One Image Summary of Quantum Mechanics

A 'Spherical Cow' in a Simple Harmonic Oscillator Potential is pretty much Quantum Mechanics in a nutshell.

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Image generated using ChatGPT-5 on 2025-Oct-19.

Random musing about Classical Projections

Assume I have a measuring device to measure the electric field intensity $\vec{E}$ in a specific direction, say, $z$. If $\vec{E}$ is oriented at an angle $\theta$ to the measuring device (or, equivalently to the $z$ direction), then the measuring device will register an intensity of $E\cos\theta$,  $E$ being the magnitude of $\vec{E}$. So far this is in line with classical physics and the parallelogram law of vector addition.

However, no measuring device is instantaneous. Every measuring device takes a finite amount of time to interact with the quantity-under-measurement before settling to a final value. This may not seem terribly relevant as the combined system comprising the measuring device and quantity-under-measurement converges to a steady state in short order. In classical physics, this steady state is real and the measuring device is truly measuring the underlying quantity.

A different perspective

One could posit a different take on the above measurement process: Since the measurement takes a finite time, one could argue that even in the steady state a measuring device is measuring a range of values - in rapid fashion - and is only able to present an average value as (macroscopic) measuring devices simply cannot respond fast enough.

Here's how one might do this. Note that

$$E\cos\theta = E\left[\cos^2(\frac{\theta}{2}) - \sin^2(\frac{\theta}{2})\right] \\ = E\cos^2(\frac{\theta}{2}) + (-E) \sin^2(\frac{\theta}{2})$$

So far this is just mathematical manipulation. However, this new form lets us look at the measurement process from a different perspective.

Perhaps the measuring device only ever measures $+E$ or $-E$ but with respective probabilities $P(+E) = \cos^2(\frac{\theta}{2})$ and $P(-E) = \sin^2(\frac{\theta}{2})$. Note that the individual probabilities add up to $1$, and the average value of $E$

$$ \langle E \rangle = E\cos^2(\frac{\theta}{2}) + (-E) \sin^2(\frac{\theta}{2}) = E\cos\theta$$

is exactly what's presented as the measured value by the measuring device.

Your point being ... ?

It may seem silly to look at it this way but it forms a nice bridge to the discussion of spin-$\frac{1}{2}$ in quantum physics. The only difference being that an individual spin-$\frac{1}{2}$ measurement yields a specific value (either $+\hbar/2$ or $-\hbar/2$) as it's the result of an instantaneous interaction between a Stern-Gerlach apparatus and the spin-$\frac{1}{2}$ particle under measurement. However, over an ensemble, the individual measurements yield the same average (‘expectation value’) as the classical case above.

Uniqueness of Ladder Operators in the Harmonic Oscillator

Background

In introductory Quantum Mechanics textbooks, the algebraic method of determining the energy eigenstates of the harmonic oscillator involves factorising the Hamiltonian as $$H=\hbar\omega(a^\dagger a + \frac{1}{2})$$ and showing that the $a^\dagger$ and $a$ operators respectively raise and lower the energy by $\hbar\omega$. Thus, the spectrum of the harmonic oscillator is determined to be a ladder with a step-size of $\hbar\omega$.

However, there is rarely any discussion on why this is the end all and be all for the harmonic oscillator spectrum. How do we know that there aren't other operators or, equivalently, a different factorisation of $H$, that would yield a new ladder with a possibly different step-size. This question had been nagging me for a long time. Now, with the help of ChatGPT-5, I understand why.

Uniqueness of ladder operators

The harmonic oscillator Hamiltonian is $$H = \frac{p^2}{2m} + \frac{1}{2}m \omega^2 x^2$$ with corresponding operator identities $$[x,p] = i\hbar, \quad [H,x] = -\frac{i \hbar}{m} p, \quad [H,p] =  i m \omega^2 \hbar x.$$

Assume a general operator $$B = \alpha x + \beta p$$ in the linear span of $\{x,p\}$ and demand that it be an eigenoperator of the adjoint action $ad_H(\cdot) = [H,\cdot]$ of $H$. In other words $$[H,B] = \lambda B$$ which implies $B$ acts a ladder with fixed energy step $\lambda$.

Evaluating the commutator relationship on the left hand side gives $$[H,B] = i\hbar(\beta m \omega^2 x - \frac{\alpha}{m}p).$$

Equating coefficients on the right hand side of the above two equations we get $$i \hbar \beta m \omega^2 = \lambda \alpha, \quad -\frac{i\hbar}{m}\alpha = \lambda\beta.$$

Nontrivial solutions, with $\alpha$ and $\beta$ not both $0$, exist only if $\lambda^2 = (\hbar\omega)^2$, so $\lambda = \pm\hbar\omega$. Thus, at least within the linear span of $\{x,p\}$, there are exactly two nontrivial eigenoperators of the adjoint action $ad_H(.)$ with eigenvalues $\pm\hbar\omega$. The corresponding normalised operators, unique up to a phase, turn out to be the canonical ladder operators $$a^\dagger = \sqrt{\frac{m\omega}{2\hbar}}x - i\frac{p}{\sqrt{2m\hbar\omega}}$$ and $$a = \sqrt{\frac{m\omega}{2\hbar}}x + i\frac{p}{\sqrt{2m\hbar\omega}}$$ and they shift energies by $\pm\hbar\omega$. 

(Justification for the above claim: for $\lambda = \pm\hbar\omega$, we get $\beta = \mp\frac{i}{m\omega}\alpha$ and hence $B = \alpha(x \mp \frac{i}{m\omega}p)$. Equivalently, $B=\alpha\sqrt{\frac{2\hbar}{m\omega}}a^{\dagger}$ or $B=\alpha\sqrt{\frac{2\hbar}{m\omega}}a.$  To wit, $B$ is proportional to $a^\dagger$ and $a$.)

Multiplying a ladder $B$ with any function $f(H)$, which commutes with $H$, produces another operator with the same step size $\lambda$: $$ [H,f(H)B] = \lambda f(H)B. $$