1
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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.) $\endgroup$ Jan 20, 2014 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, 2014 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. $\endgroup$ Jan 21, 2014 at 19:08

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
13
  • $\begingroup$ 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, 2014 at 18:11
  • $\begingroup$ Sorry I don't understand. What is your $f$ respect to my $\psi$? $\endgroup$ Jan 20, 2014 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, 2014 at 18:13
  • $\begingroup$ I now understand. YES, provided in the first couple of integrals you wrote, $|f(x)|^2$ appears in place of $f(x)$. $\endgroup$ Jan 20, 2014 at 18:15
  • $\begingroup$ I mean: The probability density is $|\psi(x)|^2$. OK? $\endgroup$ Jan 20, 2014 at 18:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.