3.6 Commutators and the generalised uncertainty principle

The question left over from the previous section: with so many observables around, can two of them be measured sharply at once?

The answer is surprisingly crisp: check whether they care about the order.

The commutator

[\hat A,\hat B]\equiv \hat A\hat B-\hat B\hat A\tag{3.6.1}

If $[\hat A,\hat B]=0$ , the two are said to commute.

Ordinary numbers always commute, so this concept simply does not exist in classical mechanics. It is a structure peculiar to quantum mechanics.

Computation: the canonical commutation relation [x̂, p̂]basic~4 min

A commutator is an operator; to find out what it equals, let it act on an arbitrary wavefunction.

[\hat x,\hat p]\psi=\hat x\hat p\psi-\hat p\hat x\psi\tag{3.6.2}

Term by term (with $\hat p=-\ii\hbar\,\dd/\dd x$ ):

\hat x\hat p\psi=x\cdot(-\ii\hbar\psi')=-\ii\hbar x\psi'\tag{3.6.3}

\hat p\hat x\psi=-\ii\hbar\frac{\dd}{\dd x}(x\psi)=-\ii\hbar(\psi+x\psi')\tag{3.6.4}

Watch the product rule in the second line — that extra $\psi$ is the whole story. Subtract:

[\hat x,\hat p]\psi=-\ii\hbar x\psi'+\ii\hbar\psi+\ii\hbar x\psi'=\ii\hbar\psi\tag{3.6.5}

This holds for arbitrary $\psi$ , so

\boxed{\ [\hat x,\hat p]=\ii\hbar\ }\tag{3.6.6}

This one equation carries half of quantum mechanics

$[\hat x,\hat p]=\ii\hbar$ is called the canonical commutation relation. Its standing is hard to overstate:

Heisenberg’s matrix mechanics (section 1.7) starts from it. Historically it was written down before the Schrödinger equation.
The algebraic solution of the harmonic oscillator (section 2.9) uses nothing but this one relation, without solving a differential equation even once.
The uncertainty principle follows directly from it — see below.
Canonical quantisation, the general recipe for building a quantum theory out of a classical one, is precisely the replacement of Poisson brackets by $\frac{1}{\ii\hbar}[\ ,\ ]$ .

As $\hbar\to0$ it degenerates to $[\hat x,\hat p]=0$ and everything commutes again — the classical limit is the limit in which everything commutes.

The picture

Working rules

[\hat A,\hat B]=-[\hat B,\hat A]\tag{3.6.7}

[\hat A,\hat B+\hat C]=[\hat A,\hat B]+[\hat A,\hat C]\tag{3.6.8}

[\hat A,\hat B\hat C]=[\hat A,\hat B]\hat C+\hat B[\hat A,\hat C]\tag{3.6.9}

The third works like the Leibniz rule for derivatives — mind the operator ordering.

Corollaries (apply the third rule repeatedly):

[\hat x,\hat p^n]=\ii\hbar\, n\hat p^{\,n-1}\tag{3.6.10}

[\hat x,f(\hat p)]=\ii\hbar\,\frac{\partial f}{\partial\hat p}\tag{3.6.11}

The mathematics

A few worth memorising

[\hat x_i,\hat x_j]=0,\qquad [\hat p_i,\hat p_j]=0\tag{3.6.12}

[\hat x_i,\hat p_j]=\ii\hbar\,\delta_{ij}\tag{3.6.13}

Position and momentum along different axes commute — $x$ and $p_y$ can be sharp simultaneously.

Angular momentum (chapter 5):

[\hat L_x,\hat L_y]=\ii\hbar \hat L_z\tag{3.6.14}

The three components fail to commute pairwise, yet

[\hat L^2,\hat L_z]=0\tag{3.6.15}

so $l$ and $m$ can be fixed together while the three components cannot. The shapes of atomic orbitals trace back to this.

