# Exact solution to 1D SHO + $(a S_+ + a^{\dagger}S_-)$ [closed]

Let $$H = \hbar\omega(a^{\dagger}a + \frac{1}{2}) + \frac{\delta}{2}(aS_+ + a^{\dagger}S_-)$$ be the hamiltonian of a spin-$\frac{1}{2}$ particle. I am asked to find the exact eigenstates of this hamiltonian. However, the problem statement says that it suffices to find eigenstates that turn to $|0,\uparrow\rangle$, and $|1, \downarrow\rangle$ as $\delta\rightarrow 0$.

My attempt

My first idea is to write everything in spin representation so that

\begin{align*} H &= \begin{pmatrix} \hbar\omega(a^{\dagger}a + \frac{1}{2}) & 0 \\ 0 & \hbar\omega(a^{\dagger}a + \frac{1}{2}) \end{pmatrix} + \frac{\hbar\delta}{2} \begin{pmatrix} 0 & a \\ a^{\dagger} & 0 \end{pmatrix}\\ &=\begin{pmatrix} \hbar\omega(a^{\dagger}a + \frac{1}{2})& \frac{\hbar\delta}{2}a\\ \frac{\hbar\delta}{2}a^{\dagger} & \hbar\omega(a^{\dagger}a + \frac{1}{2}) \end{pmatrix}. \end{align*}

However, I don't seem to be able to able to have any further insights. Any hints or links to solving problems of this kind would be greatly appreciated.

• I am working on problem 2 part (b) of this physics.uci.edu/sites/default/files/Spring_2017_1.pdf. – InertialObserver Sep 4 '17 at 2:46
• The question you are asking is not *exactly * part b). In part b) the statement is that the complete set of states is the set $\{\vert n\uparrow \rangle, \vert n\downarrow \rangle\}$. Clearly you Hamiltonian can only connect $\vert n\uparrow\rangle$ as one vector, and $\vert n+1,\downarrow\rangle$ as another vector, as per part c). – ZeroTheHero Sep 4 '17 at 2:53
• What do you think that the correct answer to part (b) is supposed to look like then? – InertialObserver Sep 4 '17 at 2:59
• You appear overwhelmed by symbols. @ZeroTheHero isolated the 2x2 block of the infinite-dimensional Hamiltonian in Fock/spin space of relevance to your problem. Define $\Delta=\delta/\omega$. How does the Hamiltonian act on the state $|0\uparrow\rangle+ (1-\sqrt{1+\Delta^2} )/\Delta |1\downarrow\rangle$? Is $\hbar \omega (2-\sqrt{1+\Delta^2})/2$ an eigenvalue? – Cosmas Zachos Sep 4 '17 at 16:26
• Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. – ACuriousMind Sep 4 '17 at 17:16

Written a bit differently one has (implicitly summing over $n, s$) $$H = \hbar\omega n|ns\rangle\langle n s| + \frac\delta2 \sqrt{n+1}~\Big( |n\uparrow\rangle\langle (n+1) \downarrow| + |(n+1)\downarrow\rangle\langle n \uparrow|\Big).$$ Now suppose you have a state $|\Psi\rangle = \sum_{ns} c_{ns} |ns\rangle$ such that $H |\Psi\rangle = E |\Psi\rangle.$ Operating on this from the left with $\langle m\uparrow|$ gives:$$E c_{m\uparrow} = \hbar\omega m c_{m\uparrow} + \frac\delta2 \sqrt{m+1}~c_{(m+1)\downarrow}$$and we see we need a formula for the $(m+1)\downarrow$ component so we act on this with $\langle (m+1) \downarrow|$ to give,$$E c_{(m+1)\downarrow} = \hbar\omega (m+1)c_{(m+1)\downarrow} + \frac{\delta}{2}\sqrt{m+1} ~c_{m\uparrow}.$$ We find that we've actually only got two equations in two unknowns, $c_{m\uparrow}$ and $c_{(m+1)\uparrow},$ rather than an interesting recurrence. In fact solving the latter for $c_{(m+1)\downarrow}$ and substituting into the former we find that the $c_{m\uparrow}$ divides away and we're left with a complicated expression for $E$ in terms of $m$. This should strike you as "well that's really odd" -- the equations are pushing back at you to tell you that with this Hamiltonian, your eigenvectors actually have a really simple form, $$|\Psi_n\rangle = \alpha_n|n\uparrow\rangle + \beta_n |(n+1)\downarrow\rangle,$$with $\alpha_0 = 0,\beta_0=1$ though there is I suppose a "bonus" $|\Psi_{-1}\rangle = |0\downarrow\rangle.$ (Note that $V |0\downarrow\rangle = 0$ because $S_- |\downarrow\rangle = a |0\rangle = 0.$)
The other answer is saying that once you observe this, you see that your Hamiltonian acting on these $(\alpha_n, \beta_n)$ vectors is $$H_n = \begin{bmatrix}\hbar\omega(n + 1/2)&(\delta/2)\sqrt{n+1}\\ (\delta/2)\sqrt{n+1}&\hbar\omega(n+3/2)\end{bmatrix},$$which is precisely the form that they give you an explicit solution for.
Your Hamiltonian can only connect $\vert n\uparrow\rangle$ and $\vert n+1,\downarrow\rangle$. If you work with these two basis states then your Hamiltonian will be
\begin{align*} H &= \begin{pmatrix} \hbar\omega(n + \frac{1}{2}) & 0 \\ 0 & \hbar\omega(n+1 + \frac{1}{2}) \end{pmatrix} + \frac{\hbar\delta}{2} \begin{pmatrix} 0 & \sqrt{n+1} \\ \sqrt{n+1} & 0 \end{pmatrix} \end{align*} (or something like this). The part in $a^\dagger a$ is clear, but the other part goes like $$aS_+\vert n+1, \downarrow\rangle = \sqrt{n+1}\vert n,\uparrow\rangle\, ,\qquad a^\dagger S_-\vert n\uparrow\rangle = \sqrt{n+1}\vert n+1,\downarrow\rangle\, .$$
• well... $a$ and $a^\dagger$ must connect state that differ by $\Delta n=\pm 1$, and the spin operators basically flip the spin up to down and down to up, so try working the matrix elements of your perturbation between states $\vert n\uparrow\rangle$ and $\vert n+2,\downarrow\rangle$... you get $0$ since the creation and destruction operators kick you up $\pm 1$... – ZeroTheHero Sep 4 '17 at 3:24