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.

$H(t)$ 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.

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**.

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.

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