Heisenberg Uncertainty Principle

Consider a long string which contains a wave that moves with a fairly well-defined wavelength across the whole length of the string. The question, "where is the wave" does not seem to make much sense, since it is spread throughout the length of string. A quick snap of the wrist and the string produces a small bump-like wave which has a well defined position. Now the question, "what is the wavelength" does not make sense, since there is no well defined period. This example illisturates the limitation on measuring the wavelength and the position simultaneously. Relating the wavelength to momentum yields the de Broglie equation, which is applicable to any wave phenomenon, including the wave equation:

$p={\frac {h}{\lambda }}={\frac {2\pi \hbar }{\lambda }}$

Now that there is a relation between momentum and position, the uncertainty of the measurement of either momentum or position takes mathematical form in the Heisenberg Uncertainty relation:

$\Delta x\Delta p\geq {\frac {\hbar }{2}}$

where the $\Delta \ A$ of each operator represents the positive square root of the variance, given generally by:

$\langle (\Delta A)^{2}\rangle =\langle A^{2}\rangle -\langle A\rangle ^{2}.$

Although both momentum and position are measurable quantities that will yield precise values when measured, the uncertainty principle states that the deviation in one quantity is directly related to the other quantity. This deviation in the uncertainty principle is the result of identically prepared systems not yielding identical results.

A generalized expression for the uncertainty of any two operators A and B is:

$\Delta A\Delta B={\frac {1}{2i}}\langle [A,B]\rangle .$

And thus, there exists an uncertainty relation between any two observables which do not commute.

More generally the uncertainty principle states that two canonically conjugated variables cannot be determined simultaneously with a precision exceeding the relation:

$\Delta A\Delta B=\hbar$

Canonically conjugated variables are those which are related by the Fourier Transform. More specifically, they are variables that when you take the Fourier Transform of a function that is dependent on one, you get a function that depends on the other. For example, position and momentum are canonically conjugated variables:

$\Phi (p,t)={\frac {1}{2\pi \hbar }}\int _{-\infty }^{+\infty }e^{\frac {-ipx}{\hbar }}\Psi (x,t)dx$; $\Psi (x,t)={\frac {1}{2\pi \hbar }}\int _{-\infty }^{+\infty }e^{\frac {ipx}{\hbar }}\Phi (p,t)dp$.

Another example of canonically conjugated variables are energy and time. It is precisely this relationship that leads to the uncertainty principle. The reader has probably noticed that the relation above – i.e. $\Delta A\Delta B=\hbar$ – is not the familiar uncertainty principle we all know, the one where $\hbar$ is divided by two. It turns out that the above relation is more general; we only get the more familiar version when the wave-packet is Gaussian.

A generalised proof of the Uncertainty Principle is as follows.

For any observable A, we have

$\sigma _{A}^{2}=<({\hat {A}}-<A>)\Psi |({\hat {A}}-<A>)\Psi >=<f|f>$

where $f\equiv ({\hat {A}}-<A>)\Psi$. Likewise, for any other observable, B,

$\sigma _{B}^{2}=<g|g>$ where $g\equiv ({\hat {B}}-<B>)\Psi$.

Now we invoke the Schwarz inequality. Recall that this is just

$|\int _{a}^{b}f(x)^{*}g(x)dx|\leq {\sqrt {(}}\int _{a}^{b}|f(x)^{2}|dx\int _{a}^{b}|g(x)^{2}|dx$

Invoking this expression,

$\sigma _{A}^{2}\sigma _{B}^{2}=<f|f><g|g>\geq |<f|g>|^{2}$.

Now for any complex number z

$|z|^{2}=[Re(z)]^{2}+[Im(z)]^{2}\geq [Im(z)]^{2}=[{\frac {1}{2i}}(z-z^{*})]^{2}$.

Letting $z=<f|g>$,

$\sigma _{A}^{2}\sigma _{B}^{2}\geq ({\frac {1}{2i}}[<f|g>-<g|f>])^{2}$.

