# Amplitude Damping of a Harmonic Oscillator

Exercise 8.21 of Nielsen and Chuang asks us to show that the operation elements for a harmonic oscillator (system) coupled to another harmonic oscillator (environment) is

$$E_k = \sum_n \sqrt{(^n_k)}\sqrt{(1-\gamma)^{n-k}\gamma^k} |n-k\rangle\langle n|$$ (1)

with $$\gamma = 1- cos^2(\chi\Delta t)$$

The Hamiltonian is $$H = \chi(a^\dagger b+b^\dagger a)$$ (2)

$$E_k$$ should be found using $$E_k = \langle k_b|U|0_b\rangle$$ where subscript b denotes environment, and $$U = e^{-iH\Delta t}$$

This question has been asked here but has not been answered. The steps that I know:

Using $$\langle k_b| = \langle 0_b| \frac{b^k}{\sqrt{k!}}$$ (3)

$$E_k = \langle k_b|e^{-i\chi(a^\dagger b+b^\dagger a)\Delta t}|0_b\rangle = \langle 0_b|\frac{b^k}{\sqrt{k!}} e^{-i\chi(a^\dagger b+b^\dagger a)\Delta t}|0_b\rangle$$

$$= \langle 0_b|\frac{b^k}{\sqrt{k!}} \sum_n \frac{[-i\chi \Delta t(a^\dagger b+b^\dagger a)]^n}{n!} |0_b\rangle$$

$$= \langle 0_b| \sum_n \sum_{k=0}^n \frac{b^k}{\sqrt{k!}}\frac{[-i\chi \Delta t(a^\dagger b+b^\dagger a)]^{n-k} [-i\chi \Delta t(a^\dagger b+b^\dagger a)]^k}{n!} |0_b\rangle$$

$$= \langle 0_b| \sum_n \sum_{k=0}^n \frac{1}{\sqrt{k!}}\frac{[-i\chi \Delta t(a^\dagger b+b^\dagger a)]^{n-k} [-i\chi \Delta t(a^\dagger b^2+bb^\dagger a)]^k}{n!} |0_b\rangle$$ (4)

Considering $$b|0_b\rangle = 0$$ the above becomes

$$= \langle 0_b| \sum_n \sum_{k=0}^n \frac{1}{\sqrt{k!}}\frac{[-i\chi \Delta t(b^\dagger a)]^{n-k} [-i\chi \Delta t(bb^\dagger a)]^k}{n!} |0_b\rangle$$

Using $$[b,b^\dagger] = bb^\dagger - b^\dagger b = 1$$, and $$bb^\dagger = 1+ b^\dagger b$$ the above becomes

$$= \langle 0_b| \sum_n \sum_{k=0}^n \frac{1}{\sqrt{k!}}\frac{[-i\chi \Delta t(b^\dagger a)]^{n-k} [-i\chi \Delta ta]^k}{n!} |0_b\rangle$$ (5)

The Binomial expansion is

$$(A + B)^n = \sum_{k=0}^n (^n_k) A^{n-k}B^k$$

To make notation simpler, setting $$A = -i\chi \Delta t(b^\dagger a)$$ and $$B = -i\chi \Delta ta$$

(5) becomes

$$\langle 0_b| \frac{1}{\sqrt{k!}}\sum_n \frac{1}{n!(^n_k)} (A+B)^n |0_b\rangle$$ (6).

A simple calculation gives $$A+B = (-i\chi \Delta t)^n (b^\dagger +1)^n a^n$$

So (6) is

$$\langle 0_b| \frac{1}{\sqrt{k!}}\sum_n \frac{1}{n!(^n_k)} [(-i\chi \Delta t) (b^\dagger +1) a]^n |0_b\rangle$$ (7)

At this point we should trace out the environment/b terms. I understand that we can write (7) as an exponential again, and get sines and cosine terms; but they will not be squared, as required by (1). I'd appreciate any help with this. Thanks.

