A particle moves in a potential \(V(x)=\begin{cases}\frac{1}{2}m\omega^2x^2 & \text{for } x\gt 0\\ \infty & \text{for } x\le 0\end{cases}\) Using the WKB quantization condition \(\frac{1}{\hbar}\int_{x_1=0}^{x_2}p(x)\,dx+\frac{\pi}{4}=n\pi\) where n is a positive integer, \(p(x)=\sqrt{2m\left(E-V(x)\right)}\) and x1 = 0 and x2 are the classical turning points, the ground state energy of the particle is given by
\(\left(\frac{3}{2}\right)\hbar\omega\)
The outer turning point is where E = V(x2): \(x_2=\sqrt{\frac{2E}{m\omega^2}}\).
The phase integral over the quarter-ellipse in phase space is
\(\int_0^{x_2}\sqrt{2m\left(E-\tfrac{1}{2}m\omega^2x^2\right)}\,dx=\frac{\pi}{4}\,x_2\,\sqrt{2mE}=\frac{\pi}{4}\sqrt{\frac{2E}{m\omega^2}}\sqrt{2mE}=\frac{\pi E}{2\omega}\)
Substituting into the given condition: \(\frac{\pi E}{2\hbar\omega}+\frac{\pi}{4}=n\pi\;\Rightarrow\;E_n=\left(2n-\frac{1}{2}\right)\hbar\omega\), n = 1, 2, 3, …
The ground state corresponds to n = 1: \(E_1=\left(2-\frac{1}{2}\right)\hbar\omega=\frac{3}{2}\hbar\omega\).
This agrees with the exact result: the infinite wall at x = 0 forces ψ(0) = 0, so only the odd oscillator states (3/2, 7/2, … ħω) survive; the lowest of them is (3/2)ħω, not the full-oscillator value (1/2)ħω.
Hence, the answer is \(\left(\frac{3}{2}\right)\hbar\omega\).
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.................... is a client server application that identifies each host on the internet with a unique user-friendly name.
Every OS has a ................ or .......... which primarily resides in the main memory of the computer to perform some of the basic functions of the OS.
(i) Kernel
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(iii) Nucleus
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