Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Heisenberg's uncertainty principle states that: if the x-component of the momentum of a particle is measured with an uncertainty

$$\Delta \vec p_x$$

then its x-position cannot, at same time, be measured more accurately than $$\Delta\vec x=\frac {\hbar}{2\Delta\vec p_x},$$

$$\Delta\vec x\Delta\vec p_x \ge \frac {\hbar}{2}.$$

What is the scientific proof of this principle? Operators Uncertainty

share|improve this question
3  
If you want a mathematical proof, I think the simplest thing you can look at is the Heisenberg-Gabor inequality (Gabor's original paper: univie.ac.at/NuHAG/FEI/finalps/CLASSICS/gabor46.pdf, esp. look at the explanation on pg. 432), which essentially states that a function and its Fourier transform cannot both be bounded domain and is easy to prove. –  recipriversexclusion Apr 20 '12 at 18:39
1  
Possible duplicate: physics.stackexchange.com/q/10362/2451 –  Qmechanic Apr 21 '12 at 23:05
1  
@Qmechanic not a dupe: the OP is looking for experimental proof as you can deduce from physics.stackexchange.com/questions/24116/… –  Sklivvz Dec 26 '12 at 11:53
    
Further discussion has been moved here –  Manishearth Jan 2 '13 at 21:27
add comment

migrated from theoreticalphysics.stackexchange.com Apr 20 '12 at 16:16

This question came from our site for scientific theorists and academic scholars interested in theoretical, research-level physics.

4 Answers

up vote 9 down vote accepted

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.

share|improve this answer
    
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. –  Sklivvz Dec 26 '12 at 11:57
    
@Sklivvz: Out of curiosity, which of OP's comments are you referring to? –  Qmechanic Dec 26 '12 at 13:33
    
    
@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. –  Dilaton Dec 26 '12 at 22:45
    
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. –  Dilaton Dec 26 '12 at 23:07
show 1 more comment

You can find a proof of the Heisenberg Uncertainty principle in our preprint:

Faycal Ben Adda and Helene Porchon, Deterministic Elaboration of Heisenberg's Uncertainty Relation, arXiv:1204.1877 [math.GM].

share|improve this answer
1  
Dear Dr. Ben Adda, if you cite yourself it is best if you say so explicitly in your answer, cf. Physics.SE policy and SE policy. –  Qmechanic Jun 6 '12 at 18:39
    
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. –  Sklivvz Dec 26 '12 at 11:57
add comment

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 wavelength with the momentum 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.

share|improve this answer
add comment

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.

share|improve this answer
    
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. –  rubenvb Jan 2 '13 at 15:23
add comment

protected by Qmechanic Jan 2 '13 at 20:34

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?