Incompatible observables and the general uncertainty relation

2026-10-19

# Incompatible observables

↳ No simultaneous eigenkets $\to$ proof by contradiction.

If we did have $[\hat A, \hat B] \neq 0$ and $\hat A, \hat B$ still have simultaneous eigenkets, then

$$ \begin{aligned} \hat A\hat B|a',b'\rangle &= \hat A b'|a',b'\rangle = a'b'|a',b'\rangle \\ &= \hat B\hat A|a',b'\rangle \end{aligned} $$

$$ \implies \hat A\hat B|a',b'\rangle = \hat B\hat A|a',b'\rangle $$

or $\hat A\hat B - \hat B\hat A = 0$ $\to$ which is a contradiction.

There is a subspace in the ket space such that $|a',b'\rangle$ holds / makes sense even when $[\hat A, \hat B] \neq 0$.

For example in the $l = 0$ state (s-state), $L_x$ and $L_z$ don’t commute, but $l = 0$ is a simultaneous eigenket (all the components give 0 there). More on $l$ in the angular momentum notes.

# Sequential measurements

          ┌─────┐   a'    ┌─────┐   b'    ┌─────┐   c'
   ─────► │  A  │ ──────► │  B  │ ──────► │  C  │ ──────►
          │     │ ──► ■   │     │ ──► ■   │     │ ──► ■
          └─────┘  rest   └─────┘  rest   └─────┘  rest
                 blocked         blocked         blocked

What’s the prob. of obtaining $|c'\rangle$?

We need $|a'\rangle \to |b'\rangle$ and $|b'\rangle \to |c'\rangle$:

$$ \therefore \text{total prob. of } |c'\rangle = |\langle c'|b'\rangle|^2|\langle b'|a'\rangle|^2 $$

Now, we had blocked some $|b'\rangle$ routes. What if we consider them one by one (and add up all the routes)?

$$ \sum_{b'}|\langle c'|b'\rangle|^2|\langle b'|a'\rangle|^2 = \sum_{b'}\langle c'|b'\rangle\langle b'|a'\rangle\langle a'|b'\rangle\langle b'|c'\rangle $$

But if we remove the $B$ filter entirely, the prob. becomes $|\langle c'|a'\rangle|^2$:

$$ |\langle c'|a'\rangle|^2 = \left|\sum_{b'}\langle c'|b'\rangle\langle b'|a'\rangle\right|^2 $$

(same imagined $|b'\rangle$, we just inserted closure).

Note the two are different! So going through all routes $\neq$ as if not going through it at all.

In both cases the pure $|a'\rangle$ beam can be regarded as

$$ |a'\rangle = \sum_{b'}|b'\rangle\langle b'|a'\rangle $$

(not an eigenvalue equation). This is what we do in the second case, as $\langle c'|a'\rangle = \sum_{b'}\langle c'|b'\rangle\langle b'|a'\rangle$. By actually recording the $|\langle b'|a'\rangle|^2$ prob. values, we change things!!

When are these expressions equal? When $[\hat A, \hat B] = 0$ or $[\hat B, \hat C] = 0$, in absence of degeneracy.

$\therefore$ for incompatible measurements only does “looking” f*ck it up.

# The general uncertainty relation

We did the dispersion in the uncertainty notes. The general uncertainty relation (insane) is:

$$ \boxed{\langle(\Delta A)^2\rangle\langle(\Delta B)^2\rangle \geq \frac{1}{4}\left|\langle[\hat A, \hat B]\rangle\right|^2} $$

Why? 3 lemmas:

Cauchy-Schwarz
$\langle A\rangle$ is real if $\hat A$ is Hermitian
if $\hat A = -\hat A^\dagger$ (anti-Hermitian), the expectation value is purely imaginary.

For 2 and 3, use $\langle\beta|\hat X|\alpha\rangle = \langle\alpha|\hat X^\dagger|\beta\rangle^{\ast}$ from the operators notes:

$$ \hat A = \hat A^\dagger \implies \langle\alpha|\hat A|\alpha\rangle = \langle\alpha|\hat A|\alpha\rangle^{\ast} \implies \text{real} $$

$$ \hat A = -\hat A^\dagger \implies \langle\alpha|\hat A|\alpha\rangle = -\langle\alpha|\hat A|\alpha\rangle^{\ast} \implies \text{purely imaginary} $$

Cauchy-Schwarz is:

Now we use Cauchy-Schwarz and the following definitions:

$$ |\alpha\rangle = \Delta A|\ \rangle, \qquad |\beta\rangle = \Delta B|\ \rangle $$

where $|\ \rangle$ is any ket.

$$ \implies \langle(\Delta A)^2\rangle\langle(\Delta B)^2\rangle \geq |\langle\Delta A\Delta B\rangle|^2 $$

Why? Since $\Delta A$ is Hermitian,

↳ the definition of expectation value. The same goes for $\langle\beta|\beta\rangle$, and $\langle\alpha|\beta\rangle = \langle\Delta A\Delta B\rangle$.

Now to evaluate $|\langle\Delta A\Delta B\rangle|^2$ we use the fact (from the commutator notes):

$$ \Delta A\Delta B = \frac{1}{2}[\Delta A, \Delta B] + \frac{1}{2}\lbrace\Delta A, \Delta B\rbrace $$

The commutator $[\Delta A, \Delta B] = [A, B]$, since $\langle A\rangle$ and $\langle B\rangle$ are just numbers.
$[A, B]$ is anti-Hermitian: $[A, B]^\dagger = (AB - BA)^\dagger = BA - AB = -[A, B]$
$\lbrace\Delta A, \Delta B\rbrace$ is Hermitian (due to the positive sign).

$$ \implies \langle\Delta A\Delta B\rangle = \underbrace{\frac{1}{2}\langle[A, B]\rangle}_{\text{purely imaginary}} + \underbrace{\frac{1}{2}\langle\lbrace\Delta A, \Delta B\rbrace\rangle}_{\text{purely real}} $$

$$ |\langle\Delta A\Delta B\rangle|^2 = \frac{1}{4}|\langle[A, B]\rangle|^2 + \frac{1}{4}|\langle\lbrace\Delta A, \Delta B\rbrace\rangle|^2 $$

↳ sum of squares, hence. Throwing away the second (non-negative) term can only make the right side smaller, and this proves the inequality we saw.

← The uncertainty relation | Next: Change of basis and the transformation matrix →

Incompatible observables and the general uncertainty relation

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