trying to solve isn't even a differential equation as such but we just want to speak more abstractly about the dynamics of the system.

This should all be screaming "Lie groups" to you. 

One may imagine that time evolution is the flow under a certain vector field. 

More precisely, suppose the state of a system is given by some $\psi=(x, p)$, which represents the position and momentum of a particle at time $t$. Then the "position at time $t$" is an observable given by the function $X(\psi)=x$, the projection onto the $x$ co-ordinate -- but we may also think about the observable "position at time $t+\Delta t$", which is some function $\Phi^{\Delta t}X$ that projects onto a different co-ordinate system.

This co-ordinate transformation on the phase space is a type of canonical transformation -- we will soon define this generally, but it is important to realize that motion/time-evolution is a type of canonical transformation. 

You may imagine that such canonical transformations form a Lie group, and as is standard the Lie algebra to this Lie group consists of vector fields (flows) on the phase space. 

Noether and symplectic geometry

The question, then, is what sort of transformations count as "canonical transformations", or equivalently, what sort of vector fields are we interested in. What kind of "co-ordinate transformations" are acceptable?

We want to be as general as possible, so we do not wish to restrict what sort of paths are predicted on the phase space -- rather, we want to restrict the relationships between the predicted paths, i.e. how a distribution evolves under time evolution. We won't go into details of why this is true (because I do not fully understand it yet -- apparently it is called Liouville's theorem and has to do with "conservation of information", which has to do with probabilities inferred from some symmetries, see a priori probability), but we expect the vector fields to be solenoidal, i.e. have zero divergence. Analogous to how irrotational vector fields are precisely those that are gradients of functions, solenoidal vector fields are precisely those that are the symplectic gradients of functions (see reddit for a proof):

$$\nabla\cdot \vec{v}=0\leftrightarrow \exists g, \vec{v}=\frac{\partial g}{\partial y}\hat{x}-\frac{\partial g}{\partial x}\hat{y}$$

This "symplectic gradient" always points along the contours of $g$ -- thus, these flows conserve the quantity $g$. More generally, the change in some observable $f$ over the flow that is the symplectic gradient of $g$ is given by $\vec{v}\cdot\nabla g$, and is called the Poisson bracket of the two observables:

$$\{f,g\}=\frac{\partial g}{\partial y}\frac{\partial f}{\partial x}-\frac{\partial g}{\partial x}\frac{\partial f}{\partial y}$$

It should be intuitively clear that this Poisson bracket corresponds to the Lie bracket of their symplectic gradients -- if the vector fields commute, they must be constant under the flow of the other, etc.

The Hamiltonian and conservation of energy

The evolution of a classical Newtonian state is given by 

$$\dot{x}=p/m$$

$$\dot{p}=F$$

We wish to find the observable $H$ such that this evolution is represented by the symplectic gradient of $H$, i.e. so that:

$$p/m=\{x, H\} = \frac{\partial H}{\partial p}\frac{\partial x}{\partial x}-\frac{\partial H}{\partial x}\frac{\partial x}{\partial p}=\frac{\partial H}{\partial p}$$

$$F = \{p, H\}=-\frac{\partial H}{\partial x}$$

Integrating, the quantity we want is

$$H=\frac{1}{2m}p^2-\int F dx$$

This quantity is called the "energy" of the system -- often we call $p^2/2m$ the "kinetic energy" and $U=-\int F dx$ the "potential energy". In particular, it is immediately clear that energy is conserved. In general, the physics of a system can be completely specified by some "Hamiltonian" $H$, this rather generalized definition of energy, whose symplectic gradient represents time translations.

Time evolution, Schrodinger and Heisenberg pictures, Noether's theorem

So far, we have discussed quantum mechanics without any reference to changes across time. You might think we could just upgrade $\psi(x)$ to $\psi(x,t)$ and e.g. an observable $Q_t$ measuring a value $q$ at time $t$ would have eigenvectors whose cross-section at $t$ are of the form $\delta(x-a)$. But this would mean the entire $\psi(x,t)$ is the state of the object, rather than there being a state at each value of $t$, and time would be an observable. This is clearly not what we want (right now -- although to be consistent with special relativity we will need to treat space and time on an equal footing later in this series).

