Quantum Foundations

It's ok they're just vectors. Written 4 months ago.

Find the fundamental principles of quantum mechanics are stored in quote blocks.

States

Quantum states are stored in bra’s $\bra{a} = (a_1, a_2, \ldots, a_n)$ and kets

\ket{b} = \begin{pmatrix}b_1\\b_2\\ \vdots \\ b_n\end{pmatrix}.

This elegant notation comes from Dirac himself. It’s the quantum analogue to vector spaces.

States also may have eigenvalues and eigenvectors. Given a system $\bold M$ , if

\bold M \ket{\lambda} = \lambda \ket{\lambda}

Then $\lambda$ is the eigenvalue and $\ket{\lambda}$ is the eigenvector.

Observability

Observables in quantum mechanics are represented by linear operators that equal their own Hermitian conjugate

\bold L = \bold L^{\dagger}.

Hermitian operators are tied to observability because of a special property. Lets take $\bold L \ket{\lambda} = \lambda \ket{\lambda}$ to be a standard Hermitian operation on $\ket{\lambda}$ . We know that

\begin{align*} \bold L \ket{\lambda} &= \lambda \ket{\lambda} \\ \bra{\lambda} \bold L^{\dagger} &= \bra{\lambda} \lambda^*. \end{align*}

Now multiplying both of these equations but the missing bra and ket vector

\begin{align*} \bra{\lambda}\bold L \ket{\lambda} &= \lambda \braket{\lambda|\lambda} \\ \bra{\lambda} \bold L^{\dagger} \ket{\lambda} &= \lambda^* \braket{\lambda|\lambda} . \end{align*}

We already $\bold L = \bold L^{\dagger}$ and therefore it must be true that $\lambda = \lambda^*$ if both equations are to be equal. Hence $\lambda$ is guaranteed to be real number — an observable quantity.

Observable and measurable quantities are represented by linear operator $\bold L$ , the Hermitian.

Measurements

The possible results of a measurements are the eigenvalues $\lambda_i$ of the observable $\bold L$ . If a system is in eigenstate $\ket{\lambda_i}$ then the result of a measurement is guaranteed to be $\lambda_i$ .

Given a quantum state $\ket{\Psi}$ at $t=0$ , can we infer what it be at any value of $t$ ? Quantum mechanics doesn’t offer that kind of deterministic guarantee we’re used to in classical physics.

However we can compute a probability of what future measurements will be, or more formally the linear operator $\bold U(t)$ can

\ket{\Psi(t)} = \bold U(t) \ket{\Psi(0)}.

$\bold U(t)$ is a special operator because of the principle of unitarity

\bold U^{\dagger}\bold U = I.

Having $\bold U(t)$ be a unitary linear operator will make state transitions easier to deal with.

This is quasi-determinism. $\bold U(t)$ can act on $\ket{\Psi(0)}$ to produce a new state $\ket{\Psi(t)}$ . However a new state does not guarantee a new measurement. This is where we draw possible the largest distinction between quantum and classical physics.

Classical	No difference between a state and a measurement.
Quantum	The difference is profound.

So if we have a quantum state $\ket{A}$ how do we measure it? First it needs to be operated on by an observable $\bold L$

\bold L \ket{A} = \lambda_i \ket{\lambda_i}.

Once we know possible $\lambda_i$ values we can then calculated their probability of being measured

P(\lambda_i) = \braket{A|\lambda_i}\braket{\lambda_i|A}.

If $\ket{A}$ is the state of a system after the observable $\bold L$ is measured the probability of measuring $\lambda_i$ is $P(\lambda_i) = \braket{A|\lambda_i}\braket{\lambda_i|A}.$

Energy

Instead of analysing time change in large clunky intervals of $t$ , lets use an infitesimal quantity $\epsilon$ which will allows us to construct the differential model for quantum time evolution,

\bold U^{\dagger}(\epsilon)\bold U(\epsilon) = I.

Lets make some arbitrary decisions with no content at all. These formulations will simply make more sense later on,

\begin{align*} \bold U(\epsilon) &= I - i\epsilon\bold H\\ \bold U^{\dagger}(\epsilon) &= I + i\epsilon \bold H^{\dagger}. \end{align*}

Now lets convert the standard time-evolution $\ket{\Psi(t)} = \bold U(t) \ket{\Psi(0)}$ to

\ket{\Psi(\epsilon)} = \bold (I - i\epsilon\bold H) \ket{\Psi(0)}.

After some light algebra we arrive at a very familiar formulation

\dfrac{\ket{\Psi(\epsilon)}- \ket{\Psi(0)}}{\epsilon} = -i\bold H \ket{\Psi(0)}.

This is the time-dependant Schrödinger’s equation

\hbar\dfrac{\partial \Psi}{\partial t} = -i\bold H \ket{\Psi}

with $\hbar$ added for dimensional consistency ( $\bold H$ has energy units on right).