Is any quantum state $|\psi\rangle$ possible such that the standard deviation $\sqrt{\langle\psi|(\Delta\hat{x})^2|\psi\rangle}$ of the position operator $\hat{x}$ is zero? If not, why?

  • $\begingroup$ What does the uncertainty principle tell you about the spread in momentum for such a state? (Hint: The delta function $\delta(x)$ is the Fourier transform of the constant function.) $\endgroup$ – lionelbrits Jan 20 '14 at 17:35
  • $\begingroup$ But, Uncertainity principle holds only when momentum and position are measured simultaneously. The question would not measure momentum - thus would not have to obey the uncertainity principle, right ? $\endgroup$ – teja4477 Jan 20 '14 at 17:51
  • $\begingroup$ The uncertainty principle is not a statement about measurements but a property of the state itself. Basically, your state is a superposition of waves of all momenta, with equal amplitudes. Besides the fact that at some point the highest momentum modes will put you beyond currently known physics, such a state will quickly smear out as the fastest waves simply "run away". So it is just a delta function for an instant in time, and then no more. $\endgroup$ – lionelbrits Jan 21 '14 at 19:08

The answer is negative. If there were such a state, $\psi$, it would satisfy $$\langle \psi| X^2 \psi \rangle - \langle \psi| X \psi \rangle^2= 0 \quad (1)$$ namely $$\langle \psi| X^2 \psi \rangle = \langle \psi| X \psi \rangle^2$$ that is, since $X=X^\dagger$ and $||\psi||=1$: $$\langle \psi| X \psi \rangle^2 = \langle X\psi| X \psi \rangle \langle \psi |\psi \rangle\:.\quad (2)$$ On the other hand, Cauchy-Schwarz' inequality implies $$\langle \psi| X \psi \rangle^2 \leq \langle X\psi| X \psi \rangle \langle \psi |\psi \rangle\:,$$ where, as is known $=$ holds if and only if the two vectors in the LHS scalar product are proportional. Since it is the case due to (2), we conclude that: $$X\psi = \lambda \psi$$ for some $\lambda \in \mathbb C$. However, as is known,the operator $X$ has no eigenvectors, its spectrum being purely continuous: $\sigma(X)= \mathbb R$. Therefore $\psi$ does not exist.

Nevertheless, formally speaking, you could think of distributions $\psi_{x_0}(x) =\delta(x-x_0)$ as formal eigenvectors of $X$, with eigenvalue $x_0$. These formal states, indicated by $|x_0\rangle$ with the suggestive (and very useful) Dirac notation, do not belong to the Hilbert space of the theory $L^2(\mathbb R)$, since they are not (equivalence classes of) functions.

It is possible to extend the formalism (as done by Gelfand) to include these distributions as defining generalized states into a rigorous mathematical sense. In that case, however, (1) cannot be intepretated literally for these generalized states, since it would involve products of $\delta$ functions which cannot be defined.

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  • $\begingroup$ Then this would mean that, $[\int_{- \propto }^{ \propto } f (x)x dx]^{2} = \int_{- \propto }^{ \propto } f(x) x^{2} dx $ If, $\int_{- \propto }^{ \propto } f(x)^{2} dx$ = exists Thus, would only a trivial solution to the above equation exist ? Apologies - I am severly handicapped with TeX. The alphas are infinity limits. $\endgroup$ – teja4477 Jan 20 '14 at 18:11
  • $\begingroup$ Sorry I don't understand. What is your $f$ respect to my $\psi$? $\endgroup$ – Valter Moretti Jan 20 '14 at 18:12
  • $\begingroup$ $f$ functions the same way as $\psi$ does. I believe then you would point at $\psi$ being a complex function ? $\endgroup$ – teja4477 Jan 20 '14 at 18:13
  • $\begingroup$ I now understand. YES, provided in the first couple of integrals you wrote, $|f(x)|^2$ appears in place of $f(x)$. $\endgroup$ – Valter Moretti Jan 20 '14 at 18:15
  • $\begingroup$ I mean: The probability density is $|\psi(x)|^2$. OK? $\endgroup$ – Valter Moretti Jan 20 '14 at 18:19

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