Heisenberg's uncertainty principle states that:

$$\sigma(x)\sigma( p_x )\ge \frac {\hbar}{2}.$$

What is the scientific proof of this principle? Operators Uncertainty


The uncertainty principle, in the variance formulation, states that in any quantum state $|\rangle$, the quantity

$$\langle (p-<p>)^2 \rangle \langle (x-\langle x\rangle)^2\rangle \ge {\hbar^2 \over 4} $$

To understand why shifting p and x by their expected value and squaring gives the squared uncertainty, see this answer.

The proof is by noting the following

$$ |\langle \psi | \eta \rangle| \le \sqrt{ ||\psi||^2 ||\eta||^2}$$

This is the statement that the dot-product of two vectors is less than the product of their lengths. It is called the "Cauchy Schwartz inequality". For the special case above, defining the operators $P= p-\langle p\rangle$ and $Q=x-\langle x\rangle$ (and squaring both sides),

$$ ( \langle P Q \rangle )^2 \le \langle PP\rangle\langle QQ\rangle $$

Where to see that the above is an instance of Cauchy Schwarz, take:

$$ |\psi\rangle = P|\rangle$$ $$ |\eta\rangle = Q|\rangle$$ While the product PQ can be decomposed into a real and imaginary part

$$ PQ = {1\over 2} (PQ+QP) + {1\over 2} (PQ-QP) $$

The first part is imaginary, because if you take the Hermitian conjugate, it changes sign. The second part is real (this is ultimately because P and Q are real, i.e. Hermitian). The expected value of PQ squared is the square of the imaginary and real parts separately

$$ (\langle P Q \rangle)^2 = {1\over 4} (\langle [P,Q]\rangle)^2 + {1\over 4}(\langle PQ+QP)\rangle)^2 $$

Since both square things are positive, this means that the left hand side is bigger than one quarter the square of the commutator. The commutator is unchanged by the shifting,

$$ [P,Q] = [p,x] = \hbar $$

So that

$$ \langle P^2 \rangle \langle Q^2\rangle \ge (\langle PQ \rangle)^2 \ge {1\over 4} (\langle [P,Q] \rangle)^2 = {\hbar^2 \over 4} $$

The proof is usually given in one line, as directly above, where the Cauchy Schwarz step (first inequality), the imaginary/real part decomposition (second inequality) and the shifted canonical commutation relations (last equality) are assumed internalized by the reader.

This proof appears on Wikipedia, it is used in all QM books, but perhaps this explanation is clearer.

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    $\begingroup$ This is not an answer anymore, after the OP clarified (in a comment) that he's looking for experimental proof. I've changed the question accordingly, but I am not really sure if this answer is appropriate anymore. $\endgroup$ – Sklivvz Dec 26 '12 at 11:57
  • $\begingroup$ @Sklivvz: Out of curiosity, which of OP's comments are you referring to? $\endgroup$ – Qmechanic Dec 26 '12 at 13:33
  • $\begingroup$ @Qme physics.stackexchange.com/questions/24116/… $\endgroup$ – Sklivvz Dec 26 '12 at 16:48
  • $\begingroup$ @Qmechanic Obviously the OP was satisfied with this answer, and he has accepted it. So I dont know what Slivvz is up to by saying it is not an answer anymore. Please leave it as it is. Maybe the change of the question Sklivvz has done is not what the OP wanted to? I thought editing someone else's question such that it changes the meaning of the question is considered rude or at least inappropriate at Physics SE? By no means should this answer be deleted because of this. $\endgroup$ – Dilaton Dec 26 '12 at 22:45
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    $\begingroup$ Since the OP was a member of TP.SE and seems not to be active here anymore, it is not possible to ask him about his intention. So if anything should be done at all, I would rather roll back the question such that it fits again with the accepted answer. $\endgroup$ – Dilaton Dec 26 '12 at 23:07

A wide variety of experiments, of which the Double Slit experiment is the most dramatic, can be used to establish that matter is best represented as a wave on microscopic scales. Once you represent matter as a wave, then it is natural to associate its position with the spread of the wave, and its momentum with the wavelength of the wave. Once you do this, however, it should be clear that there is a tradeoff between a well defined ''location'' of the wave, and a well defined ``wavelength'' of the wave. Therefore, one cannot simultaneously precisely define a particle's position and momentum. Extra precision in one must come with a lost in precision in the other.

