# Heisenberg Uncertainty Principle

The **Heisenberg Uncertainty Principle** is a relationship between certain types of physical variables like position and momentum, which roughly states that you can never simultaneously know both variables exactly. Informally, this means that both the position and momentum of a particle in quantum mechanics can never be exactly known.

Mathematically, the Heisenberg uncertainty principle is a lower bound on the product of uncertainties of a pair of conjugate variables. The most well-known expression takes the position and momentum to be the conjugate variables:

$\sigma_x \sigma_p \geq \frac{\hbar}{2}.$

Suppose the position of a particle is known to very high precision, so that $\sigma_x$ is very small. Then the uncertainty principle shows that $\sigma_p$ must be large, i.e. the momentum is not known precisely. Furthermore, neither uncertainty can vanish, so neither position nor momentum can ever be exactly measured.

States that saturate the inequality in the Heisenberg uncertainty principle are called **squeezed states**, of which coherent states like the photon states in a laser are a subclass.

#### Contents

## Conjugate Variables

The uncertainty principle relates the standard deviations of two **conjugate variables**, which are any two variables related to each other by the Fourier transform. For example, the wavefunction of a free particle is:

$\Psi(x,t) = \frac{1}{\sqrt{2\pi \hbar}}\int_{-\infty}^{\infty}{\phi(p)}{e}^{i(px-Et)/ \hbar}dp,$

where the $\phi (p)$ are states of definite momentum, i.e. eigenstates of the momentum operator.

If we let $\Phi(p,t)={\phi(p)}{e}^{iEt/ \hbar}$, the integral becomes

$\Psi(x,t) = \frac{1}{\sqrt{2\pi \hbar}}\int_{-\infty}^{\infty}{\Phi(p,t)}{e}^{ipx/ \hbar}dp.$

Therefore, the position-space wavefunction $\Psi(x,t)$ is indeed the Fourier transform of another wavefunction that is dependent on momentum instead. This wavefunction, $\Phi(p,t)$, is called the momentum-space wavefunction.

The momentum-space wavefunction can be recovered given the position-space wavefunction by taking the inverse Fourier transform:

$\Phi(p,t) = \frac{1}{\sqrt{2\pi \hbar}}\int_{-\infty}^{\infty}{\Psi(x,t)}{e}^{-ipx/ \hbar}dx.$

This Fourier analysis achieves the interesting result of displaying the direct relationship between two variables $x$ and $p$ that describe the same wave-packet in different spaces. In this example, a free particle is measured in the position space by $\Psi(x,t)$, but measured in the momentum space by $\Phi(p,t)$. Regardless of which space one chooses to measure the particle, the physics doesn't change. Therefore, this relationship of conjugate variables allows one to measure particles in physical experiments in two ways. In fact, much of crystallography and solid-state physics relies on measurements made in momentum space in lieu of position space.

## Derivation of the Uncertainty Principle

The standard deviation $\sigma_x$ of some variable $x$ is defined by

$\sigma_{x}^{2} = \left\langle{(x-\langle x \rangle)}^{2}\right\rangle = \left\langle x^2 \right\rangle - \langle x \rangle^2.$

That is, $\sigma_x^2$ gives the average squared difference from the mean $\langle x \rangle$.

The uncertainty principle holds for any quantum mechanical state $|\Psi\rangle$. To derive the uncertainty principle thus requires consideration of the expectation value of $(x - \langle x \rangle)^2$ in some arbitrary state $|\Psi\rangle$. This expectation value is given by:

${\sigma}_{x}^{2}=\left\langle\Psi\left|\left(\hat{x}-\langle\hat{x}\rangle\right)^2\right|\Psi\right\rangle.$

Since $x$ is Hermitian, setting $f=(\hat{x}-\langle\hat{x}\rangle)\Psi$ gives ${\sigma}_{x}^{2}=\langle f|f\rangle$.

Similarly, the momentum uncertainty is written:

${\sigma}_{p}^{2}=\left\langle\Psi \left|\left(\hat{p}-\langle\hat{p}\rangle\right)^2 \right|\Psi\right\rangle =\langle g|g\rangle,$

where $g=(\hat{p}-\langle\hat{p}\rangle)\Psi$.

