0
$\begingroup$

I've found in an article that it is possible to prepare experimentally the superposition of two coherent (quasi-classical) states to obtain the Schroedinger cat state: $$ \left|\psi_{\pm}(t)\right\rangle=\frac{1}{C_{\pm}} e^{-i \omega t}(|\alpha(t)\rangle \pm|-\alpha(t)\rangle) $$ with $\alpha(t)=\alpha \mathrm{e}^{-i \omega t}$ and the normalization constant $C_{\pm} .$

  1. How can I visualize these states? Does this superposition have any physical meaning and why do we need it?

  2. Also I was thinking that as a state of the harmonic oscillator it should be a solution of the Schroedinger equation, although it depends only on time t, but not on x, so how can I write the Schroedinger equation for it?

  3. Am I right that, as all states, $|\psi_{+}(t)\rangle$ and $|\psi_{-}(t)\rangle$ can be represented in the basis of the eigenstates of the Hamiltonian $\hat{H}=\frac{\hat{p}^{2}}{2 m}+\frac{1}{2} m \omega^{2} \hat{x}^{2}$? I mean if I want to get them in this "usual" basis, i should just write $\psi = \sum\ C_a\psi_a $ with $$C_a=\int \psi_a^*\psi \,dx$$, where $\psi_a$ - the eigenstates of the Hamiltonian of the harmonic oscillator $$\psi_{a}(x) =\frac{1}{\sqrt{2^{n} n ! \sqrt{\pi}}} e^{-x^{2} / 2} H_{n}(x) \\ \\ H_{n}(x) =(-1)^{n} e^{x^{2}} \frac{d^{n}}{d x^{n}} e^{-x^{2}} $$? Again it looks weird because $\psi_{\pm}$ has no dependency on x.

Thanks in advance for helping me figure it out.

$\endgroup$
5
  • $\begingroup$ Please provide a link to the article. $\endgroup$ – my2cts Jun 6 at 13:55
  • $\begingroup$ Indeed $|\psi_\pm\rangle$ has no dependence on $x$ because it is a state (a vector in the Hilbert space ${\mathcal H}$) not a wavefunction. The wavefunction is the set of components $\psi_\pm(x)=\langle x|\psi_\pm\rangle$ of the state vector in the position basis. $\endgroup$ – mike stone Jun 6 at 13:57
  • $\begingroup$ This is a linear combination of a state with itself. So one of the combinations is identically zero. $\endgroup$ – my2cts Jun 6 at 13:58
  • $\begingroup$ Have you reviewed your coherent states? Read up on your cat states? $\endgroup$ – Cosmas Zachos Jun 6 at 15:25
  • $\begingroup$ Toss the cat states. Just review Schroedinger's wave packets, their evolution in phase-space, and the conversion of wave functions from the position to the excitation number representation. Cat states are self-evident once you appreciate coherent states. $\endgroup$ – Cosmas Zachos Jun 6 at 15:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.