Uncertainty Principle

The uncertainty principle, simply put, states that certain measurements cannot be simultaneously taken of a system such that there is absolute certainty in every measurement. Perhaps the most well known of all uncertainty relations is Heisenberg's between momentum and position measurements. To derive the relations that dictate the general uncertainty relation and the Heisenberg uncertainty relation, we'll need to do some playing around using the matrix interpretation of quantum mechanics. Let's start with deriving some basic stuff.

Expectation Value

Let's derive an equation for the expectation value, $\left\langle\hat{A}\right\rangle$, the average measurement value when infinite measurements of $\hat{A}$ are taken on the state vector $\lvert\psi\rangle$. Recall that if $\hat{A}$ has eigenvalues $a_n$ with corresponding eigenstates $\lvert a_n\rangle$, then $|\langle a_n\mid\psi\rangle|^2$ gives the probability of having $a_n$ returned as the measurement value when applying the operator $\hat{A}$ to $\lvert\psi\rangle$. Therefore, the expectation value must be given by:

\begin{equation}
\left\langle\hat{A}\right\rangle = \sum_n\left|\langle a_n\mid\psi\rangle\right|^2a_n
\end{equation}
\begin{equation}
\left\langle\hat{A}\right\rangle = \sum_n\langle\psi\mid a_n\rangle\langle a_n\mid\psi\rangle a_n
\end{equation}
\begin{equation}
\left\langle\hat{A}\right\rangle = \sum_n\langle\psi\rvert\hat{A}\lvert a_n\rangle\langle a_n\mid\psi\rangle
\end{equation}

Because of the orthonormality of $\lvert a_n\rangle$, $\sum_n \lvert a_n\rangle\langle a_n\rvert=I$, where $I$ is the identity matrix. Therefore, the expectation value, $\left\langle\hat{A}\right\rangle$, is derived to be:

\begin{equation}
\left\langle\hat{A}\right\rangle = \langle\psi\rvert \hat{A}\lvert \psi\rangle
\end{equation}

Standard Deviation

Recall that standard deviation is calculated by averaging the squares of differences between measurement values and the mean value, then taking the square root of that average. Quantum mechanically, $\left|\langle a_n\mid\psi\rangle\right|^2$ gives the probability of getting the measurement value $a_n$ and $\left(a_n - \left\langle \hat{A}\right \rangle\right)^2$ gives the square of the difference between $a_n$ and the mean measurement value. Therefore, the standard deviation, $\sigma_\hat{A}$, of the measurement $\hat{A}$ can be written as:

\begin{equation}
\sigma_\hat{A} = \left[\sum_n|\langle a_n\mid\psi\rangle|^2\left(a_n - \left\langle\hat{A} \right\rangle\right)^2\right]^{1/2}
\end{equation}
\begin{equation}
\sigma_\hat{A} = \left[\sum_n\langle\psi\rvert\left(a_n - \left\langle\hat{A}\right\rangle\right)^2 \lvert a_n\rangle\langle a_n\mid\psi\rangle\right]^{1/2}
\end{equation}
\begin{equation}
\sigma_\hat{A} = \left[\sum_n\langle\psi\rvert\left(\hat{A} - \left\langle\hat{A}\right\rangle \right)^2\lvert a_n\rangle\langle a_n\mid\psi \rangle\right]^{1/2}
\end{equation}
\begin{equation}
\sigma_\hat{A} = \left\langle\left(\hat{A} - \left\langle\hat{A}\right\rangle\right)^2 \right\rangle^{1/2}
\end{equation}

General Uncertainty Relation

Let's consider two measurements, defined by Hermitian operators $\hat{A}$ and $\hat{B}$, with standard deviations of $\sigma_\hat{A}$ and $\sigma_\hat{B}$ respectively. We can say that:

\begin{equation}
{\sigma_\hat{A}}^2{\sigma_\hat{B}}^2 = \left\langle\left(\hat{A} - \left\langle\hat{A}\right\rangle\right)^2 \right\rangle\left\langle\left(\hat{B} - \left\langle\hat{B}\right\rangle \right)^2\right\rangle
\end{equation}