The generalised uncertainty relation

The full derivationadvanced~10 min

Let $\hat A$ and $\hat B$ be Hermitian and the system be in a normalised $\ket{\psi}$ . Define the mean-subtracted operators

\Delta\hat A=\hat A-\langle A\rangle,\qquad \Delta\hat B=\hat B-\langle B\rangle\tag{3.6.16}

( $\langle A\rangle=\bra{\psi}\hat A\ket{\psi}$ is a number.) The variance is then

\sigma_A^2=\bra{\psi}(\Delta\hat A)^2\ket{\psi}=\braket{f}{f},\qquad \ket{f}\equiv\Delta\hat A\ket{\psi}\tag{3.6.17}

and likewise $\sigma_B^2=\braket{g}{g}$ with $\ket{g}\equiv\Delta\hat B\ket{\psi}$ . (This step used the Hermiticity of $\Delta\hat A$ .)

Step 1: the Cauchy-Schwarz inequality.

\braket{f}{f}\braket{g}{g}\ge|\braket{f}{g}|^2\tag{3.6.18}

i.e. $\sigma_A^2\sigma_B^2\ge|\braket{f}{g}|^2$ .

Step 2: keep only the imaginary part. For any complex number $z$ , $|z|^2\ge(\operatorname{Im}z)^2=\left(\frac{z-z^*}{2\ii}\right)^2$ . Take $z=\braket{f}{g}$ and note $z^*=\braket{g}{f}$ :

|\braket{f}{g}|^2\ge\left(\frac{\braket{f}{g}-\braket{g}{f}}{2\ii}\right)^2\tag{3.6.19}

Step 3: trade it for a commutator.

\braket{f}{g}=\bra{\psi}\Delta\hat A\,\Delta\hat B\ket{\psi}\tag{3.6.20}

Expand (note that $\langle A\rangle,\langle B\rangle$ are numbers and move around freely):

=\langle \hat A\hat B\rangle-\langle A\rangle\langle B\rangle\tag{3.6.21}

Likewise $\braket{g}{f}=\langle \hat B\hat A\rangle-\langle A\rangle\langle B\rangle$ . Subtract, and the products of means cancel entirely:

\braket{f}{g}-\braket{g}{f}=\langle[\hat A,\hat B]\rangle\tag{3.6.22}

Step 4: assemble.

\boxed{\ \sigma_A^2\sigma_B^2\ge\left(\frac{1}{2\ii}\langle[\hat A,\hat B]\rangle\right)^2\ }\tag{3.6.23}

or, equivalently,

\sigma_A\,\sigma_B\ge\frac12\left|\langle[\hat A,\hat B]\rangle\right|\tag{3.6.24}

Plug in x and p

Take $\hat A=\hat x$ , $\hat B=\hat p$ , $[\hat x,\hat p]=\ii\hbar$ :

\sigma_x\sigma_p\ge\frac12|\ii\hbar|=\frac{\hbar}{2}\tag{3.6.25}

This is chapter 2’s inequality — now proved.

At the time, section 2.6 could only say “the rigorous proof needs the generalised uncertainty relation, see section 3.6” and offer an intuitive origin from Fourier analysis. The two roads now meet: the Fourier road explained why, and this road delivers the rigorous, general form.

Moreover, the inequality applies to any pair of observables. Angular momentum, for instance:

\sigma_{L_x}\sigma_{L_y}\ge\frac{\hbar}{2}|\langle \hat L_z\rangle|\tag{3.6.26}

Note that the right-hand side is not a constant — it depends on the state. If $\langle L_z\rangle=0$ , then $L_x$ and $L_y$ can in principle be sharp together — a vanishing lower bound does not guarantee it can be achieved, but at least this inequality no longer stands in the way.

Theorem and proofadvanced~6 min

Theorem: two Hermitian operators $\hat A,\hat B$ commute if and only if they share a common orthonormal complete eigenbasis.