A clever application of the Cauchy-Schwarz inequality gives the rough form of the Heisenberg uncertainty principle:

${\sigma}_{x}^{2}{\sigma}_{p}^{2} = \langle f|f\rangle\langle g|g\rangle \ge {|\langle f|g\rangle |}^{2}.$

To obtain the final expression, compute the right-hand side $\langle f|g\rangle$ above:

$\begin{aligned} \langle f|g\rangle &=\langle(\hat{x}-\langle\hat{x}\rangle)\Psi|(\hat{p}-\langle \hat{p}\rangle)\Psi\rangle \\ &=\langle\Psi|(\hat{x}-\langle\hat{x}\rangle)(\hat{p}-\langle\hat{p}\rangle)\Psi\rangle \\ &=\langle\Psi|(\hat{x}\hat{p}-\hat{x}\langle\hat{p}\rangle)-\hat{p}\langle\hat{x}\rangle+\langle\hat{x}\rangle\langle\hat{p}\rangle)\Psi\rangle \\ &=\langle\Psi|\hat{x}\hat{p}\Psi\rangle-\langle p\rangle\langle\Psi|\hat{x}\Psi\rangle-\langle x\rangle\langle\Psi|\hat{p}\Psi \rangle+\langle x\rangle\langle p\rangle\langle\Psi|\Psi\rangle\\ &=\langle\hat{x}\hat{p}\rangle-\langle\hat{p}\rangle\langle\hat{x}\rangle-\langle\hat{x}\rangle\langle\hat{p}\rangle+\langle\hat{x}\rangle\langle\hat{p}\rangle\\ &=\langle\hat{x}\hat{p}\rangle-\langle\hat{p}\rangle\langle\hat{x}\rangle. \end{aligned}$

Taking the complex conjugate,

$\langle g|f\rangle=\langle\hat{p}\hat{x}\rangle-\langle\hat{x}\rangle\langle\hat{p}\rangle.$

Using the following property of complex numbers: ${z}^{*}z \ge {\left[\frac{1}{2i}(z-{z}^{*})\right]}^{2},$

replacing $z$ with $\langle f|g\rangle$ and ${z}^{*}$ with $\langle g|f\rangle$ yields

${\sigma}_{x}^{2}{\sigma}_{p}^{2}\ge {\left[\frac{1}{2i}(\langle f|g\rangle-\langle g|f\rangle)\right]}^{2}.$

Substituting back in for all of the inner products previously computed, this translates to

${\sigma}_{x}^{2}{\sigma}_{p}^{2} \ge {\left|\frac{1}{2i}\left\langle [\hat{x},\hat{p}]\right\rangle \right|}^{2}.$

This is the generalized Heisenberg uncertainty principle. Note that no part of the above derivation uses the fact that $x$ and $p$ represent position and momentum, only the fact that they are conjugate variables. Therefore, the above inequality holds for any two conjugate variables.

The canonical commutation relation gives $[\hat{x},\hat{p}]=i\hbar$. Substituting in above and taking the square root yields the most well-known form of the Heisenberg uncertainty principle:

${\sigma}_{x}{\sigma}_{p} \ge \frac{\hbar}{2}.$

Which of the following gives the correct lower bound $B$ in the uncertainty relation for the operators $\hat{x}$ and $\hat{p}^2$?

The relevant uncertainty relation is $\sigma_x \sigma_{p^2} \geq B.$

## Generalization: Energy-Time Uncertainty

Another uncertainty relation which is often referenced in discussion of quantum mechanics is the **energy-time uncertainty principle**,

$\sigma_E \sigma_t \geq \frac{\hbar}{2}.$

It is tempting to interpret this equation as the statement that a system may fluctuate in energy by an arbitrarily large amount over a sufficiently short time scale. This explanation is often given as a description for particle-antiparticle production and annihilation, where a particle and its antiparticle appear spontaneously from the vacuum briefly via "borrowed" energy before colliding and returning to vacuum. However, this explanation is not very precise and the given inequality is not so well-defined in quantum mechanics despite the nice physical interpretation. The reason it is not well-defined is because there is no operator in quantum mechanics corresponding to the measurement of time, although the Hamiltonian is the operator corresponding to energy. Nevertheless, as explained in this section, there is some way to make sense of an energy-time uncertainty principle by considering how the measurement of an arbitrary operator changes in time.

