Past exam of the quantum theory and quantum computers course of the University of Oxford / 2011 / 1

In the scalar two-particle model, the fermionic exchange symmetry and bosonic exchange symmetry conditions areFor particles with spin, exchange acts on all their labels; the stated coordinate-only model fixes the spatial symmetry sector. Define $S\Psi (x_1,x_2)=\Psi (x_2,x_1)$. Equal masses make the kinetic operator invariant under exchange, and the assumed potential is invariant too, so $HS=SH$. Therefore $H\Psi =E\Psi$ implies $H(S\Psi )=E(S\Psi )$. The componentsare respectively symmetric and antisymmetric eigenfunctions whenever nonzero. Their sum is the original nonzero quantum state, so at least one is nonzero. An eigenspace may contain both sectors if the energy is degenerate.

The normal modes of two harmonically coupled identical particles provide the separation. The coordinate transformation to $y_{+},y_{-}$ is orthogonal, with absolute Jacobian one. Exchanging the particles leaves $y_{+}$ fixed and changes $y_{-}$ to $-y_{-}$. Hence bosonic quantum states are even and fermionic quantum states odd in the relative coordinate. The chain rule givesAlso $x_1^2+x_2^2=y_{+}^2+y_{-}^2$ and $(x_1-x_2{)}^2=2y_{-}^2$. Combining the two trapping potentials with the interaction therefore gives the separated Hamiltonian operatorThese are independent oscillators of frequencies $\omega$ and $\omega \cos \alpha$. Both frequencies are positive in the stated range. Their product eigenstates have energiesThe parity of an oscillator level is $(-1{)}^n$. Consequently bosons allow even $n_{-}$ and fermions odd $n_{-}$, while $n_{+}$ is unrestricted.

The bosonic ground state uses $n_{+}=n_{-}=0$, and the fermionic ground state uses $n_{+}=0,n_{-}=1$. The oscillator length in the relative mode isUsing the normalized Gaussian for each mode givesIndeed ${\varphi }_{\tau }(y)=\sqrt{2}y{\psi }_{\tau }(y)/\tau$ by differentiating the Gaussian, so it is the normalized odd first excited state. The ground energies are $E_{\mathrm{B}}=\hslash \omega (1+\cos \alpha )/2$ and $E_{\mathrm{F}}=\hslash \omega (1+3\cos \alpha )/2$.

Choose normalized factors for a normalized product quantum state; this can always be done by reciprocal rescaling of its factors. Since the coordinate Jacobian has absolute value one,If the supplied factors are not individually normalized, the general expression has an additional factor $\Vert \psi {\Vert }^2$; choosing normalized factors is the convention in the displayed formula.

For the ground states, put $\rho (y)=|{\psi }_{\tau }(y){|}^2=e^{-y^2/{\tau }^2}/(\sqrt{\pi }\tau )$. Then $|{\varphi }_{\tau }(y){|}^2=2y^2\rho (y)/{\tau }^2$ and ${\rho }^{\prime }=-2y\rho /{\tau }^2$. Thus, for differentiable $f$ with suitable Gaussian integrability and vanishing boundary terms,This is integration by parts, and every polynomial satisfies the required endpoint conditions. The relative-coordinate moments in even and odd oscillator ground states follow by substituting $f(y)=y^n$:For odd $n$, both expectations vanish by parity, so this is an equality of zeros rather than an increase in signed mean separation. The even moments show the physical difference. In particular ${\mathbb{E}}_{\mathrm{B}}[y_{-}^2]={\tau }^2/2$ and ${\mathbb{E}}_{\mathrm{F}}[y_{-}^2]=3{\tau }^2/2$, so the fermionic root-mean-square particle separation is $\sqrt{3}$ times the bosonic value. The fermionic probability density vanishes at coincidence $y_{-}=0$, expressing Pauli exclusion principle in this spatial symmetry sector. This is an exchange constraint on the wave function, not an additional classical force. As $\alpha$ approaches $\pi /2$, the relative oscillator softens and its width diverges; the excluded endpoint would no longer have this normalizable oscillator ground state.