User:Guy vandegrift/btvs/Effort 2

Calculation The dotted line denotes equilibrium. The lowest and highest modes interact the the central "atom"

To keep the prerequisite level of this calculation as low as possible, we avoid quantum field theory and issues involving the creation of new particles by considering three weakly coupled oscillators as shown in the figure. All three oscillators have the same frequency, but the central miss is double that of the other two, implying it has twice the spring constant. While this system has three normal modes, we are primarily concerned with the two that are symmegric about the central mass (modes 1 and 3). The coupling can be modeled as weak springs connecting the masses, or string of low tension.

Further simplification is achieved by avoiding perturbation theory by finding an exact solution by decomposing the classical system into normal modes, and also by avoiding the canonical transformations associated with classical Hamiltonian mechanics. Instead we write Schrödinger's equation for each mass using the potential energy stored in the springs and connecting string. After that, a change of variables is used to rewrite the partial differential equation into separable form. This change of variables is inspired by writing Newton's laws of motion for the three masses in matrix form. In this way, we produce an exact solution accessible to students who possess only a knowledge of matrix algebra and the linear partial differential equation associated with the simple harmonic oscillator. Although an exact solution is found, creation and destruction operators are used to construct an approximate solution that resembles three uncoupled oscillators.

At time, t=0, the central mass is placed in the second excited state, and it is shown that after that there is a non-zero probability of finding each of the outer masses in the first excited state. There is also a finite probability that only one of the outer masses is in the second excited state. But if the outer masses are observed to be both excited, and if the simple harminic oscillators are allowed to vibrate in two perpendicular directions, then the states are entangled in a way that permits Bell's inequality to be violated. To verify that this inequality is violated, it is neccessary to repeat the experiment many times, recording the data only on those occassions when each of the outer mass is observed to be in the first excieted state.

Isolated two dimensional harmonic oscillator

References:caltech unm

We begin by neglecting the forces caused by the horizontal connecting springs. This is analogous by ignoring radiation effects whey finding the energy states of a single atom (the two outer masses vaguely resemble the quantized field). For our puposes, we need consider only the three lowest energy levels:

For a harmonic oscillator with mass, $m$ and natural frequency, $\omega$ , we difine,

$\alpha ={\frac {m\omega }{\hbar }}$ ,

and write the first three energy levels as:

$u_{0}(x)=\left({\frac {\alpha }{\pi }}\right)^{1/4}e^{-\alpha x^{2}/2}$ $u_{1}(x)=\left({\frac {4\alpha ^{3}}{\pi }}\right)^{1/4}xe^{-\alpha x^{2}/2}$ $u_{2}(x)=\left({\frac {\alpha }{4\pi }}\right)^{1/4}\left(2\alpha x^{2}-1\right)e^{-\alpha x^{2}/2}$ Useful for the calculations in the appendix is:

$u_{n}(x)={\frac {(a^{\dagger })^{n}}{\sqrt {n!}}}u_{0}(x)$ ,

where the "creation operator" can be used to generate,

$a^{\dagger }={\sqrt {m\omega \over 2\hbar }}\left(x-{i \over m\omega }{\hat {p}}\right)={\sqrt {m\omega \over 2\hbar }}\left(x+{\hbar \over m\omega }{\frac {\partial }{\partial x}}\right)$ With two dimensions, we may write solutions as composites of the nth excited state in the x-direction and the mth excited state in the y-direction. Since there are two dimensions for each of the three masses, each mass has a ground state on (½

Separability of a collection of independent simple harmonic oscillators

If we neglect the interconnecting strings, we have a collection of isolated simple harmonic oscillators, and Schrödinger's equation is:

${\frac {\hbar }{i}}{\frac {\partial \Psi }{\partial t}}=-\sum _{j=1}^{6}{\frac {\hbar ^{2}}{2m_{j}}}{\frac {\partial ^{2}\Psi }{\partial \xi _{j}^{2}}}+{\frac {1}{2}}\sum _{j=1}^{6}k_{j}\xi _{j}^{2}\,\Psi$ The ground state of a two dimensional oscillator

Leting ${\vec {r}}=x{\hat {x}}+y{\hat {y}}$ be the two dimensional coordinates, we have:

$u_{0}({\vec {r}})=u_{0}(x)u_{0}(y)=\left({\frac {\alpha }{\pi }}\right)^{1/2}e^{-\alpha r^{2}/2}$ This state is invariant under rotations about the z-axis, as well as parity (i.e., when both x and y are replaced by −x and −y).

