I have encountered a large number of cute proofs and exercises, too cute in fact to be left unwritten down. I will collect them in this blog post. It will span a lot of different areas of physics, but a certain emphasis will be on elementary quantum physics. Some of the exercises are elementary, others are quite nontrivial.

No degenerate Eigenstates in 1 dimension.

**Virial Theorem.**

We derived a Virial Theorem in the section Mechanical Similarity in Classical Mechanics. Turns out there’s a QM analog for expectation values

**Proof**:

Plug in and

$latex \frac{d}{dt}<xp>= 2<T>- <xdV/dx>$

For stationary states, the LHS is zero (expectation values are constant). , which means

**Coherent States for Harmonic Oscillator**

Claim: Coherent states of the harmonic oscillator are eigenfunctions of the lowering operator. (there are no normalizable eigenfunctions of the raising operator)

**Proof:**

Start from answer: suppose

Calculate: , by expressing:

using

WKB

**Generalized Uncertainty Principle**:

The uncertainty principle for the variance of 2 noncompatible observables is a direct consequence of the Cauchy Schwarz Inequality (CS). Recall CS states:

and the equal applies when f is a multiple of g. Furthermore, we can make it even more restrictive by considering the square of the magnitude of the imaginary part of the LHS:

Consider 2 observables , and a state Then

where and similarly for .

Note if we call , and , we can plug them back into the CS equation. On the LHS, we have

On the RHS, we have:

from which we obtain:

We note that equality is achieved when f is a purely imaginary multiple of g. So in the case of A = X, B = P, we can solve for the wavefunctions that hit the uncertainty limit (coherent states):

which yields that f is a gaussian. No surprise there…

**Momentum Operator in Quantum Mechanics**

I know of 2.5 ways to get the form of the momentum operator in quantum mechanics. I say 2.5 because it is actually 2 ways, with one of the ways having variant versions.

**1st**:** Via generator of Translation**

We view momentum as the generator of translations, the same way classical momentum is the part of the generating function of the classical canonical transformation for translations.

First let’s define a translation in QM:

where K is a hermitian operator. Note that

and

Since the generator of infinitesimal translation (X = x + dx, P = p) in classical mechanics is :

where the first part is generator if identity, second generate translations, we suppose that K is related to the momentum operator in QM. Problem is K has units of , so instead, we set:

,

and

from which we obtain the commutation relations:

We now have a good idea of what the operator looks like for infinitesimal translation. How about finite translations? We just compound a lot of small translations:

To get the p representation in x basis we apply a little trick:

Now we can taylor expand the ket:

**Method 2: Nifty guess or powerful math trick**

We use the fact that the commutator [x,p] = ihI, and the fact that [a,bc]= [a,b]c + b[a,c].

Then we compute

If we keep expanding, we note that we will obtain something like:

Since any “nice” functions f(x) can be taylor expanded into a polynomial in x, we can conclude that p must acts like a derivative operator on f(x), in fact that

$<x| p =- i\hbar \partial_x$