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While looking at an intuitive explanation for the Heisenberg Uncertainty Principle (related question below), there was a mention of an axiomatic approach to establishing the uncertainty principle. Could someone please point out a source with detailed steps and explanations from first principles?

Related Question

Can the Heisenberg Uncertainty Principle be explained intuitively?

Some (re)-search will reveal the below proof (and others) which are perhaps not immediately grasped by people unfamiliar with certain terminology / concepts.

Related Proof

Heisenberg Uncertainty Principle scientific proof

In the above answer, It is not clear

  1. How the product of vectors, $PQ$, is decomposed into real and imaginary parts?

  2. How the expected value of $PQ$ squared is the square of the imaginary and real parts separately?

  3. How both square things are positive since a complex portion is involved?

  4. How square things being positive means that the left hand side is bigger than one quarter the square of the commutator?

  5. Why the commutator is unchanged by the shifting $[P,Q]=[p,x]=ℏ$ ?

Please note, I was a decent physics student (perhaps not, but am still deeply interested) who has wandered off into the social sciences for graduate studies. Hence, I am a bit rusty on the notation and terminology. Any pointers to brush up the concepts and fill the gaps in existing explanations would be much appreciated. I understand my questions might seem very trivial or obvious to experts, hence please pardon my ignorance of any basic concepts.

EDIT: Please note, this is not a duplicate question. The question to which the reference is made to mark this question a duplicate is linked into this question. This present question originated as a result of a few doubts regarding the other question. I will remove this edit once this is clear.

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  • $\begingroup$ Axiomatic proofs belong to mathematics. The need for the HUP as a principle comes from the data: the need to model the wave nature of the probabilities for measurements in the microcosm. All the mathematical proofs can tell you is : that the mathematical model fits the data successfully. $\endgroup$ – anna v Jul 26 '16 at 5:07
  • $\begingroup$ I will repeat: The HUP is called a principle but it could have been called a "law", as in the "gravitational laws", or a "rule" as in "Born rule". In physics, "principles", "laws", "postulates" are words used to emphasize that extra constraints on the mathematics and the solutions of the mathematics have to be imposed, to select the subset of solutions which FIT THE DATA AND OBSERVATIONS. There is no way of "proving data". That is why physical theories are validated and not proven. The consistency of the mathematical predictions to the data validates the physics theory. Does not prove it $\endgroup$ – anna v Aug 7 '16 at 4:14
  • $\begingroup$ Therefore the answers you get showing that the HUP emerges from the mathematics used, is not proof of the HUP. It is a proof of the consistency of the mathematical model to the data. When something is correctly proven in mathematics, it cannot change. Physical postulates/principles/laws etc are subject to change to fit observations, Example: for high velocities galilean postulates are discarded and one gets special relativity postulates that fit the data. $\endgroup$ – anna v Aug 7 '16 at 4:17
  • $\begingroup$ @annav Thanks for your clarifications. I understand what you saying. My edit is based on Qmechanic marking this question as duplicate, which it is not since it refers and seeks clarifications on the question based on which he /she marked it as a duplicate. $\endgroup$ – texmex Aug 7 '16 at 5:13
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1) It is a product of operators. And they are not so much decomposed into real and imaginary parts, but rather into self-adjoint and antiself-adjoint parts.

If we take self-adjoint $A,B$ linear operators on some suitable Hilbert space, it is clear that $$ AB=\frac{1}{2}(AB+BA)+\frac{1}{2}(AB-BA), $$ since $$ \frac{1}{2}(AB+BA)+\frac{1}{2}(AB-BA)=\frac{1}{2}AB+\frac{1}{2}BA+\frac{1}{2}AB-\frac{1}{2}BA=2\frac{1}{2}AB=AB. $$ Now, since $A$ and $B$ are self-adjoint, $$ (AB+BA)^\dagger=B^\dagger A^\dagger+A^\dagger B^\dagger=BA+AB=AB+BA, $$ so this "anticommutator", $AB+BA$, is self-adjoint if $A$ and $B$ are.

However, if we look at the commutator, $AB-BA$, $$ (AB-BA)^\dagger=B^\dagger A^\dagger-A^\dagger B^\dagger=BA-AB=-(AB-BA), $$ the commutator of self-adjoint operators $A,B$ is antiself-adoint.

Now, the reason he called them "real" and "imaginary", is because in the space of all linear operators of a unitary vector space, self-adjoint operators are analogous to real numbers within the complex number field, and antiself-adjoint operators are analogous to imaginary numbers.

