# Heisenberg Uncertainty Derivation

The Heisenberg uncertainty principle seems like a principle that is so fundamentally experimental but it can actually be derived through theory. This requires the consideration of three concepts: the Cauchy-Shwarz inequality, measurement operators, and commutators.

1. The Cauchy-Shwarz inequality

$|\langle&space;v|w\rangle|^2\leq&space;\langle&space;v|v\rangle\langle&space;w|w\rangle$

This can be proved through the completeness relation and treating $\inline&space;\frac{|w\rangle}{\sqrt{\langle&space;w|w&space;\rangle}}$ as one of the orthonormal basis vectors. Look at the proof below where $\inline&space;i$ is the set of orthonormal basis vectors.

$\langle&space;v|v\rangle\langle&space;w|w\rangle=\sum_i&space;\langle&space;v|i\rangle\langle&space;i|v\rangle\langle&space;w|w\rangle$
$\geq&space;\frac{\langle&space;v|w\rangle\langle&space;w|v\rangle\langle&space;w|w\rangle}{\langle&space;w|w\rangle}$
$=&space;\langle&space;v|w\rangle\langle&space;v|w\rangle^*$
$=&space;|\langle&space;v|w\rangle|^2$

2. Measurement Operators

Observables or, in linear algebra terms, bases have different basis states that they are comprised of as well as a set of measurement operators. A measurement operator for some state $\inline&space;|s\rangle$ can be expressed as $\inline&space;M_s=|s\rangle\langle&space;s|$ such that $\inline&space;\langle&space;\psi|M_s|\psi\rangle$ is the probability that the measurement of $\inline&space;|\psi&space;\rangle$ yields $\inline&space;|s\rangle$. This can be easily be seen by substitution: $\inline&space;\langle&space;\psi|M_s|\psi\rangle=\langle\psi|s\rangle\langle&space;s|\psi\rangle=&space;|\langle\psi|s\rangle|^2$. Each state also has some associated energy $\inline&space;E_s$. Using this, one can multiply the probability of each state by its energy and sum over all states to arrive at an expected value for the energy. The mathematical representation of this is shown below.

$\sum_i&space;E_i\langle\psi|M_i|\psi\rangle$

This wavefunctions can be factored out to arrive at the following form.

$\langle\psi|\left&space;(&space;\sum_i&space;E_iM_i&space;\right&space;)|\psi\rangle$

The sum of scaled operators on the inside can be treated as a single operator $\inline&space;S&space;\equiv&space;\sum_i&space;E_i&space;|i\rangle\langle&space;i|$. This is basically the decomposition of some observable or basis. Note that this can be done with different observables that have different states. Now, it can be seen that $\inline&space;\langle\psi|S|\psi\rangle$ is the expected energy of the observable, often notated $\inline&space;\langle&space;S\rangle$ for simplicity.

One can use this average value to measure standard variation $\inline&space;V_s$ by defining it as the average square of the difference between the measured value and the expected value. The square is used so that positive and negative deviations from the mean don’t cancel out. This can be put in mathematical terms, in some sense, as $\inline&space;\left&space;\langle&space;(S-\langle&space;S\rangle)^2&space;\right&space;\rangle$. This is simplified below.

$V_ S\equiv\left&space;\langle&space;(S-\langle&space;S\rangle)^2&space;\right&space;\rangle$

$=\left&space;\langle&space;S^2-2S\langle&space;S\rangle&space;+\langle&space;S\rangle&space;^2&space;\right&space;\rangle$

$=\langle&space;S^2\rangle&space;-2\langle&space;S\rangle^2&space;+\langle&space;S\rangle&space;^2$

$=\langle&space;S^2\rangle&space;-\langle&space;S\rangle^2$

3. Commutators

If $\inline&space;A$ and $\inline&space;B$ are two operators, the following describes measures of commutativity and anti-commutativity.

$[A,B]=AB-BA$

$\{A,B\}=AB+BA$

Assume $\inline&space;\langle\psi|AB|\psi\rangle=a+bi$. Then, $\inline&space;\langle\psi|[A,B]|\psi\rangle=2a$ and $\inline&space;\langle\psi|\{A,B\}|\psi\rangle=2bi$. This allows for the relationship shown below.

$|\langle\psi|\{A,B\}|\psi\rangle|^2+|\langle\psi|[A,B]|\psi\rangle|^2=4|\langle\psi|AB|\psi\rangle|^2$

The Derivation

Through the Cauchy-Shwarz inequality, the two vectors $\inline&space;A|\psi\rangle$ and $\inline&space;B|\psi\rangle$ can be related in the following way.

$|\langle\psi|AB|\psi\rangle|^2\leq\langle\psi|A^2|\psi\rangle\langle\psi|B^2|\psi\rangle$

Using earlier knowledge about commutators, this can be rewritten as it is done below.

$|\langle\psi|AB|\psi\rangle|^2 \leq \langle\psi|A^2|\psi\rangle\langle\psi|B^2|\psi\rangle$
$|\langle\psi|\{A,B\}|\psi\rangle|^2+|\langle\psi|[A,B]|\psi\rangle|^2 \leq 4\langle\psi|A^2|\psi\rangle\langle\psi|B^2|\psi\rangle$
$|\langle\psi|[A,B]|\psi\rangle|^2 \leq 4\langle\psi|A^2|\psi\rangle\langle\psi|B^2|\psi\rangle$

This can be put in the nicer notation described earlier.

$|\langle[A,B]\rangle|^2\leq 4\langle&space;A^2\rangle\langle&space;B^2\rangle$

Now assume $\inline&space;A\equiv&space;C-\langle&space;C\rangle$ and $\inline&space;B\equiv&space;D-\langle&space;D\rangle$ where $\inline&space;C$ and $\inline&space;D$ are observables. These new definitions are substituted  into the above inequality.

$|\langle[C-\langle&space;C\rangle,D-\langle&space;D\rangle]\rangle|^2\leq 4&space;\left&space;\langle&space;(C-\langle&space;C\rangle)^2&space;\right&space;\rangle\left&space;\langle&space;(D-\langle&space;D\rangle)^2&space;\right&space;\rangle$

$|\langle&space;CD-C\langle&space;D\rangle&space;-D\langle&space;C\rangle+\langle&space;C\rangle\langle&space;D\rangle-DC+D\langle&space;C\rangle+C\langle&space;D\rangle+\langle&space;D\rangle\langle&space;C\rangle\rangle|^2\leq 4&space;V_C V_D$

$|\langle&space;CD-DC\rangle|^2\leq 4&space;V_C V_D$

$|\langle&space;[C,D]\rangle|^2\leq 4&space;V_C V_D$

If one treats $\inline&space;C$ as position and $\inline&space;D$ as momentum, then can make use of the canonical commutation relation in quantum mechanics which states that the commutator between position and momentum is $\inline&space;i\hbar$.

$|\langle[x,p]\rangle|^2\leq&space;4V_x&space;V_p$

$|\langle i\hbar\rangle|^2\leq&space;4V_x&space;V_p$

$\hbar^2\leq&space;4V_x&space;V_p$

Because variance is the merely the square of deviation, one can take the square root of both sides to arrive at at Heisenberg uncertainty principle.

$\hbar^2\leq&space;4V_x&space;V_p$

$\hbar\leq&space;2\Delta&space;x&space;\Delta&space;p$

$\frac{\hbar}{2}\leq&space;\Delta&space;x&space;\Delta&space;p$

If you want to know more or see where I learned it from, read “Quantum computation and information” by Isaac Chuang and Michael Nielsen. It should be in the “Books” section of this site.