• Notice that when expanding the binomial, the terms where $a^\dagger b$ is on the rightmost part of the expression will give $0$ because $b | 0_b \rangle = 0$. Also check commutator before doing binomial formula. – AHusain Jul 21 '19 at 20:33
• @AHusain Thanks. I did some edits. I'm still stuck at (7) though... – Bashir Jul 21 '19 at 23:18
• If we let (5) act on a combined system-environment state: $\langle 0_b| \sum_n \sum_{k=0}^n \frac{1}{\sqrt{k!}}\frac{[-i\chi \Delta t(b^\dagger a)]^{n-k} [-i\chi \Delta ta]^k}{n!} |0_bn\rangle$, we can use $(b^\dagger)^{n-k} |0_b\rangle = \sqrt{(n-k)!}|n-k\rangle$ and $a^n |n\rangle = \sqrt{n!}$ ... – Bashir Jul 22 '19 at 1:10

## 2 Answers

The way that I approach the maths is very different from the way that you do it (I imagine it can be approached as you're trying, it's just not how I think about it).

I start by introducing $$|n,m\rangle$$ as a basis denoting the excitation states of $$a,b$$ respectively. We recognise that $$H$$ preserves $$n+m$$, so the whole thing decomposes into a bunch of subspaces determined by the total $$n+m$$. This has several consequences, but for one, it means that what we want to calculate is the amplitude $$\langle n-k,k|e^{-iH\Delta t}|n,0\rangle.$$ So, we can focus specifically on the $$n^{th}$$ subspace, which is described by an $$(n+1)\times (n+1)$$ matrix of the form $$H_n=\chi\sum_{i=0}^{n-1}\sqrt{(i+1)(n-i)}(|i,n-i\rangle\langle i+1,n-i-1|+|i+1,n-i-1\rangle\langle i,n-i|).$$ These are matrices you might recognise, being the spin-$$n/2$$ rotation matrix $$J_x$$. Or, as in my case, you recognise it from perfect state transfer. Depending on where you recognise it from, and what level of detail you want the proof, it might be acceptable to quote the result $$\langle n-k,k|e^{-iH\Delta t}|n,0\rangle=\sqrt{\binom{n}{k}}\cos^{n-k}(\chi\Delta t)\sin^{k}(\chi\Delta).$$ (I'm doing this from memory. I have the feeling there are suppose to be some phase factors like $$e^{i\pi k/2}$$.) Or you might take the eigenvalues/eigenvectors of the matrix and prove it. I know they can be found here, but require manipulation of hypergeometric functions.

Alternatively, there is a proof that doesn't require any hard maths, it's just a bit annoying. Consider a system of $$n$$ qubits, with a Hamiltonian $$H'=\chi\sum_{i=1}^nX_i,$$ and let $$|\psi_i\rangle$$ be a uniform superposition of all states where $$i$$ of those qubits are in the $$|1\rangle$$ state, with the rest in $$|0\rangle$$. You can show that $$H'|\psi_i\rangle=\chi\sqrt{i(n-i+1)}|\psi_{i-1}\rangle+\chi\sqrt{(i+1)(n-i)}|\psi_{i+1}\rangle,$$ Which is entirely equivalent to $$H_n$$ acting on $$|i,n-i\rangle$$. So, if we can work out the evolution of $$H'$$ acting on $$|0\rangle^{\otimes n}$$, evolving to $$|\psi_k\rangle$$, we're done! But this evolution is easy because it's actually just $$n$$ independent qubits, so $$e^{-iH'\Delta t}|0\rangle^{\otimes n}=\left((\cos(\chi\Delta t)|0\rangle-i\sin(\chi\Delta t)\right)^{\otimes n}.$$ We can rewrite this as $$\sum_{i=0}^n(-i)^k\sqrt{\binom{n}{k}}\cos^{n-i}(\chi\Delta t)\sin^i(\chi\Delta t)|\psi_i\rangle.$$ The origin of the binomial term is the normalisation of the state $$|\psi_i\rangle$$.

