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The standard derivation

But now suppose the space is a ring of length $L$, it seems the derivation could work out exactly the same and we get $$\Delta p \Delta x \geq \hbar/2.$$

But since $\Delta x$ is limited, and if we choose an eigenstate of momentun, we can actually achieve $$\Delta p \Delta x = 0.$$

What's wrong here?


marked as duplicate by ACuriousMind, AccidentalFourierTransform, John Rennie quantum-mechanics May 4 '16 at 16:21

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  • $\begingroup$ I think you have to very careful when the state is an eigenstate of one of the operators in the uncertainty relation. You run into the same problem in the continuum if you consider plane waves, or delta-functions in space. $\endgroup$ – Mikael Fremling May 4 '16 at 7:13
  • 1
    $\begingroup$ No, Mikael, you don't run into this problem if $x$ is not periodic. The delta-functions have the infinite uncertainty of the other quantity, the product is an indeterminate form, and may be said to obey the inequality when properly regulated. $\endgroup$ – Luboš Motl May 4 '16 at 9:40
  • $\begingroup$ I'd suggest changing the duplicated question's title, because "How can I solve this quantum mechanical “paradox”" doesn't allow it to be searched easily. $\endgroup$ – seilgu May 5 '16 at 6:58

The point is that the domain $D(P)$ of $P$ must be such that $P$ is (essentially) self-adjoint thereon. Otherwise it does not represent an observable. I am assuming that $D(X)= L^2([0,L],dx)$ instead, where $X$ is automatically self-adjoint.

The vector $\psi$ you use to prove Heisenberg inequality has to belong to $D(PX) \cap D(XP)$ as you see by direct inspection proving the wanted inequality.

In $L^2(\mathbb R, dx)$ we have $D(P)\cap D(X) \supset {\cal S}(\mathbb R)$ the subspace space of Schwartz functions which is also invariant under $X$ and $P$ so that any $\psi \in {\cal S}(\mathbb R)$ can be used in Heisenberg inequalities for instance.

In the present case there is a natural definition of $D(P)$ on the ring of length $L$, making $P$ essentially self-adjoiont: $D(P)$ is the space of $C^1$ functions such that $\psi(0)= \psi(L)$.

In this case we find $D(PX)= \{0\}$.

(The regularity condition could be weakened using a weak notion of derivative, but this weaker version of the initial definition of $P$ does not change the final conclusion.)

With the said definition $P$ turns out to be essentially self-adjoint with pure point spectrum made of the points $\frac{2\pi n}{L}$ where $n \in \mathbb Z$ and I assume $\hbar=1$. The corresponding eigenvectors are

$$\psi_n(x) = \frac{e^{i 2\pi n x/L}}{\sqrt{L}}$$

you see that $\psi_n \not \in D(PX)$ as $[0, L] \ni x \mapsto x\psi_n(x)$ does not belong to $D(P)$. Therefore the conclusion $$\Delta P_{\psi_n} \Delta X_{\psi_n}=0$$ does not violate Heisenberg inequality just because it is not valid with the said definition of $X$ and $P$.

On the other hand, it is disputable if these $X$ and $P$ are correct representations of the position and momentum operators because they do not satisfy the canonical commutation relations because, again, $D(PX)= \{0\}$ and these relations require that $D(PX)\cap D(XP)$ is not trivial to have any physical sense since they are valid thereon.

Actually there is a more fundamental overall obstruction to obtain natural physically meaningful operators satisfying canonical commutation relations on the ring.

Suppose that we are so clever to find out a (dense) domain of (sufficiently regular) functions $D\subset L^2([0,L], dx)$ such that our initial $X$ and $P= -i d/dx$ are well defined, at least Hermitian, and satisfy the standard commutation relations thereon. A known theorem by Nelson (actually already obtained by Dixmier in this particular case) joined with the famous Stone-von Neumann theorem establishes that, if $X^2+P^2$ is essentially self-adjoint on $D$ then there exist a unitary operator $U : L^2([0,L], dx) \to L^2(\mathbb R, dx)$ such that (the closures of) $UXU^{-1}$ and $UPX^{-1}$ are the standard self-adjoint position and momentum operators on $L^2(\mathbb R, dx)$ (*).

This is impossible because $UXU^{-1}$ would be unbounded whereas our $X$ is bounded. An essentially equivalent absurdum is that $X$ would have $\mathbb R$ and not $[0,L]$ as spectrum.

In principle you may find some common domain where $X$ (defined on the whole $L^2([0,L],dx)$) and $P$ (initially defined as a $-i$ times a derivative) satisfy canonical commutation relations and thus also Heisenberg inequality, but you have to renounce to some physical relevant property. In particular either $X$ or $P$ or $X^2+P^2$ cannot represent observables.

(*) Actually in general, in the presence of common closed invariant subspaces, $L^2(\mathbb R,dx)$ could be replaced by a direct orthogonal sum of severl copies of it leaving unchanged the final result.


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