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A particle of mass \(m\) in potential
\(V(x,y)=\frac{1}{2}m\omega^2(4x^2 + y^2)\) is in the energy eigen-state
with \(E=\frac{5}{2}\hbar \omega\), then is eigenfunction will be
- \(y \exp\Big(-\frac{m\omega}{2\hbar}(2x^2+y^2)\Big)\)
- \(x \exp\Big(-\frac{m\omega}{2\hbar}(2x^2+y^2)\Big)\)
- \(y \exp\Big(-\frac{m\omega}{2\hbar}(x^2+y^2)\Big)\)
- \(xy \exp\Big(-\frac{m\omega}{2\hbar}(x^2+y^2)\Big)\)
RECALL IMPORTANT FACTS
For a harmonic oscilaltor in one dimension
\( H = \frac{p^2}{2m} + V(x,y) = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2\)
\(E =(n+1/2) \hbar \omega\)
\( u_n(x) = H_n(x) \exp\Big(\frac{m\omega}{2\hbar} x^2\Big)
NOTE IMPORTANT POINTS
- Two dimensional oscillator
- Frequencies \(\omega_1=2\omega, \qquad \omega_2=\omega\),
- Total energy \(E=(n_1+1/2) \hbar \omega_1 + (n_2+1/2)\hbar \omega2 = \hbar\omega(2n_1+n_2 +3/2)\)Therefore (2n_1+n2)=1 This has only one solution in integers \(n_1=0, n_2=1\)
- Hermite polynomial \(H_1(y)\) will give \(y\) factor no \(x\) factor for \(H_0(x)\) ,So choice is between Option 1 and Option3.
- The \(x\) part of the exponential factor is correct in Option (a) because we need \((m\omega_1 /2\hbar) x^2\).
ANSWER
Option (a)
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