# Heisenberg's Uncertainty Principle

1. May 29, 2014

### velo city

I was watching a YouTube video about heisenberg's uncertainty principle and was wondering if it only works for position and momentum or does it work for any other two measurable quantities?

2. May 29, 2014

### Staff: Mentor

It works for any pair of non-commuting measurable quantities (but say "observables" instead of "measurable quantities" if you want to sound like a pro).

Unfortunately there's no intuitive/easy/non-mathematical definition of "non-commuting observable", so if you aren't going to dig into the math you'll have to take someone else's word for which observables commute and which don't.

3. May 29, 2014

### velo city

So what is the mathematical explanation.?

4. May 29, 2014

### UltrafastPED

Heisenberg's Uncertainty Principle (HUP) works for any two conjugate variables; the terminology comes from the Hamiltonian/Lagrangian methods of analytical mechanics, which use energy instead of forces to solve mechanical systems.

You first pick a set of variables to describe the positions of the moving parts ... for example, an angle for a pendulum, or a displacement for a spring. Then you form the Lagrangian for the system, and determine the "momentum conjugate to" each of the position variables; these pairs are the "conjugate pairs".

This description is fairly technical - but there it is. Also see http://en.wikipedia.org/wiki/Conjugate_variables

5. May 29, 2014

### velo city

Oh so the momentum conjugate is $\frac{∂L}{∂\dot{q}}$ ?

6. May 29, 2014

### velo city

Wait but the coordinates are canonical if their poisson bracket is 0 right?

7. May 29, 2014

### velo city

So you can only measure two observables at the same time if the commutant is 0?

8. May 29, 2014

### Staff: Mentor

Every observable corresponds to a Hermitian operator and any measurement of that observable must yield an eigenvalue of that operator applied to the state of the system under measurement; these are postulates of QM. If the operators do not commute they cannot have common eigenfunctions, so if the state is an eigenstate of one operator it cannot be in an eigenstate of the other. Google for "Born Rule" to see what happens when you apply an operator to a state that is not an eigenstate for that operator.

9. May 29, 2014

### Staff: Mentor

Pretty much yes, although it would be more formally correct to say that only if A and B commute can you set up a system in such a way that a measurement of A will yield a given value with 100% certainty and a measurement of B will also yield a given value with 100% certainty.

Loosely speaking, measuring A and getting the value $a$ leaves the system in an eigenstate of A with eigenvalue $a$; the only way that we can be sure that a subsequent measurement of B will yield the specific value $b$ is if that state is also an eigenstate of B with eigenvalue $b$; and that is only possible if A and B commute.

10. May 29, 2014

### Matterwave

I'm not very well versed in geometrical quantization, but the normal quantization in QM involves really the Cartesian coordinates and their conjugate momenta. Heisenberg and Schroedinger's formulation of Quantum Mechanics, unlike the Bohr-Sommerfeld quantization rules, are distinctly not canonical coordinate transformation invariant. The Canonical quantization of Heisenberg and Schroedinger applies only to the Cartesian canonical coordinates and not to curvilinear coordinates.

Perhaps in the past years someone has been able to figure out a Geometrical quantization procedure that keeps the symplectic and geometric structure of Hamiltonian mechanics which agrees with experiment, but I am not aware of any such successes.

11. May 29, 2014

### Meson080

The following paragraph is extracted from wikipedia-complementarity (See: http://en.wikipedia.org/wiki/Complementarity_(physics)).

The more accurate one property is measured, the less accurate the complementary property is measured, according the Heisenberg uncertainty principle. Further, a full description of a particular type of phenomenon can only be achieved through measurements made in each of the various possible bases — which are thus complementary. The complementarity principle was formulated by Niels Bohr, a leading founder of quantum mechanics.[1]
Examples of complementary properties:

Position and momentum
Spin on different axis
Wave and particle
Value of a field and its change (at a certain position)

I hope this answered your question.

12. May 30, 2014

### tom.stoer

Even so Heisenberg introduced and explained his uncertainty principle in terms of "measurement" of conjugate variables, the principle is independent of any measurement! It's a property of the underlying pure mathematical formalism of quantum mechanics, namely operators representating observables on Hilbert spaces. For any pair of non-commuting observables such an uncertainty can be derived.

In the case of position and momentum the uncertainty simply follows from the Fourier transformation. It is not unique to quantum mechanics but applies to all wave phenomena.

Therefore formally the Heisenberg uncertainty principle says that there do not exist quantum states violating the inequality.

This formulation is independent of any "measurement", and you can avoid the difficult question how "measurement" in quantum mechanics can be defined.

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