(⟸) The easy direction. Suppose $\{\ket{n}\}$ is a common eigenbasis, $\hat A\ket{n}=a_n\ket{n}$ , $\hat B\ket{n}=b_n\ket{n}$ . Then

\hat A\hat B\ket{n}=a_nb_n\ket{n}=\hat B\hat A\ket{n}\tag{3.6.27}

This holds for every basis vector, hence by linearity for every state, so $[\hat A,\hat B]=0$ .

(⟹) The other direction. Suppose $[\hat A,\hat B]=0$ and $\hat A\ket{a}=a\ket{a}$ . Look at $\hat B\ket{a}$ :

\hat A\big(\hat B\ket{a}\big)=\hat B\hat A\ket{a}=a\big(\hat B\ket{a}\big)\tag{3.6.28}

So $\hat B\ket{a}$ is also an eigenvector of $\hat A$ , with the same eigenvalue $a$ .

If $a$ is non-degenerate, $\hat B\ket{a}$ can only be proportional to $\ket{a}$ , i.e. $\hat B\ket{a}=b\ket{a}$ — $\ket{a}$ is automatically an eigenvector of $\hat B$ too, and we are done.

If $a$ is degenerate, then $\hat B$ maps the eigenspace $\mathcal H_a$ into itself. Restricted to $\mathcal H_a$ , $\hat B$ is still Hermitian and can therefore be diagonalised inside it — re-choose the basis within the degenerate subspace so that it is simultaneously an eigenbasis of $\hat B$ . $\blacksquare$

That last step is exactly where the “freedom to choose a basis under degeneracy” from section 3.4 earns its keep.

Compatible observables and complete commuting sets

Commuting observables are called compatible: there exist states that are eigenstates of both, and measuring one does not scramble the other.

If an observable is degenerate, it cannot label states uniquely on its own. So you bring in another observable that commutes with it to tell the states apart. Keep adding until the combination of eigenvalues pins down the state uniquely — such a set of operators is a complete set of commuting observables (CSCO).

Hydrogen is the standard example: $\hat H$ alone is heavily degenerate (four-fold at $n=2$ ). Add $\hat L^2$ and $\hat L_z$ — the three operators commute pairwise — and the eigenvalue combination $(n,l,m)$ fixes the state uniquely. Add spin $\hat S_z$ and you get $(n,l,m,m_s)$ — exactly the four quantum numbers from chemistry class.

They are not memorised facts; they are the outcome of “find a set of pairwise-commuting observables that fully resolves the degeneracy”.

Key formulas

Commutator
$[\hat A,\hat B]=\hat A\hat B-\hat B\hat A$
Identically zero in classical mechanics; purely quantum
Canonical commutation relation
$[\hat x,\hat p]=\ii\hbar$
The starting point of matrix mechanics; ħ→0 is the classical limit
Leibniz rule
$[\hat A,\hat B\hat C]=[\hat A,\hat B]\hat C+\hat B[\hat A,\hat C]$
Mind the operator ordering
Generalised uncertainty relation
$\sigma_A\sigma_B\ge\frac12\left|\langle[\hat A,\hat B]\rangle\right|$
Cauchy-Schwarz + keeping only the imaginary part
Angular momentum commutators
$[\hat L_x,\hat L_y]=\ii\hbar\hat L_z,\qquad[\hat L^2,\hat L_z]=0$
Hence l and m can be fixed together, but not all three components
Compatibility criterion
$[\hat A,\hat B]=0\iff\text{a common eigenbasis exists}$
The origin of the CSCO and the four quantum numbers

Self-check4 questions

What comes next

By this point, most of chapter 2’s “decrees” have turned into conclusions.

The next section compresses the whole theory into five postulates — drawing a clean line between what must be assumed and what can be derived.

Commutators and the generalised uncertainty principle

Recommended first

After this section you should be able to

The commutator

The picture

The mathematics

The generalised uncertainty relation

Key formulas

Self-check4 questions

What comes next

The commutator

◑The picture

∑The mathematics

The generalised uncertainty relation

Commuting ⟺ sharing a common eigenbasis

∎Key formulas

What comes next

The picture

The mathematics

Key formulas