As an important theorem in quantum theory, the uncertainty principle is almost always presented alongside its physical significance. However, from a mathematical perspective, abstracting the mathematical form of the uncertainty principle helps us discover "uncertainty principles" in many other fields.
In this article, we will discuss the n-dimensional matrix form of the uncertainty principle. First, it should be explained that the uncertainty principle is essentially an inequality concerning "two Hermitian operators and a unit vector." In quantum mechanics, Hermitian operators correspond to infinite-dimensional Hermitian matrices. A Hermitian matrix is a matrix that is equal to its own conjugate transpose. Here, we discuss a simpler case: n-dimensional real matrices. In the context of n-dimensional real matrices, a Hermitian matrix is simply what we call a real symmetric matrix.
Let \boldsymbol{x} be an n-dimensional unit vector, i.e., |\boldsymbol{x}|=1, and let \boldsymbol{A} and \boldsymbol{B} be n \times n real symmetric matrices. In quantum mechanics, \boldsymbol{x} represents the wavefunction, but here it is merely a unit real vector. Let \boldsymbol{I} denote the n \times n identity matrix.
Consider: \bar{A}=\boldsymbol{x}^{T}\boldsymbol{A}\boldsymbol{x}, \quad \bar{B}=\boldsymbol{x}^{T}\boldsymbol{B}\boldsymbol{x} From these notations, it can be seen that these quantities correspond to the expectation values of observables. Of course, if you are not familiar with quantum mechanics, you can simply focus on the matrix forms above.
Next, we 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 \times n real symmetric matrices, it naturally follows that: \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 have: \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, it can be proven 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}\right)\boldsymbol{x}-\boldsymbol{x}^{T} \left( \boldsymbol{B}\boldsymbol{A}\right)\boldsymbol{x} \right|^2 \end{aligned}
Therefore: \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. We will encounter this expression frequently. Note that the definition of the commutator here differs slightly from the standard definition in quantum mechanics (which often includes an i\hbar factor).
Finally, we arrive at the inequality concerning 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 provides us with an "uncertainty principle." What quantum mechanics does is utilize a similar inequality generalized to the complex domain and infinite dimensions, while imbuing it with statistical significance.
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