Proof of Heisenberg's Uncertainty Principle 📂Quantum Mechanics

Proof of Heisenberg's Uncertainty Principle

Theorem¹

For two operators $A$ and $B$, the following holds.

$$ \sigma_{A}^{2}\sigma_{B}^{2} \ge \left( \dfrac{1}{2\i} \braket{[A, B]} \right)^{2} $$

Here $\sigma_{A}^{2}$ is the variance of $A$, and $[A, B]$ is the commutator of $A$ and $B$.

Explanation

From the theorem above, we can see that the physical quantities corresponding to two operators that do not commute cannot be measured exactly at the same time. That the physical quantity $B$ is measured exactly means that the value of $\sigma_{B}^{2}$ decreases, but if $[A, B] \ne 0$, then a minimum value is fixed for the left-hand side of the inequality, which means that as $\sigma_{B}^{2}$ decreases, the value of $\sigma_{A}^{2}$ increases. Therefore, the more precisely the physical quantity of $B$ is measured, the more uncertain the physical quantity of $A$ becomes.

Proof

By definition, $\sigma_{A}^{2}$ is as follows.

$$ \sigma_{A}^{2} = \braket{\psi | (A - \braket{A})^{2} | \psi} = \braket{(A - \braket{A})\psi | (A - \braket{A})\psi} $$

For convenience, let $f = (A - \braket{A})\psi$.

$$ \sigma_{A}^{2} = \braket{f | f} $$

Similarly, letting $g = (B - \braket{B})\psi$,

$$ \sigma_{B}^{2} = \braket{g | g} $$

By the Cauchy–Schwarz inequality, we obtain the following.

$$ \sigma_{A}^{2}\sigma_{B}^{2} = \braket{f | f} \braket{g | g} \ge \left| \braket{f | g} \right|^{2} $$

Also, for any complex number $z$, the following holds.

$$ \left| z \right|^{2} = \Re{z}^{2} + \im{z}^{2} \le \im{z}^{2} = \left( \dfrac{1}{2\i} (z - z^{\ast}) \right)^{2} $$

The last equality can be easily verified by setting $z = x + \i y$ and computing directly. Now, from the two inequalities above, we obtain the following.

$$ \sigma_{A}^{2}\sigma_{B}^{2} \ge \left| \braket{f | g} \right|^{2} \ge \left( \dfrac{1}{2\i} (\braket{f | g} - \braket{g | f}) \right)^{2} \tag{1} $$

Now let us compute $\braket{f | g}$.

$$ \begin{align*} \braket{f | g} &= \braket{(A - \braket{A})\psi | (B - \braket{B}) | \psi} \\ &= \braket{\psi | (A - \braket{A})(B - \braket{B}) | \psi} \\ &= \braket{\psi | AB - A\braket{B} - \braket{A}B + \braket{A}\braket{B} | \psi} \\ &= \braket{\psi | AB | \psi} - \braket{B}\braket{\psi | A | \psi} - \braket{A}\braket{\psi | B | \psi} + \braket{A}\braket{B} \braket{\psi | \psi} \\ &= \braket{AB} - \braket{B}\braket{A} - \braket{A}\braket{B} + \braket{A}\braket{B} \\ &= \braket{AB} - \braket{A}\braket{B} \end{align*} $$

In the same way, $\braket{g | f} = \braket{BA} - \braket{A}\braket{B}$ holds. Therefore, we obtain the following.

$$ \begin{align*} \braket{f | g} - \braket{g | f} &= (\braket{AB} - \braket{A}\braket{B}) - (\braket{BA} - \braket{A}\braket{B}) \\ &= \braket{AB} - \braket{BA} \\ &= \braket{AB - BA}\\ &= \braket{[A, B]} \end{align*} $$

The reason the third equality in the equation above holds is that the expectation is linear. Substituting this result into (1), we obtain the following.

$$ \sigma_{A}^{2}\sigma_{B}^{2} \ge \left( \dfrac{1}{2\i} \braket{[A, B]} \right)^{2} $$

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