Instead, a more appropriate approach is to say that the state is a function of time $|\psi(t)\rangle$ and the evolution of the state is given by some operation $|\psi(t)\rangle=U[|\psi(0)\rangle]$.

How do we know that $U$ is a linear operator? What does it mean for $U$ to be a linear operator anyway? The only sense in which such a linearity can be tested is by looking at a state in a superposition. So suppose $|\psi(0)\rangle=|\psi_1(0)\rangle+|\psi_2(0)\rangle$. Now $|\psi(t)\rangle=U[|\psi_1(0)\rangle+|\psi_2(0)\rangle]$ -- this is from the perspective of some observer Alice.

But if another observer Bob had previously observed and collapsed the system to $|\psi_1(0)\rangle$ at time 0, then according to him, the state should evolve to $U[|\psi_1(0)\rangle]$, and if he had observed the system in $|\psi_2(0)\rangle$, his knowledge of the system would evolve to $U[|\psi_2(0)\rangle]$.

So according to Alice, who doesn't know what Bob has observed (she has not observed him), her knowledge of the system can also be written as $U[|\psi_1(0)\rangle]+U[|\psi_2(0)\rangle]$. Thus

$$U[|\psi_1(0)\rangle+|\psi_2(0)\rangle]=U[|\psi_1(0)\rangle]+U[|\psi_2(0)\rangle]$$
I.e. $U$ is linear, so we can write it as a linear operator as in $U|\psi(0)\rangle$. (The above scenario is called Wigner's friend)

$U$ is also clearly a unitary operator, as it must preserve all lengths.

We can consider infinitesimal time evolutions $U_t(dt)$ representing evolution of the state from $t$ to $t+dt$. Then:

$$|\psi(t)\rangle=U_0(dt)\dots U_{t-dt}(dt)|\psi(0)\rangle$$
This product integral can be written alternatively as:

$$|\psi(t)\rangle=\mathcal{T}\left\{e^{\int \ln U_t(dt)}|\psi(0)\rangle\right\}$$

$\mathcal{T}$ is the time-ordering operator which orders a product like $H(t_1)H(t_2)$ in order of ascending $t$ in an expansion. Can you see why this is necessary (hint: $e^{AB}\ne e^{A+B}$ for noncommuting $A,B$).

Oh, and it's not actually an operator -- not even in the math sense, it's a "formal operation", one that takes a form or sentence (rather than its value) -- in this case the $\exp$ Taylor expansion -- and changes it some way.

$\ln U_t(dt)$ is an infinitesimal, and it's easy to see that it is equal to $U_{t}'(0)dt$ -- a member of the Lie algebra. We know, of course, that the Lie Algebra of the unitary group is comprised of anti-Hermitian operators (this can be checked without Lie Algebra, of course), and so $iU_{t}'(0)$ is Hermitian. From Lie Algebra, we can tell that this represents a generator of time translations -- and from a little experience of classical mechanics, we want this to represent energy. So for dimensional consistency with energy, we write:

$$H(t)=i\hbar U_{t}'(0)$$
(Why $\hbar$ and not $h$? Because $U_{t}'(0)$ is basically already in "radians per second".) This is called the Hamiltonian operator. It determines $U_t$, and thus describes the time evolution of a state. How exactly? Since:

$$|\psi(t+dt)\rangle=U_t(dt)|\psi(t)\rangle$$
We can write re-arranging:

$$\frac{\partial |\psi(t)\rangle}{\partial t}=-\frac{i}{\hbar}H(t)|\psi(t)\rangle$$
This is the most general form of the Schrodinger equation. Note that the earlier exponential equation is the "general solution" to this equation -- obviously not very useful, rewritten as what is known as the Dyson series:

$$|\psi(t)\rangle=\mathcal{T}\left\{e^{-i/\hbar\int H(t) dt}|\psi(0)\rangle\right\}$$
It's easy to show from this that the evolution of a density matrix $\rho(t)$ is similarly:

$$\frac{\partial\rho(t)}{\partial t}=-\frac{i}\hbar [H, \rho]$$
Which is the von Neumann equation, whose solution is given by:

$$\rho(t)=\mathcal{T}\left\{e^{-i/\hbar\int H(t) dt}\rho(0)e^{i/\hbar \int H(t) dt}\right\}$$
These should all appear as obvious special cases of Lie theoretic results.