Triple degeneracy in the second excited state

$u_{1}({\vec {r}})=C_{xx}\,u_{2}(x)u_{0}(y)+C_{yy}\,u_{0}(x)u_{2}(y)+C_{xy}\,u_{1}(x)u_{1}(y)$ Or setting $\alpha =1$ and ignoring the normalizaiton factors:

$u_{2}(x)u_{0}(y)\propto (2x^{2}-1)e^{-r^{2}/2}$ $u_{0}(x)u_{2}(y)\propto (2y^{2}-1)e^{-r^{2}/2}$ $u_{1}(x)u_{1}(y)\propto xye^{-r^{2}/2}$ States that are symmetric about the z axis

The wavefunctions shown above for basis vectors for our wavefunction. But since our Bell's theorem thought experiment places the initial in a highly symmetric state, we can reduce the number of required functions. One such symmetry is with respect to rotations about the z-axis. This is accomplished by having the initial state be either the ground state (with zero photons), or with two photons in what is analogous to the 2S state of the hydrogen atom. Setting $C_{xy}=0$ and $C_{xx}=C_{yy}=1/{\sqrt {2}}$ creates the normalized wavefunction that achieves this feat:

$u_{2}(x)u_{0}(y)+u_{0}(x)u_{2}(y)\propto (2r^{2}-2)e^{-r^{2}/2}\propto (r^{2}-1)e^{-r^{2}/2}$ See page 5 of File:MIT 2D harmonic oscillator.pdf or page 11 of File:2DharmonicOscillator.pdf.

${\frac {u_{2}(x)u_{0}(y)+u_{0}(x)u_{2}(y)}{\sqrt {2}}}={\sqrt {\frac {\alpha }{\pi }}}(\alpha r^{2}-1)e^{-\alpha r^{2}/2}$ Double degeneracy first excited state

The first excited state can be a superposition of two independent states:

$u_{1}({\vec {r}})=C_{x}\,u_{1}(x)u_{0}(y)+C_{y}\,u_{0}(x)u_{1}(y)$ To visualize this, take $\alpha =1$ , and ignore the normalization constants:

$u_{1}(x)u_{0}(y)\propto xe^{-(x^{2}+y^{2})/2}=r\cos \theta e^{-r^{2}/2}$ $u_{0}(x)u_{1}(y)\propto ye^{-(x^{2}+y^{2})/2}=r\sin \theta e^{-r^{2}/2}$ This can be viewed as a single photon "in" the isolated two dimensional harmonic oscillator. Polarization can be linear, circular, or elliptical. Normalization demands that

$|C_{x}|^{2}+|C_{y}|^{2}=1.$ Rotational invariance for a pair of oscillators in the first excited state

Neither $u_{1}(x)u_{0}(y)$ nor $u_{0}(x)u_{1}(y)$ are invariant under rotations, i.e., where $(x,y)$ are expressed in rotated coordinates $(x',y')$ :

${\begin{pmatrix}x'\\y'\end{pmatrix}}={\begin{pmatrix}\cos \theta &\sin \theta \\-\sin \theta &\cos \theta \end{pmatrix}}\cdot {\begin{pmatrix}x\\y\end{pmatrix}}$ However a pair of oscillators can achieve such in variance. For reasons that will become apparent later, we shall label this pair of oscillators with the subscripts one and three: $(x_{1},y_{1})$ and $(x_{3},y_{3})$ . The seperable nature of this system permits us to write solutions as products. It is easy to show that,

$u_{1}(x_{1})u_{0}(y_{1})\cdot u_{0}(x_{3})u_{1}(y_{3})+u_{0}(x_{1})u_{0}(y_{1})\cdot u_{1}(x_{3})u_{0}(y_{3})$ is invariant because

$x_{1}y_{3}+y_{1}x_{3}=(x'_{1}\cos \theta +y'_{1}\sin \theta )(-x'_{3}\sin \theta +y'_{3}\cos \theta )+(-x'_{1}\sin \theta +y'_{1}\cos \theta )(x'_{3}\cos \theta +y'_{3}\sin \theta )$ 