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  • $\begingroup$ There is no tradeoff when you derive a slit pattern or any edge pattern with a photon/ particle solution. Matter is best represented as a particle because a wave can only account for some of the phenomena. Not only that but a wave cannot realistically be explained or accounted for. No one even offers a description. For example how would you physical describe the wave action of a single photon traveling from a star to Earth? Does it fill the whole universe and if not why not? $\endgroup$ – Bill Alsept Aug 7 '16 at 16:06
  • $\begingroup$ @BillAlsept: en.wikipedia.org/wiki/Wave_packet $\endgroup$ – Jerry Schirmer Aug 7 '16 at 20:17
  • $\begingroup$ The first paragraph and the illustration pretty much describe a photon. I do not see how this could describe a wave considering there are no ends to a waves. Are you saying your wave has some kind of boundary limit? It spreads out but not too far? That does not sound like a wave it sounds more like a local packet of energy like a photon. $\endgroup$ – Bill Alsept Aug 7 '16 at 20:23
  • $\begingroup$ @BillAlsept: it's a localized wave. The article describes how you can localize a wave. The whole point of the hiesenberg uncertainty principle is that there's a trade-off between the localization of a particle, and having a well-definied momentum of a wave. Proper particles in QFTs are neither waves nor particles. $\endgroup$ – Jerry Schirmer Aug 7 '16 at 21:18
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    $\begingroup$ and "a real packet of energy" IS A WAVELET. A WAVELET IS A TYPE OF WAVE. We write examples using monochromatic plane waves for simplicity, not because anyone believes that they are the sole solutions to wave equations, or are physically realizable states. $\endgroup$ – Jerry Schirmer Aug 7 '16 at 22:33

Not sure what you mean by scientific proof. A hypotheses can be validated by scientific method. Its not proof as in mathematics. Because physics does not deal with abstract mathematical ideas which can be proved by following some predefined axioms and rules. Which has nothing to do with observation.

If uncertainty principle is taken as truth then physical phenomenon regarding subatomic particles can be explained and to some extent predicated by quantum mechanical framework.

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  • $\begingroup$ In this case though, the Uncertainty Principle is a physical consequence of the mathematics used to describe physics (operators of position and momentum and their relation: the Fourier transform). So one could argue that within the framework of Quantum theory, there is definite proof of the uncertainty principle. The theory is then validated by various experiments, which can only go as far as to (not) contradict theory. $\endgroup$ – rubenvb Jan 2 '13 at 15:23

For an observable $A$, write $\langle A\rangle$ for the expected value of $A$ and $\Delta A$ for its standard deviation (so that $\langle A\rangle$ and $\Delta A$ both depend on the current state $\phi$). If $\langle A\rangle=0$ then $\Delta A=\langle A^2\rangle$.

Now given two observables $A$ and $B$, adjust so $\langle A\rangle=\langle B\rangle=0$. Let $\phi$ be the current state and $x$ an arbitrary real number.

Then $$\eqalign{ 0\phantom{3}\le\phantom{3}&\Big\langle (A+ixB)\phi,(A+ixB)\phi\Big\rangle \cr &= \overline{\phi} A^2 \phi + x^2 \overline{\phi} B^2 \phi -ix\overline{\phi} BA \phi +ix \overline{\phi} AB\phi \cr &= \langle A^2\rangle +x \langle i[A,B]\rangle + x^2 \langle B^2\rangle \cr}$$ This holds for all real $x$ so the quadratic in the last line has either no real zeros or one double zero; either way, the discriminant is non-positive: $$\langle i[A,B]\rangle ^2 - 4\langle A^2\rangle \cdot \langle B^2\rangle \phantom{3} \le\phantom{3}0$$ as needed.

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