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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.

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  • $\begingroup$ I am working on problem 2 part (b) of this physics.uci.edu/sites/default/files/Spring_2017_1.pdf. $\endgroup$ – InertialObserver Sep 4 '17 at 2:46
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    $\begingroup$ 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). $\endgroup$ – ZeroTheHero Sep 4 '17 at 2:53
  • $\begingroup$ What do you think that the correct answer to part (b) is supposed to look like then? $\endgroup$ – InertialObserver Sep 4 '17 at 2:59
  • $\begingroup$ 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? $\endgroup$ – Cosmas Zachos Sep 4 '17 at 16:26
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    $\begingroup$ 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. $\endgroup$ – ACuriousMind Sep 4 '17 at 17:16
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The other answer is correct but it looks like you might be having some trouble understanding it so let me give my take on it.

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.

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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 $$ aS_+\vert n+1, \downarrow\rangle = \sqrt{n+1}\vert n,\uparrow\rangle\, ,\qquad a^\dagger S_-\vert n\uparrow\rangle = \sqrt{n+1}\vert n+1,\downarrow\rangle\, . $$

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  • $\begingroup$ What do you mean by "connecting" two states?, and why can this hamiltonian only do it for those two specific states? Is there any literature you could refer me to that could explain what is going on here in this problem? $\endgroup$ – InertialObserver Sep 4 '17 at 3:19
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    $\begingroup$ 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 $\pm 1$... $\endgroup$ – ZeroTheHero Sep 4 '17 at 3:24

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