Question

A particle is in the ground state of the harmonic oscillator with classical frequency $\omega$, when suddenly the spring constant quadruples, so $\omega \text{'}=2\omega$, without initially changing the wave function (of course, $\Psi$will now evolve differently, because the Hamiltonian has changed). What is the probability that a measurement of the energy would still return the value $\frac{\hslash \omega }{2}$? What is the probability of getting $\hslash \omega$?

Short answer

The probability of getting $\frac{1}{2}\hslash \omega$is zero, and the probability of $\hslash \omega$is 0.943.

Step-by-step solution

The given values of the question

The particle is in ground state and its classical frequency is $\omega$. when the spring becomes entangled in a never-ending tangle $\omega \text{'}=2\omega$.

The probabilities of ℏω2 and ℏω

The new allowed energies are ${E_n}^{\text{'}}=(n+\frac{1}{2})\hslash {\omega }^{\text{'}}=2(n+\frac{1}{2})\omega =\hslash \omega ,3\hslash \omega ,5\hslash \omega ,....$

so the probability of getting$\frac{1}{2}\hslash \omega$ is zero.

The probability of getting $\hslash \omega$is $P_0={\left|c_0\right|}^2$, where $c_0=\int \Psi (x,0){\psi }_0^{\text{'}}dx$with

Substitute the known values in$P_0={\left|c_0\right|}^2$

Thus, the probability of $\hslash \omega$is 0.943.