QM M00 1

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Consider a spin-\frac{1}{2} particle constrained to move on a 1D line with a harmonic oscillator potential and a magnetic field so that the Hamiltonian is:

H=\frac{1}{2m}p^{2}+\frac{1}{2}m\omega^{2}x^{2}+\omega S_{z}

The first energy level is not degenerate but all the other levels are doubly degenerate.

Now add a small magnetic field in the \hat{x} direction with a magnitude proportional to x . The Hamiltonian is:

H=\frac{1}{2m}p^{2}+\frac{1}{2}m\omega^{2}x^{2}+\omega S_{z}+\alpha xS_{x}

Calculate the energy difference in the levels to lowest order.

Define the original the Hamiltonian:

H_{0}=\frac{1}{2m}p^{2}+\frac{1}{2}m\omega^{2}x^{2}+\omega S_{z}

And add the perturbation:

H^{\prime}=\alpha xS_{x}

We define the basis of eigenstates of H0 to be |n_{k},\pm\rangle , where nk is the quantum number for the kinetic energy, and \pm defines either the + or - spin state. The energy from the state is n\hbar\omega.

Since we know:

S_{x}=\frac{\hbar}{2}\left(\begin{matrix} 0 &amp; 1\\ 1 &amp; 0\end{matrix}\right);\qquad S_{z}=\frac{\hbar}{2}\left(\begin{matrix} 1 &amp; 0\\ 0 &amp; -1\end{matrix}\right)

so that the properties of Sx are:

S_{x}|+\rangle_{z}=\frac{\hbar}{2}|-\rangle_{z};\qquad S_{x}|-\rangle_{z}=\frac{\hbar}{2}|+\rangle_{z}

We can also decompose the x operator:

\hat{x}=\frac{1}{\sqrt{2}}\left(a^{\dagger}+a\right)\sqrt{\frac{\hbar}{m\omega}}

where a and a^{\dagger} are the annhiliation and creation operators. The energy perturbation to first order is given by:

E_{n}\left(\lambda\right)=E_{n}^{0}+\sum_{m}\left\langle \varphi_{m}\right|H^{\prime}\left|\varphi_{m}\right\rangle

Where we must diagonalize the states:

\begin{matrix} \left|\varphi_{m+}\right\rangle &amp; = &amp; \frac{1}{\sqrt{2}}\left(\left|m-1,+\right\rangle + \left|m,-\right\rangle\right)\\ \left|\varphi_{m-}\right\rangle &amp; = &amp; \frac{1}{\sqrt{2}}\left(\left|m-1,+\right\rangle - \left|m,-\right\rangle\right) \end{matrix}

So the change in energy is given by:

\Delta E_{m}\left(\lambda\right)=\left\langle \varphi_{m}\right|H^{\prime}\left|\varphi_{m}\right\rangle

Now it simply remains to calculate the values of \left\langle \varphi_{m\pm}\right|H^{\prime}\left|\varphi_{m\pm}\right\rangle:

\left\langle \varphi_{m\pm}\right|H^{\prime}\left|\varphi_{m\pm}\right\rangle =\left\langle \varphi_{m\pm}\right|\alpha\frac{1}{\sqrt{2}}\left(a^{\dagger}+a\right)\sqrt{\frac{\hbar}{m\omega}}S_{x}\left|\varphi_{m\pm}\right\rangle

So that for \left|\varphi_{m+}\right\rangle:

\frac{1}{\sqrt{2}}\left(\left\langle m-1,+\right| + \left\langle m,-\right|\right) \alpha\frac{1}{\sqrt{2}}\left(a^{\dagger}+a\right)\sqrt{\frac{\hbar}{m\omega}}S_{x} \frac{1}{\sqrt{2}}\left(\left|m-1,+\right\rangle + \left|m,-\right\rangle\right) =\alpha\sqrt{\frac{\hbar}{2m\omega}}\frac{\hbar}{2}\sqrt{n}

And for \left|\varphi_{m-}\right\rangle:

\frac{1}{\sqrt{2}}\left(\left\langle m-1,+\right| - \left\langle m,-\right|\right) \alpha\frac{1}{\sqrt{2}}\left(a^{\dagger}+a\right)\sqrt{\frac{\hbar}{m\omega}}S_{x} \frac{1}{\sqrt{2}}\left(\left|m-1,+\right\rangle - \left|m,-\right\rangle\right) =-\alpha\sqrt{\frac{\hbar}{2m\omega}}\frac{\hbar}{2}\sqrt{n}

So that the total energy difference is just given by:

\Delta E\left(\left|\varphi_{m+}\right\rangle \right)=\sqrt{\frac{\hbar^{3}\alpha^{2}n}{8m\omega}}; \qquad \Delta E\left(\left|\varphi_{m-}\right\rangle \right)=-\sqrt{\frac{\hbar^{3}\alpha^{2}n}{8m\omega}}

We must treat the ground state separately, since it is nondegenerate. We must take it out to second order:

\Delta E\left(\left|0\right\rangle \right)=\sum_{p\neq0}\frac{\left|\left\langle \varphi_{p}^{i}\right|H^{\prime}\left|0\right\rangle \right|^{2}}{-E_{p}^{0}}

Finding the nonzero values of this, we get:

\left\langle 1,+\right|H^{\prime}\left|0,-\right\rangle =\left\langle 1,-\right|\alpha\frac{1}{\sqrt{2}}\left(a^{\dagger}+a\right)\sqrt{\frac{\hbar}{m\omega}}S_{x}\left|0,-\right\rangle =\alpha\sqrt{\frac{\hbar}{2m\omega}}\frac{\hbar}{2}

as the only one. Thus:

\Delta E\left(\left|0\right\rangle \right)=\frac{\left|\alpha\sqrt{\frac{\hbar}{2m\omega}}\frac{\hbar}{2}\right|^{2}}{-\hbar\omega}=-\alpha^{2}\frac{\hbar^{2}}{8m\omega^{2}}

At last we get:

\begin{matrix}\Delta E_{n} &amp; = &amp; -\alpha^{2}\frac{\hbar^{2}}{8m\omega^{2}};\qquad n=0 \\ &amp; = &amp; \left\{\begin{matrix}\alpha\sqrt{\frac{\hbar^{3}n}{8m\omega}}\\ -\alpha\sqrt{\frac{\hbar^{3}n}{8m\omega}} \end{matrix} \right.&nbsp;;\qquad n&gt;0 \end{matrix}

where the pair for n > 0 indicates the splitting of a degenerate pair.

This page was recovered in October 2009 from the Plasmagicians page on Prelim_M00_QM1 dated 22:28, 23 December 2005.

Retrieved from "https://qed.princeton.edu/main/PlasmaWiki/Old_Prelims/QM_M00_1"
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