Question

16.2 Harmonic-oscillator number operator Using the definitions $a^\dagger = \sqrt{\frac{M}{2\hbar\omega}} \left(\omega x - \frac{ip}{M}\right)$ and $a = \sqrt{\frac{M}{2\hbar\omega}} \left(\omega x + \frac{ip}{M}\right)$, evaluate $a^\dagger a$, noting also that $px - xp = h/i$. Then manipulate the result to see that the Hamiltonian $(p^2/(2M) + kx^2/2)$ is $\hbar\omega(a^\dagger a + 1/2)$.

          16.2 Harmonic-oscillator number operator
Using the definitions
$a^\dagger = \sqrt{\frac{M}{2\hbar\omega}} \left(\omega x - \frac{ip}{M}\right)$
and
$a = \sqrt{\frac{M}{2\hbar\omega}} \left(\omega x + \frac{ip}{M}\right)$, 
evaluate $a^\dagger a$, noting also that $px - xp = h/i$. Then manipulate the result to see that the
Hamiltonian $(p^2/(2M) + kx^2/2)$ is $\hbar\omega(a^\dagger a + 1/2)$.

16.2 Harmonic-oscillator number operator
Using the definitions
a^† = √((M)/(2ħω))(ω x - (ip)/(M))
and
a = √((M)/(2ħω))(ω x + (ip)/(M)),
evaluate a^† a, noting also that px - xp = h/i. Then manipulate the result to see that the
Hamiltonian (p^2/(2M) + kx^2/2) is ħω(a^† a + 1/2).

Added by Robert L.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Sri K

04/30/2022

Step 1

We have the definitions: a^(†) = \sqrt{\frac{M}{2\hbar\omega}}(\omega x - \frac{ip}{M}) a = \sqrt{\frac{M}{2\hbar\omega}}(\omega x + \frac{ip}{M}) Show more…

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16.2 Harmonic-oscillator number operator Using the definitions a^(†)=sqrt(((/bar (M)))/(2ℏomega ))(omega x-(ip)/(M)) and a=sqrt((M)/(2ℏomega ))(omega x+(ip)/(M)), evaluate a^(†)a, noting also that px-xp=(h)/(i). Then manipulate the result to see that the Hamiltonian ((p^(2))/(2M)+k(x^(2))/(2)) is homega (a^(†)a+(1)/(2)). 16.2 Harmonic-oscillator number operator Using the definitions M and M 2hw evaluate afa , noting also that px - xp = h/i. Then manipulate the result to see that the Hamiltonian (p2/(2M) + kx2/2 ) is hw(aa + 1/2)