But

$<f|g>=<({\hat {A}}-<A>)\Psi |({\hat {B}}-<B>)\Psi >=<\Psi |({\hat {A}}-<A>)({\hat {B}}-<B>)\Psi >$

$=<\Psi |{\hat {A}}{\hat {B}}-{\hat {A}}<B>-{\hat {B}}<A>+<A><B>)\Psi >$

$=<\Psi |{\hat {A}}{\hat {B}}\Psi >-<B><\Psi |{\hat {A}}\Psi >-<A><\Psi |{\hat {B}}\Psi >+<A><B><\Psi |\Psi >$

$<{\hat {A}}{\hat {B}}>-<B><A>-<A><B>+<A><B>=<{\hat {A}}{\hat {B}}>-<A><B>$.

Similarly,

$<g|f>=<{\hat {A}}{\hat {B}}>-<A><B>$

so

$<f|g>-<g|f>=<{\hat {A}}{\hat {B}}>-<{\hat {B}}{\hat {A}}>=<[{\hat {A}},{\hat {B}}]>$

where

$[{\hat {A}},{\hat {B}}]\equiv {\hat {A}}{\hat {B}}-{\hat {B}}{\hat {A}}$

is the commutator of the two operators.

Conclusion:

$\sigma _{A}^{2}\sigma _{B}^{2}\geq ({\frac {1}{2i}}<[{\hat {A}},{\hat {B}}]>)^{2}$.

This is the generalised uncertainty principle.

So, here is a true physical constraint on a wave packet. If we compress it in one variable, it expands in the other! If it is compressed in position (i.e., localized) then it must must be spread out in wavelength. If it is compressed in momentum, it is spread out in space.

So now let's see what is the physical meaning of Heisenberg’s inequalities?

1. Suppose we prepare N systems in the same state. For half of them, we measure their positions x; for the other half, we measure their momenta $p_{x}$. Whatever way we prepare the state of these systems, the dispersions obey these inequalities.

2. These are intrinsic properties of the quantum description of the state of a particle.

3. Heisenberg uncertainty relations have nothing to do with the accuracy of measurements. Each measurement is done with as great an accuracy as one wishes. They have nothing to do with the perturbation that a measurement causes to a system, inasmuch as each particle is measured only once.

4. In other words, the position and momentum of a particle are defined numerically only within limits that obey these inequalities. There exists some “fuzziness” in the numerical definition of these two physical quantities. If we prepare particles all at the same point, they will have very different velocities. If we prepare particles with a well-defined velocity, then they will be spread out in a large region of space.

5. Newton’s starting point must be abandoned. One cannot speak simultaneously of x and p. The starting point of classical mechanics is destroyed. Some comments are in order.

A plane wave corresponds to the limit $\triangle p$. Then $\triangle x$ is infinite. In an interference experiment, the beam, which is well defined in momentum, is spread out in position. The atoms pass through both slits at the same time.We cannot “aim” at one of the slits and observe interferences.

The classical limit (i.e. how does this relate to classical physics) can be seen in a variety of ways that are more or less equivalent. One possibility is that the orders of magnitude of x and p are so large that ${\frac {\hbar }{2}}$ is not a realistic constraint. This is the case for macroscopic systems. Another possibility is that the accuracy of the measuring devices is such that one cannot detect the quantum dispersions $\triangle x$ and $\triangle p$.

See also Generalized Heisenberg Uncertainty Relation

As a trivial example, suppose the first observable is position $({\hat {A}}=x)$, and the second is momentum ${\hat {B}}={\frac {\hbar }{i}}{\frac {d}{dx}}$.

The commutation relation between these two observables is just

$[{\hat {x}},{\hat {p}}]=i\hbar$.

So,

$\sigma _{A}^{2}\sigma _{B}^{2}\geq ({\frac {1}{2i}}i\hbar )^{2}=({\frac {\hbar }{2}})^{2}$.

A worked problem showing the uncertainty in the position of different objects over the lifetime of the universe: Uncertainty Relations Problem 1

A problem about how to find kinetic energy of a particle, a nucleon specifically, using the uncertanity principle : Uncertainty Relations Problem 2

Another problem verifying Uncertainty relation: Verifying the Relation