• Thanks for the answer. I still need to wrap my head around it. In your second last step, you get $cos^{n-i}(\chi \Delta t)$ and $sine^i(\chi \Delta t)$ because $e^{-iH^\prime \Delta t}$ acts on $|0\rangle ^{\otimes n}$. Do we get the same using $\langle 0_b| \sum_n \sum_{k=0}^n \frac{1}{\sqrt{k!}}\frac{[-i\chi \Delta t(b^\dagger a)]^{n-k} [-i\chi \Delta ta]^k}{n!} |0_b\rangle$ acting on $|0\rangle ^{\otimes n}$? This is basically what my question boils down to: Where do we get the powers for the cosine and sine terms. Your answer explains that, but I want to make sure I understand. – Bashir Jul 22 '19 at 15:44
• $\langle 0_b| \sum_n \sum_{k=0}^n \frac{1}{\sqrt{k!}}\frac{[-i\chi \Delta t(b^\dagger a)]^{n-k} [-i\chi \Delta ta]^k}{n!} |0_b\rangle$ can be written as $\langle 0_b| \sum_n \frac{(-i\chi \Delta t)^n}{n!} \sum_{k=0}^n \frac{1}{\sqrt{k!}} (b^\dagger a)^{n-k} a^k |0_b\rangle$ = $(cos(\chi \Delta t) -isin(\chi \Delta t))\sum_{k=0}^n \frac{1}{\sqrt{k!}} (b^\dagger a)^{n-k} a^k |0_b\rangle$ – Bashir Jul 22 '19 at 15:49
• For further reference, on how two harmonic oscillators give a spin system, the keyword is Schwinger boson. Though you might see it used in larger systems in your preliminary search. You should find a simpler reference in the results as well. – AHusain Jul 22 '19 at 19:12
• @AHusain Thank you! I'm useless at the names of things. – DaftWullie Jul 23 '19 at 6:32
• @Bashir I didn't see a route through the sort of manipulations that you were doing that would give the right sort of answer, which is why I went in the direction I did. That said, I also didn't really see how you were getting some of your expressions. You seemed to be assuming that the operator $b$ commutes with terms, but it doesn't: $b^\dagger b$ is not the same as $bb^\dagger$, for example. – DaftWullie Jul 23 '19 at 6:36

It's been a while since the question was posted but since I have a solution that differs from the one above, I would like to put it forth. It's mathematically gory but satisfactory in my opinion. We start by considering the state $$|n, m\rangle$$ as some basis state in the combined excited state space of $$a$$ and $$b$$ respectively.

We realize that $$H$$ and consequently $$\ e^{-iH\Delta t}$$ ultimately results in a combination of states that preserve $$n+m$$. Once we make this realization, we can say that the non zero terms of the inner product $$\langle k_b|U|0_b\rangle$$ are the following:

$$\begin{equation} E_k = \langle k_b|U|0_b\rangle =\sum_{n}\langle n-k_a, k_b|U|n_a,0_b\rangle\space|n-k_a\rangle\langle n_a| \end{equation} \tag{1}$$

Consider $$\langle n-k_a, k_b|U|n_a,0_b\rangle\space$$. Using the properties of the creation and annihilation operators, we get this to be

$$\begin{equation}= \langle0_a, 0_b|\frac{a^{n-k}}{\sqrt{n-k!}}\frac{b^{k}}{\sqrt{k!}}U\frac{{a^{\dagger}}^n}{\sqrt{n!}}|0_a, 0_b\rangle\end{equation}$$

Since $$U$$ is unitary and $$U^{\dagger}U = I$$, the above expression becomes,

$$\begin{equation}=\frac{\sqrt{\binom{n}{k}}}{n!}\langle0_a, 0_b|a^{n-k}b^{k}U{a^{\dagger}}^nU^{\dagger}U|0_a, 0_b\rangle\tag{2}\end{equation}$$

Now, $$U|0_a, 0_b\rangle = |0_a, 0_b \rangle$$ since if no photon was at the input mode, none can be found at the output mode. This is not a claim we make out of thin air. It does make sense and in case one is curious, similar justifications also exist in section 7.4.2 of Nielsen and Chuang and in chapter 6 of Introductory Quantum Optics by Christopher C. Gerry and Peter L. Knight.