Remember the Cauchy–Schwarz inequality which says $\langle x\mid x\rangle\langle y\mid y\rangle \geq \left|\langle x \mid y \rangle\right|^2$ for any arbitrary $\lvert x \rangle$ and $\lvert y \rangle$? Well, if we consider $\left(\hat{A} - \left\langle\hat{A}\right\rangle\right)\lvert\psi\rangle$ as $\lvert x\rangle$ and $\left(\hat{B} - \left\langle\hat{B}\right\rangle \right) \lvert\psi\rangle$ as $\lvert y\rangle$, we can say this:

\begin{equation}
{\sigma_\hat{A}}^2{\sigma_\hat{B}}^2 \geq \left|\left\langle\left(\hat{A} - \left\langle\hat{A} \right\rangle\right) \left(\hat{B} - \left\langle\hat{B}\right\rangle\right) \right\rangle\right|^2
\end{equation}
\begin{equation}
{\sigma_\hat{A}}^2{\sigma_\hat{B}}^2 \geq \left|\left\langle\left(\hat{A}\hat{B} - \hat{A}\left\langle\hat{B}\right\rangle -\hat{B}\left\langle\hat{A} \right\rangle + \left\langle\hat{A}\right\rangle \left\langle\hat{B} \right\rangle\right) \right\rangle \right|^2
\end{equation}

Notice that the expansion of the expression:

$$\frac{1}{2}\left\{\hat{A} - \left\langle\hat{A}\right\rangle, \hat{B} - \left\langle\hat{B}\right\rangle\right\} = \frac{1}{2}\hat{A}\hat{B} + \frac{1}{2}\hat{B}\hat{A} - \hat{A}\left\langle\hat{B}\right\rangle -\hat{B}\left\langle \hat{A}\right\rangle + \left\langle\hat{A} \right\rangle\left\langle\hat{B}\right\rangle$$

is almost identical to the parenthesized expression of the above inequality. Let's try to write the inequality with regard to the anticommutator $\frac{1}{2}\left\{\hat{A} - \left\langle\hat{A}\right\rangle, \hat{B} - \left\langle\hat{B}\right\rangle\right\}$.

\begin{equation}
{\sigma_\hat{A}}^2{\sigma_\hat{B}}^2 \geq \left|\left\langle\frac{1}{2} \left[\hat{A}, \hat{B}\right] + \frac{1}{2}\left\{\hat{A} - \left\langle\hat{A}\right\rangle, \hat{B} - \left\langle\hat{B} \right\rangle\right\}\right\rangle\right|^2
\end{equation}
\begin{equation}
{\sigma_\hat{A}}^2{\sigma_\hat{B}}^2 \geq \left|\frac{1}{2} \left\langle\left[\hat{A}, \hat{B}\right]\right\rangle + \frac{1}{2}\left\langle\left\{\hat{A} - \left\langle\hat{A}\right\rangle, \hat{B} - \left\langle\hat{B} \right\rangle\right\}\right\rangle\right|^2
\end{equation}

You're probably wondering why we would want to put that anticommutator in there so bad. Well now, take a good look at what we have. $\left[\hat{A}, \hat{B}\right]$ is a commutator of Hermitian operators, which makes it anti-Hermitian. On the other hand, $\left\{\hat{A} - \left\langle\hat{A}\right\rangle, \hat{B} - \left\langle\hat{B}\right\rangle \right\}$ is the anticommutator of Hermitian operators, making itself Hermitian. The expectation value of an anti-Hermitian operator is imaginary and that of a Hermitian operator is real. Therefore, we can distribute the absolute square in the above inequality to the imaginary component, $\frac{1}{2} \left\langle\left[\hat{A}, \hat{B}\right]\right\rangle$, and the real component, $\frac{1}{2}\left\langle\left\{\hat{A} - \left\langle\hat{A}\right\rangle, \hat{B} - \left\langle\hat{B} \right\rangle\right\}\right\rangle$.