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$H(t)$ is not the same as $i\hbar\partial/\partial t$. $H(t)$ is a Hermitian operator, i.e. an observable, while $\partial/\partial t$ does not act on the Hilbert space at all. One could also see what could wrong by equating the two in the "solution to the Schrodinger equation" above. The Schrodinger equation does not say that $H$ and the time-derivative are equal in general -- rather, it says that they are the same on a valid state vector $|\psi(t)\rangle$ -- you cannot just "factor this out".

So the Hamiltonian is fundamentally what determines the dynamics of a quantum system. Give me a Hamiltonian, and you've given me a theory. The Schrodinger equation (or equivalently the von Neumann equation) above is just an axiom of quantum mechanics/of any quantum mechanical theory.

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Can we talk about the velocity and acceleration observables for a moment? Actually, we can't, because they fundamentally have to do with time evolution, and we can't have observables that depend on the time-evolution of the state -- observables must act on the Hilbert space. But we can define observables that predict how the state will evolve (like the Hamiltonian with the Schrodinger equation).

Doing this systematically is where the Heisenberg picture comes in.

What does this mean? Everything we've discussed so far is the Schrodinger picture, where the state evolves on a fixed background basis created by the observables' eigenvectors -- so observables represent active transformations. Instead, we can have a completely different picture of reality, the Heisenberg picture, where we view time-evolution as simply viewing the state in a different basis -- then the observables represent passive transformations.

OK, so how do we do this? Remember how every question in quantum mechanics can fundamentally be asked in terms of expectation values (specifically those of Hermitian projections). The expected value of an observable at time $t$ of course evolves as:

$$\langle A\rangle(t) = \langle\psi|U^*(t)AU(t)|\psi\rangle$$
In the Schrodinger picture, we attach the $U(t)$ to $|\psi(0)\rangle$ to make $\langle A\rangle(t)=\langle\psi(t)|A|\psi(t)\rangle$. In the Heisenberg picture instead, we attach the $U(t)$ to the $A(0)$, writing $\langle A\rangle(t)=\langle\psi|A(t)|\psi\rangle$.
From differentiating conjugation in $A(t)=U^*(t)A(0)U(t)$, we get:

$$\frac{dA}{dt}=\frac{i}\hbar [H, A]$$
This is the Heisenberg equation. Immediately, it yields:

$$\begin{array}{l}\frac{{dX}}{{dt}} = \frac{i}{\hbar }\left[ {H,X} \right]\\\frac{{{d^2}X}}{{d{t^2}}} =  - \frac{1}{{{\hbar ^2}}}\left[ {H,\left[ {H,X} \right]} \right]\end{array}$$
Thinking of the evolution of $X$ as a translation of the co-ordinate system, etc., what this does is give us two conditions on what the Hamiltonian should look like for a "Euclidean" system:

$$\begin{array}{l}\left[ {H,X} \right] =  - \frac{{i\hbar }}{m}P\\\left[ {H,\left[ {H,X} \right]} \right] = \frac{{{\hbar ^2}}}{m}U'(x)\end{array}$$
This gives us yet another strong reason (besides the fact that the Hamiltonian generates time-translations, that the "eigenvectors" of $\partial/\partial t$ are the energy states by the de Broglie theorem (but not really), etc.) to suspect that the Hamiltonian represents the energy of the system. Indeed if we use:

$$H=\frac1{2m}P^2+U(x)$$
We can confirm those conditions above. Well, this is certainly not the only Hamiltonian compatible with classical mechanics, so at this point, I'll just say that this is confirmed by experiment, and is an axiom of the quantum theory of Euclidean systems.

Exercise: By taking expectation values in the Heisenberg equation, show that $m\frac{d}{dt}\langle x\rangle =\langle p\rangle$ and $\frac{d}{dt}\langle p\rangle = -\langle U'(x)\rangle$ under the Euclidean Hamiltonian. This is called the Ehrenfest theorem.

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I'll discuss one final application of the Heisenberg formalism: it makes Noether's theorem completely trivial.

Indeed, $dA/dt=0$ iff $[H,A]=0$ iff $\forall\tau, H=e^{-i/\hbar A\tau}He^{i/\hbar A \tau}$. Then $A$ is a conserved quantity and conjugation with it as an infinitesimal generator represents a symmetry of the Hamiltonian.