# What causes Heisenberg's uncertainty principle?

• I
I know the equation which explains how it works, but why is it like that?

The closest thing I can think as to why is because of the wave-particle duality of matter given by λ=h/mv, which also explains itself. But I don't know exactly why.

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blue_leaf77
Homework Helper
It's because position and momentum operators do not commute. In general any two non-commuting operators will be connected by some uncertainty relation.

It's because position and momentum operators do not commute. In general any two non-commuting operators will be connected by some uncertainty relation.
Could you please explain/elaborate on that?

Nugatory
Mentor
Could you please explain/elaborate on that?
That result falls out of the mathematical formalism. If a quantum system is in a state in which some observable has a definite value, then that state must be an eigenstate of that observable. Thus, if the system state is such that two different observables both have definite values, then the state must be an eigenstate of both observables. However, there is a basic theorem that says a state can be an eigenstate of two different observables only if they commute, so if the observables don't commute then there is no state in which they both have definite values.

Of course this answer just begs the question, WHY does the universe behave according to this particular set of mathematical rules? Physics has no really satisfactory answer to that question. The universe behaves the way it does, and it's up to us to discover rules that describe that behavior.

Zafa Pi, blue_leaf77, bhobba and 1 other person
vanhees71
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2019 Award
Well, that's not entirely true. The correct statement is that, if two self-adjoint operators commute there's a common (generalized) orthogonal eigensystems of the operators. If the operators don't commute there may be special cases, where the operators have a common eigenvector. Example: Angular momentum components for ##l=0##: There's one state, which is eigenvector for all three components of angular momentum with eigenvalue 0.

blue_leaf77
Nugatory
Mentor
Well, that's not entirely true. The correct statement is that, if two self-adjoint operators commute there's a common (generalized) orthogonal eigensystems of the operators. If the operators don't commute there may be special cases, where the operators have a common eigenvector. Example: Angular momentum components for ##l=0##: There's one state, which is eigenvector for all three components of angular momentum with eigenvalue 0.
Ah - yes, zero eigenvalues do allow for a number of interesting special cases.

Orodruin
Staff Emeritus
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Ah - yes, zero eigenvalues do allow for a number of interesting special cases.
It is not only zero eigenvalue that matters. What matters is whether or not the operators have a common eigenvector. This is not necessarily related to the eigenvalue being zero.

bhobba
dextercioby
Homework Helper
What causes X = Why does X happen? Physics is rather limited when it comes to „why”, because this „why” is always a neverending chain.

victorhugo
jtbell
Mentor
Another way to look at the Heisenberg uncertainty principle for position and momentum is to consider it as a consequence of the mathematics of adding together waves that have different wavelengths:

Any wave function (not just quantum-mechanical ones) has a similar uncertainty principle. In general, it involves the wavenumber k, which in QM is related to the momentum p. Electrical engineers see this uncertainty principle in electronic signal processing. It's part of Fourier analysis.

blue_leaf77
Homework Helper
Could you please explain/elaborate on that?
Sorry for late reply, I was trying to find a lecture note that derives this but couldn't find it in time. As Nugatory said, Heisenberg uncertainty principle is a purely mathematical consequence of the formulation of QM we are using. The derivation can be found in here (section 2), in particular the end result looks like
$$(\Delta A)^2 (\Delta B)^2 \geq \left( \langle\psi|\frac{1}{2i}[A,B]|\psi\rangle\right)^2$$
If ##[A,B]=0## then the two observables would not obey the uncertainty principle.

bhobba
vanhees71
Gold Member
2019 Award
As I said you find one proof also in my old posting

https://www.physicsforums.com/threa...ly-uncertainty-principle.811326/#post-5093474

This proof even admits the derivation of the very states for which the equality sign holds, and these are the coherent and squeezed states (i.e., Gaussians).

Of course the uncertainty principle also holds for ##[A,B]=0##. Only then it's not very exciting, because in this case it just states the obvious fact that the standard deviations of the two observables in any state are not negative :-).

why is it like that?
Heisenberg found that then he could use classical equations. At least that's what The Feynman Lectures says.

Sorry for late reply, I was trying to find a lecture note that derives this but couldn't find it in time. As Nugatory said, Heisenberg uncertainty principle is a purely mathematical consequence of the formulation of QM we are using. The derivation can be found in here (section 2), in particular the end result looks like
$$(\Delta A)^2 (\Delta B)^2 \geq \left( \langle\psi|\frac{1}{2i}[A,B]|\psi\rangle\right)^2$$
If ##[A,B]=0## then the two observables would not obey the uncertainty principle.
It's called the spectral theorem, and the proof isn't particularly difficult.