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

Short Answer

Expert verified

The probability of getting 12is zero, and the probability of is 0.943.

Step by step solution

01

The given values of the question

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

02

The probabilities of ℏω2 and ℏω

The new allowed energies are En'=(n+12)'=2(n+12)=,3,5,....

so the probability of getting12 is zero.

The probability of getting is P0=c02, where c0=(x,0)0'dxwith

Substitute the known values inP0=c02

Thus, the probability of is 0.943.

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