Since time is not an operator, it is unclear how time enters quantum mechanics at all. The answer is that time is incorporated into the Schrödinger equation, where it describes the time rate of change of a wavefunction. Physically, the passage of time is recorded by noting that certain physical observables are changing over time: for instance, perhaps the position of a particle is changing, which one interprets as motion over time, or the momentum of a particle is changing, which one interprets as accelerating or decelerating over time.

To quantify this statement, consider the Ehrenfest theorem governing the dynamics of the expectation value of an operator $\hat{A}$ in terms of the commutator with the Hamiltonian:

$\frac{d}{dt}\left\langle\hat{A}\right\rangle =\frac{i}{\hbar}\left\langle\left[\hat{H},\hat{A}\right]\right\rangle +\left\langle\frac{\partial \hat{A}}{\partial t}\right\rangle.$

Assuming that $\hat{A}$ does not depend explicitly on time (which is typically the case), the second term vanishes and the expectation value of the $\left\langle \left[\hat{H} ,\hat{A}\right]\right\rangle$ is generally nonzero. Therefore, $\hat{H}$ and $\hat{A}$ satisfy the generalized Heiseinberg uncertainty relation:

$\sigma_H^2 \sigma_A^2 \geq {\left|\frac{1}{2i}\left\langle \left[\hat{H},\hat{A}\right]\right\rangle \right|}^{2} = {\left|\frac{1}{2i} \frac{\hbar}{i} \frac{d\left\langle \hat{A} \right\rangle}{dt} \right|}^{2} = \frac{\hbar^2}{4} \left| \frac{d\left\langle \hat{A} \right\rangle}{dt} \right|^2$

Notably, since the Hamiltonian is the energy operator, $\sigma_H$ corresponds to the uncertainty in energy.

Taking square roots now gives the relation:

$\sigma_H \sigma_A \geq \frac{\hbar}{2} \left|\frac{d\left\langle \hat{A} \right\rangle}{dt} \right|.$

Define $\sigma_t$ by the relation:

$\sigma_A = \left|\frac{d\left\langle \hat{A} \right\rangle}{dt} \right| \sigma_t.$

In other words, $\sigma_t$ corresponds to the time it takes the measured value of $\left\langle \hat{A} \right\rangle$ to shift by $\sigma_A$, for *any operator* $\hat{A}$.

Given this definition, substituting in yields the relation:

$\sigma_H \sigma_t \geq \frac{\hbar}{2},$

which is a more rigorous expression of the energy-time uncertainty principle. Formally, using the definition of $\sigma_t$ one can see that if the energy changes slowly, the rate of change of all expectation values of any observable operator must also be slow. This makes sense because if the energy (i.e., the Hamiltonian) changes slowly, the wavefunction itself that is evolved by the Hamiltonian changes slowly, so the expectation value of observables corresponding to measurements of that wavefunction must change slowly as well.

Suppose an exotic particle with a very large mass of $20 \pm 0.01 \text{ TeV}$ is measured at the Large Hadron Collider. What is the correct lower bound for the lifetime of this exotic particle given no additional information, in seconds? (What does this say about the detectability of very heavy particles?)

The energy of an electron in a magnetic field $B$ is, in some unit system:

$E = \pm \frac{1}{2} \mu B,$

where choice of positive or negative sign corresponds to spin-down or spin-up respectively, and $\mu$ is the spin magnetic moment of the electron.

In an adiabatic transition, the parameters of a quantum system are gradually changed to bring a system smoothly from one state to another state. Suppose an electron starts in the spin-up ground state in a magnetic field of strength $B$. The magnetic field is then reduced slowly to strength $\frac{B}{10}$ and then increased slowly again back to strength $B$. Find the minimum time for the process of tuning the magnetic field to occur for which the electron is expected to remain in the spin-up ground state after the process ends. Hint: consider the energy-time uncertainty principle.

**Note**: this is a very simple demonstration of the fact that adiabatically tuning electron spins requires relatively long time scales.

## References

[1] Griffiths, David J. *Introduction to Quantum Mechanics*. Second Edition. Pearson: Upper Saddle River, NJ, 2006.

**Cite as:**Heisenberg Uncertainty Principle.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/heisenberg-uncertainty-principle/