I have been thinking recently about game theory, and some of the reading I’ve been doing has been on connections between mixed strategies in game theory and uncertainty principles — the most famous of which is probably the Heisenberg Uncertainty Principle from quantum mechanics. This inspired me to dust off my Hilbert space theory and prove the famous principle for myself.

In quantum mechanics, a **state** of a quantum system is just an element *ψ* of a suitable Hilbert space *H* that has unit norm, i.e. one for which ‖*ψ*‖ := √〈*ψ*, *ψ*〉 = 1. Typically, *H* is a Lebesgue *L*^{2} space of complex-valued functions like *L*^{2}(ℝ; ℂ) for a single particle with one degree of freedom — moving on a line, say — in which case the inner product is

〈*ψ*, *φ*〉 := ∫_{ℝ} *ψ*(*x*) *φ̅*(*x*) d*x*.

The square of the absolute value of *ψ* is the probability distribution of the system, so you get the expected value of properties (“observables”) of the system by integrating them against |*ψ*|^{2}.

More precisely, the **average** of a linear operator *A*: *H* → *H* (in a particular state *ψ*) is denoted by 〈*A*〉 and is simply the inner product 〈*A**ψ*, *ψ*〉. The **standard deviation** of *A* (again in a particular state *ψ*) is denoted by *σ*_{A} and defined to be 〈(*A* − 〈*A*〉)^{2}〉^{1⁄2}, by analogy with the standard deviation of a random variable in probability/statistics. When the standard deviation of *A* is small, you have a very precise/certain measurement of some aspect *A* of the system. Uncertainty principles crop up when you ask the question, “Can I have high-precision measurements of two aspects of the system at the same time?”

The answer turns out to depend on whether or not the two measurement operators “commute”, i.e. whether or not you get the same answer regardless of the order in which you do them. For any two operators *A* and *B* from *H* into itself, we define their **commutator** by [*A*, *B*] := *A**B* − *B**A*, and their **anti-commutator** by {*A*, *B*} := *A**B* + *B**A*.

The best-known uncertainty principle is Heisenberg’s, which concerns the position and momentum operators *Q* and *P* respectively. The position operator *Q* acts as (*Q**ψ*)(*x*) := *x**ψ*(*x*), and the momentum operator *P* acts as (*P**ψ*)(*x*) := −*i*ℏ*ψ*′(*x*). The constant ℏ is the **reduced Planck constant**; the important point is that it’s a real number that’s strictly greater than zero. Kennard’s formulation of the Heisenberg Uncertainty Principle is the inequality

*σ*_{P} *σ*_{Q} ≥ ℏ⁄2,

from which it follows that we cannot have arbitrarily precise measurements of position and momentum at the same time, since Kennard’s inequality provides a non-zero lower bound on the product of their standard deviations. Kennard’s principle follows from the easy relationship [*P*, *Q*] = −*i*ℏ and the following uncertainty principle of Robertson for any two self-adjoint operators *A* and *B*:

*σ*_{A} *σ*_{B} ≥ | 〈[*A*, *B*]〉 ⁄ 2*i* |.

Robertson’s principle, in turn, follows from an inequality of Schrödinger:

*σ*_{A}^{2} *σ*_{B}^{2} ≥ | (〈{*A*, *B*}〉 − 2〈*A*〉〈*B*〉) ⁄ 2 |^{2} + | 〈[*A*, *B*]〉 ⁄ 2*i* |^{2}.

Schrödinger’s inequality is, it turns out, rather easy to prove. The only inequality you need is the Cauchy–Schwarz inequality (which is basically Inequality #1 in any introductory course on Hilbert spaces); the rest is a cunning change variables at the beginning and a few lines of algebraic massage.

**Proof.** Fix an arbitrary state *ψ* in *H*. The key notational trick is to introduce *f* := (*A* − 〈*A*〉)*ψ* and similarly *g* := (*B* − 〈*B*〉)*ψ*. With this notation, *σ*_{A} = ‖*f*‖ and *σ*_{B} = ‖*g*‖. Thus, by the Cauchy–Schwarz inequality,

*σ*_{A}^{2} *σ*_{B}^{2} = ‖*f*‖^{2} ‖*g*‖^{2} ≥ |〈*f*, *g*〉|^{2}.

This is actually the only inequality that we need; the rest is all algebraic massage of the right-hand side of the above inequality in *z* := 〈*f*, *g*〉 and its complex conjugate *z*̄ = 〈*g*, *f*〉. In particular, re-write the right-hand side of the above application of the Cauchy–Schwarz inequality as

|*z*|^{2} = |Re(*z*)|^{2} + |Im(*z*)|^{2} = |(*z* + *z*̄) ⁄ 2|^{2} + |(*z* − *z*̄) ⁄ 2*i*|^{2}

and substitute in the the following two easy facts:

*z* := 〈*f*, *g*〉 = 〈*A**B*〉 − 〈*A*〉〈*B*〉

and, similarly, *z*̄ = 〈*g*, *f*〉 = 〈*B**A*〉 − 〈*A*〉〈*B*〉. When you do this, the Schrödinger inequality, as stated above, drops out very nicely. ∎