2, and the rest) First we should note that we take expectation values with respect to quantum states. If our particle is in the state $|\psi\rangle$, then the expectation value of $A$ with respect to the state $|\psi\rangle$ is $\langle A\rangle_\psi=\langle\psi|A|\psi\rangle$, from which we can see that the expectation value is linear.

Now then, $$ \langle AB\rangle=\left\langle\frac{1}{2}(AB+BA)+\frac{1}{2}(AB-BA)\right\rangle=\frac{1}{2}\langle AB+BA\rangle+\frac{1}{2}\langle AB-BA\rangle .$$

We should note that the expectation value of an operator is related to its eigenvalues. The expectation value of a self-adjoint operator is real, because its eigenvalues are real, and the expectation value of an antiself-adjoint operator is imaginary, because the eigenvalues are imaginary. Also, because the commutator $[A,B]=AB-BA$ is antiself-adjoint, there exists a self-adjoint operator $C$, for which $[A,B]=iC$, since $iC$ is then antiself-adjoint ($C$ is self-adjoint, but the $i$ swaps sign).

Note now, that the post you were quoting was wrong in the sense we do not take the square of the expectation value, but the square of the absolute value of the expectation value.

But since $\langle AB-BA\rangle=\langle[A,B]\rangle=\langle iC\rangle=i\langle C\rangle$, and then we take the absolute value square of $\langle AB\rangle$: $$ |\langle AB\rangle|^2=\frac{1}{4}\langle AB+BA\rangle^2+\frac{1}{4}\langle C\rangle^2,$$ but then $$ |\langle AB\rangle|^2\ge \frac{1}{4}\langle C\rangle^2 ,$$ and everything here is less than $(\Delta A)^2(\Delta B)^2$, which means that $$(\Delta A)(\Delta B)\ge\frac{1}{2}\langle C\rangle,$$ but for $x$ and $p$, $C=-i[x,p]=-ii\hbar\mathbb{I}=\hbar\mathbb{I},$ so $\frac{1}{2}\langle C\rangle=\hbar/2$.

Do note, that the proof you linked is slightly wrong, based on my first glance, or it used some implicit algebraic manipulations I find nontrivial, but the general line of thought is the same.

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  • $\begingroup$ Thanks for this detailed answer. Could you please provide a self contained reference (as concise as possible) to get familiar with the notation, terminology and concepts discussed here.? Much appreciative of your help. $\endgroup$ – texmex May 20 '16 at 3:38
  • $\begingroup$ @user249613 I learned QM from the lecture notes released by the lecturer who held the QM course to me. It is available freely, but only in hungarian, so I doubt that'd be of any avail to you. Most of that lecture has been based on J. J. Sakurai's QM book "Modern Quantum Mechanics", though which is a great book. I can also recommend Griffiths' QM book, which is a very nice introductory book (Sakurai's is more advanced), however Griffiths uses wave functions in function notation most of the time as opposed to the more abstract state vector-braket notation approach. $\endgroup$ – Bence Racskó May 20 '16 at 12:41
  • $\begingroup$ Much appreciative of your patient clarifications. The key assumption to derive the uncertainty principle seems to be the relationship between canonical conjugate operators; $x$ and $p$, $[x,p]=i\hbar$. I have seen at this link (en.wikipedia.org/wiki/Canonical_commutation_relation) that this means the two operators are fourier transforms of each other. Could you please point out a good source for more intuition and detailed steps on these canonical conjugate operators, their fourier transforms and why they need to staisfy this commutator relationship. $\endgroup$ – texmex Jul 25 '16 at 9:07
  • $\begingroup$ @Qmechanic Please note this is not a duplicate. This questions explicitly mentions and seeks clarifications on portions of the post that you point out as the one that has the answer. $\endgroup$ – texmex Jul 26 '16 at 8:07
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The proof here shows that Hermitian operators $A,\,B$ satisfy $\sigma_A\sigma_B\geq\frac{1}{2}\left|\left\langle\left[A,\,B\right]\right\rangle\right|$. (It's a consequence of the Cauchy-Schwarz inequality on the Hilbert space.) Since $\left[x,\,p\right]=\text{i}\hbar$, $\sigma_x\sigma_p\geq\frac{\hbar}{2}$.

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  • $\begingroup$ Thanks for your time to point out this other answer. I have the seen the other proof you mention. I am not clear on how the product is decomposed and the part thereafter. I have edited the question to reflect this. Could you please elaborate those steps? $\endgroup$ – texmex Jan 17 '16 at 10:32

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