\begin{equation}
{\sigma_\hat{A}}^2{\sigma_\hat{B}}^2 \geq \frac{1}{4}\left|\left\langle \left[\hat{A} , \hat{B}\right]\right\rangle\right|^2 + \frac{1}{4}\left|\left\langle\left\{\hat{A} - \left\langle\hat{A}\right\rangle, \hat{B} - \left\langle\hat{B}\right\rangle\right\} \right\rangle\right|^2
\end{equation}

We know that $\frac{1}{4}\left| \left\langle\left\{\hat{A} - \left\langle\hat{A}\right\rangle, \hat{B} - \left\langle\hat{B} \right\rangle\right\} \right\rangle\right|^2$ must be greater than or equal to $0$, meaning we can just remove that term from the inequality and the inequality would still hold.

\begin{equation}
{\sigma_\hat{A}}^2 {\sigma_\hat{B}}^2 \geq \frac{1}{4}\left| \left\langle\left[\hat{A}, \hat{B}\right] \right\rangle\right|^2
\end{equation}

And so, for any two measurements defined by Hermitian operators, $\hat{A}$ and $\hat{B}$, the uncertainties of both measurements are defined by:

\begin{equation}
\sigma_\hat{A}\sigma_\hat{B} \geq \frac{1}{2}\left|\left\langle \left[\hat{A}, \hat{B}\right]\right\rangle\right|
\end{equation}

Heisenberg Uncertainty Relation

The Heisenberg uncertainty relation deals with the measurements of position and momentum in one dimension, so let's find the commutator of the one-dimensional position operator, $\hat{x}=x$ and one-dimensional momentum operator, $\hat{p}_x= -i\hbar\frac{\partial}{\partial x}$.

\begin{equation}
\left[\hat{x}, \hat{p}_x\right]\psi(x) = \hat{x}\hat{p}_x\psi(x) - \hat{p}_x\hat{x}\psi(x)
\end{equation}
\begin{equation}
\left[\hat{x}, \hat{p}_x\right]\psi(x) = -i\hbar x\frac{\partial\psi(x)}{\partial x} + i\hbar\frac{\partial x \psi(x)}{\partial x}
\end{equation}
\begin{equation}
\left[\hat{x}, \hat{p}_x\right]\psi(x) = -i\hbar x\frac{\partial\psi(x)}{\partial x} + \left(i\hbar x\frac{\partial\psi(x)}{\partial x} + i\hbar\psi(x)\right)
\end{equation}
\begin{equation}
\left[\hat{x}, \hat{p}_x\right]\psi(x) = i\hbar\psi(x)
\end{equation}
\begin{equation}
\left[\hat{x}, \hat{p}_x\right] = i\hbar
\end{equation}

Now that we have the commutator, all that's left to do is plug everything into equation $(16)$.

\begin{equation}
\sigma_\hat{x} \sigma_{\hat{p}_x} \geq \frac{1}{2}\left|i\hbar \langle\psi\mid\psi\rangle\right|
\end{equation}
\begin{equation}
\sigma_\hat{x} \sigma_{\hat{p}_x} \geq \frac{\hbar}{2}
\end{equation}

(If you're asking yourself where the $i$ and $\langle\psi\mid\psi\rangle$ went, remember that $|i|=1$ and that $\lvert\psi\rangle$ is normalized so that $\langle\psi\mid \psi\rangle=1$.) And that's the Heisenberg uncertainty relation. An interesting thing to note is that the only reason why there is uncertainty between $\hat{x}$ and $\hat{p}_x$ is because they don't commute. If instead we have two operators that commute, say the $y$-component position operator, $\hat{y}=y$, and $\hat{p}_x$, then it would be completely possible to simultaneously make both measurements with complete accuracy (assuming you somehow got your hands on 100% accurate instruments).