The Matrix Form of the Uncertainty Principle

不确定性原理的矩阵形式 苏剑林 · 2014-01-05 · Phy-chem inequality quantum mechanics matrix

As an important theorem of quantum theory, the uncertainty principle always shows up wrapped in physical meaning. But from a mathematical point of view, extracting the underlying mathematical form of the uncertainty principle can help us discover "uncertainty principles" in more fields.

In this post, we will discuss the n-dimensional matrix form of the uncertainty principle. First, let me explain that the uncertainty principle is essentially "an inequality involving two Hermitian operators and a unit vector." In quantum mechanics, a Hermitian operator corresponds to an infinite-dimensional Hermitian matrix; a Hermitian matrix is one that equals itself after taking both the conjugate and the transpose. Here, though, we discuss a simpler case: n-dimensional real matrices, where the Hermitian matrices are exactly what we call real symmetric matrices.

Let $\boldsymbol{x}$ be a $n$-dimensional unit vector, i.e. $|\boldsymbol{x}|=1$, and let $\boldsymbol{A}$ and $\boldsymbol{B}$ be n-th order real symmetric matrices. In quantum mechanics, $\boldsymbol{x}$ is the wave function, but here it is simply a real unit vector; and we denote by $\boldsymbol{I}$ the $n$-th order identity matrix.

Consider

$$\bar{A}=\boldsymbol{x}^{T}\boldsymbol{A}\boldsymbol{x},\bar{B}=\boldsymbol{x}^{T}\boldsymbol{B}\boldsymbol{x}$$

From this notation we can see that these quantities correspond to the expectation values of observables. Of course, if you don't know quantum mechanics, you can just look at the matrix form above. more

Next, let's consider

$$\begin{aligned} &\left(\Delta A\right)^2=\boldsymbol{x}^{T}\left( \boldsymbol{A}-\bar{A}\boldsymbol{I} \right)^2 \boldsymbol{x}\\ &\left(\Delta B\right)^2=\boldsymbol{x}^{T}\left( \boldsymbol{B}-\bar{B}\boldsymbol{I} \right)^2 \boldsymbol{x} \end{aligned}$$

Since $\boldsymbol{A}$ and $\boldsymbol{B}$ are $n$-th order real symmetric matrices, we naturally have

$$\begin{aligned} &\left(\Delta A\right)^2=\left| \left( \boldsymbol{A}-\bar{A}\boldsymbol{I} \right)\boldsymbol{x}\right|^2\\ &\left(\Delta B\right)^2=\left| \left( \boldsymbol{B}-\bar{B}\boldsymbol{I} \right)\boldsymbol{x}\right|^2 \end{aligned}$$

Using the Cauchy-Schwarz inequality, we get

$$\begin{aligned} \left(\Delta A\right)^2\left(\Delta B\right)^2 &=\left| \left( \boldsymbol{A}-\bar{A}\boldsymbol{I} \right)\boldsymbol{x}\right|^2\left| \left( \boldsymbol{B}-\bar{B}\boldsymbol{I} \right)\boldsymbol{x}\right|^2\\ &\geq \left| \boldsymbol{x}^{T} \left( \boldsymbol{A}-\bar{A}\boldsymbol{I} \right)\left( \boldsymbol{B}-\bar{B}\boldsymbol{I} \right) \boldsymbol{x} \right|^2\\ &=\left| \boldsymbol{x}^{T} \left( \boldsymbol{A}\boldsymbol{B}\right)\boldsymbol{x}-\left(\boldsymbol{x}^{T}\boldsymbol{A}\boldsymbol{x} \right)\left(\boldsymbol{x}^{T}\boldsymbol{B}\boldsymbol{x}\right)\right|^2 \end{aligned}$$

Similarly, one can show that

$$\begin{aligned} \left(\Delta A\right)^2\left(\Delta B\right)^2 \geq \left|\left( \boldsymbol{x}^{T}\boldsymbol{A}\boldsymbol{x}\right)\left(\boldsymbol{x}^{T}\boldsymbol{B}\boldsymbol{x}\right)-\boldsymbol{x}^{T} \left( \boldsymbol{B}\boldsymbol{A}\right)\boldsymbol{x} \right|^2 \end{aligned}$$

so

$$\begin{aligned} &\left(\Delta A\right)^2\left(\Delta B\right)^2 \\ \geq &\frac{1}{2}\left| \boldsymbol{x}^{T} \left( \boldsymbol{A}\boldsymbol{B}\right)\boldsymbol{x}-\left(\boldsymbol{x}^{T}\boldsymbol{A}\boldsymbol{x} \right)\left(\boldsymbol{x}^{T}\boldsymbol{B}\boldsymbol{x}\right)\right|^2\\ &+\frac{1}{2}\left|\left( \boldsymbol{x}^{T}\boldsymbol{A}\boldsymbol{x}\right)\left(\boldsymbol{x}^{T}\boldsymbol{B}\boldsymbol{x}\right)-\boldsymbol{x}^{T} \left( \boldsymbol{B}\boldsymbol{A}\right)\boldsymbol{x} \right|^2\\ \geq &\frac{1}{4}\left| \boldsymbol{x}^{T} \left( \boldsymbol{A}\boldsymbol{B}\right)\boldsymbol{x}-\boldsymbol{x}^{T} \left( \boldsymbol{B}\boldsymbol{A}\right)\boldsymbol{x} \right|^2\\ =&\frac{1}{4}\left| \boldsymbol{x}^{T} \left( [\boldsymbol{A},\boldsymbol{B}]\right)\boldsymbol{x}\right|^2 \end{aligned}$$

where

$$[\boldsymbol{A},\boldsymbol{B}]=\boldsymbol{A}\boldsymbol{B}-\boldsymbol{B}\boldsymbol{A}$$

is called the commutator (also known as the commutation bracket, etc.). We will see this expression often in the future. Note that the definition of the commutator here differs slightly from the one used in quantum mechanics.

Finally, we arrive at an inequality relating two real symmetric matrices and a unit vector:

$$\begin{aligned} \left(\Delta A\right)\left(\Delta B\right) \geq \frac{1}{2}\left| \boldsymbol{x}^{T} \left( [\boldsymbol{A},\boldsymbol{B}]\right)\boldsymbol{x}\right| \end{aligned}$$

As long as the two matrices do not commute, this inequality gives us an "uncertainty principle." What quantum mechanics does is generalize an inequality of this same form to the complex domain and to infinite dimensions, and then endow it with statistical meaning.