Moving on, consider $$U a^{\dagger}U^{\dagger}$$. We can simplify this using the following version of the Baker-Campbell-Hausdorff formula which states that,

$$\begin{equation}e^{\lambda G}Ae^{-\lambda G} = \sum_{n=0}^{n=\infty} \frac{\lambda ^n}{n!} C_n\end{equation}$$ where $$A, G$$, and $$C_n$$ are operators, and $$C_0 = A, C_1 = [G, C_0], C_2 = [G, C_1], ...C_n = [G, C_{n-1}]$$. Simplifying, we get

$$\begin{equation}U a^{\dagger}U^{\dagger} = a^{\dagger}\cos(\chi\Delta t) - ib^{\dagger}\sin(\chi\Delta t)\tag{3}\end{equation}$$

Now, $${Ua^{\dagger}}^nU^{\dagger} = UaU^{\dagger} UaU^{\dagger} UaU^{\dagger}U...UaU^{\dagger} = ({UaU^{\dagger}})^n$$. Substituting $$(3)$$ back in $$(2)$$ using this fact, we get

\begin{aligned} &=\frac{\sqrt{\binom{n}{k}}}{n!}\langle0_a, 0_b|a^{n-k}b^{k}(a^{\dagger}\cos(\chi\Delta t) - ib^{\dagger}\sin(\chi\Delta t))^n|0_a, 0_b\rangle \\ &= \sum_{r=0}^{r=n}\frac{\sqrt{\binom{n}{k}}}{n!}\binom{n}{r}(-i\sin(\chi\Delta t))^r(\cos(\chi\Delta t))^{n-r} \langle 0_a, 0_b|a^{n-k}b^{k}(a^{\dagger})^{n-r}(b^{\dagger})^r|0_a, 0_b\rangle\end{aligned}

Only the term with r $$=$$ k survives since the operators must add up to no net change in the number state for the inner product to be non-zero. This gives,

\begin{aligned} &= \frac{\sqrt{\binom{n}{k}}}{n!}\binom{n}{k}(-i\sin(\chi\Delta t))^k(\cos(\chi\Delta t))^{n-k} \langle 0_a, 0_b|a^{n-k}b^{k}(a^{\dagger})^{n-k}(b^{\dagger})^k|0_a, 0_b\rangle\\ &= \frac{\sqrt{\binom{n}{k}}}{n!}\binom{n}{k}(-i\sin(\chi\Delta t))^k(\cos(\chi\Delta t))^{n-k}\sqrt{(n-k)!(k!)(k!)(n-k!)} \langle 0_a, 0_b|0_a, 0_b\rangle \end{aligned}

Letting $$\gamma = 1-\cos^{2}\chi\Delta t$$ and substituting this result back in equation $$(1)$$, we obtain, \begin{aligned}E_k = \langle k_b|U|0_b\rangle &= (-i)^{k}\sum_{n}\sqrt{\binom{n}{k}}\sqrt{(1-\gamma)^{n-k}\gamma^{k}}|n-k_a\rangle \langle k_a| \end{aligned}

Which is what is required. The phase of $$(-i)^k$$ can be taken away and the new set of $$E_k's$$ would be unitarily related to the initial $$E_k's$$ and therefore would result in the same quantum operation by the unitary freedom in quantum operations.

• Thanks for the different answer. I have another question on Physics Stack Exchange for which I have not received an answer yet. Any thoughts about that? – Bashir Apr 14 at 15:51