I am referring to Exercise 7.18 of "Quantum Computing and Information 10th Anniversary Edition" by Nielsen and Chuang. The exercise wants me to show that the time evolution operator related to Rabi Oscillation is given by Eqn(7.77):

$$ U = e^{-iHt}\\= e^{-i\delta t}|00\rangle \langle 00| +(\cos(\Omega t)+i\frac{\delta}{\Omega}\sin(\Omega t))|01\rangle \langle 01| + \\(\cos(\Omega t)-i\frac{\delta}{\Omega}\sin(\Omega t))|10\rangle \langle 10|-i\frac{g}{\Omega}\sin\left(\Omega t\right)(|01\rangle \langle 10|+|10\rangle \langle 01|)$$

with $H$ given as Eqn(7.76):

$$H=-\begin{pmatrix} \delta&0&0\\0&\delta&g\\0&g&-\delta\end{pmatrix}$$

I tried to work this out by splitting $H$ into the sum of two parts:

$$H_{-1}=-\begin{pmatrix} \delta&0&0\\0&0&0\\0&0&0\end{pmatrix}=-H_{1}\\H_{-2}=-\begin{pmatrix} 0&0&0\\0&\delta&g\\0&g&-\delta\end{pmatrix}=-H_{2}$$

Since $H_{-1}H_{-2}=0$ and by the Baker-Campbell-Hausdorff (BCH) Formulae, $U = e^{-it(H_{-1}+H_{-2})}=e^{-itH_{-1}}e^{-itH_{-2}}=e^{itH_{1}}e^{itH_{2}}$. I then introduced $\Omega=\sqrt{\delta^2+g^2}$ such that $\big(\frac{H_{2}}{\Omega}\big)^2=\begin{pmatrix} 0&0&0\\0&1&0\\0&0&1\end{pmatrix}$. So now $U$ can be written as such:

$$U=e^{itH_{1}}e^{it\Omega\frac{H_{2}}{\Omega}}\\ =\bigg[\sum_{m=0}^{\infty} \frac{(-1)^m}{(2m)!}(it H_{1})^{2m}+\sum_{m=0}^{\infty} \frac{(-1)^m}{(2m+1)!}(it H_{1})^{2m+1}\bigg]\times \\ \bigg[\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}(it H_{2})^{2n}+\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}(it H_{2})^{2n+1}\bigg]$$

There are four terms in the above multiplication so I'll list down each of them.

$$\sum_{n=0}^{\infty}\sum_{m=0}^{\infty} \frac{(-1)^m}{(2m)!}(it H_{1})^{2m} \frac{(-1)^n}{(2n)!}(it H_{2})^{2n}\\ =\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}(it H_{2})^{2n}+\sum_{n=0}^{\infty}\sum_{m=1}^{\infty} \frac{(-1)^m}{(2m)!}(it H_{1})^{2m} \frac{(-1)^n}{(2n)!}(it H_{2})^{2n}\\ =\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}(it H_{2})^{2n}+\sum_{m=1}^{\infty} \frac{(-1)^m}{(2m)!}(it H_{1})^{2m}$$


$$\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\frac{(-1)^m}{(2m)!}(it H_{1})^{2m}\frac{(-1)^n}{(2n+1)!}(it H_{2} )^{2n+1}=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}(it H_{2} )^{2n+1}$$


$$\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\frac{(-1)^m}{(2m+1)!}(it H_{1})^{2m+1}\frac{(-1)^n}{(2n)!}(it H_{2} )^{2n}=\sum_{m=0}^{\infty}\frac{(-1)^m}{(2m+1)!}(it H_{1} )^{2m+1}$$


$$\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\frac{(-1)^m}{(2m+1)!}(it H_{1} )^{2m+1}\frac{(-1)^n}{(2n+1)!}(it H_{2} )^{2n+1}=\begin{pmatrix} 0&0&0\\0&0&0\\0&0&0\end{pmatrix}$$

Adding the right side of each of the four terms, I get:

$$U=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}(it\Omega \frac{H_{2}}{\Omega})^{2n}+\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}(it\Omega \frac{H_{2}}{\Omega})^{2n+1}\\+\sum_{m=1}^{\infty} \frac{(-1)^m}{(2m)!}(it H_{1})^{2m}+\sum_{m=0}^{\infty}\frac{(-1)^m}{(2m+1)!}(it H_{1} )^{2m+1} \\=\cos(\Omega t)\begin{pmatrix} 0&0&0\\0&1&0\\0&0&1\end{pmatrix}+i\frac{\sin(\Omega t)}{\Omega}\begin{pmatrix} 0&0&0\\0&\delta&g\\0&g&-\delta\end{pmatrix} \\+\sum_{m=1}^{\infty} \frac{(-1)^m}{(2m)!}(it H_{1})^{2m}+\sum_{m=0}^{\infty}\frac{(-1)^m}{(2m+1)!}(it H_{1} )^{2m+1} $$

The major problem here is the 2nd last term. The term that is supposed to sum to cosine starts with $m=1$. If I were to force the summation over $m$ to start from 0, it looks like the final form of $U$ has to subtract an Identity Matrix from it. Even so, the sign of some of the terms seem different than what is given by Eqn(7.77). But what troubles me most is the summation over $m$. Is anyone patient enough to look through it and point out whats going on?


2 Answers 2


You seem to be overcomplicating this somewhat!

You are right to split it up into the two terms $H_1$ and $H_2$. So, we have $$ e^{-i(H_1+H_2)t}=e^{-iH_1t}e^{-iH_2t}. $$ Now, straightforwardly, $$ e^{-iH_1t}=I+(e^{-i\delta t}-1)|00\rangle\langle 00|. $$ Next, we need to think about the $e^{-iH_2t}$ term. Of course, it maps $|00\rangle$ to $|00\rangle$. So, we only have to think about the subspace spanned by $\{|01\rangle,|10\rangle\}$. In this subspace, we have $H_2$ written as $$ \delta Z+gX=(g,0,\delta)\cdot\vec{\sigma}, $$ using the Pauli matrices. Now we can use the formula we're given (except that the formula in my version of N&C is incorrect, with sin and cos swapped. this is in the errata list) $$ e^{i\vec{n}\cdot\vec{\sigma}}=\cos(\Omega t)+i\sin(\Omega t)\frac{\delta Z+gX}{\Omega}, $$ where $\Omega=\sqrt{\vec{n}\cdot\vec{n}}=\sqrt{\delta^2+g^2}$. So, overall, we have $$ e^{-iH_2t}=|00\rangle\langle 00|+\cos(\Omega t)(|01\rangle\langle 01|+|10\rangle\langle 10|)-i\frac{\sin(\Omega t)}{\Omega}\left(\delta (|01\rangle\langle 01|-|10\rangle\langle 10|)+g(|01\rangle\langle 10|+|10\rangle\langle 01|)\right). $$ I think, overall, this gives slightly different signs to those in equation 7.77.

  • $\begingroup$ we've came across a sign problem before in that the 'definition' of the Hamilton seems to give a different sign to what I was expecting (I'm not exactly one for getting minus signs right all the time but this is the third time now) $\endgroup$
    – Mithrandir24601
    Commented Sep 23, 2019 at 7:23
  • $\begingroup$ @Mithrandir24601 I'm also not too careful with minus signs. Or, more specifically, there's the potential for a global sign that I know I haven't been careful with. But here, there also seems to be a relative minus sign between the X and Z terms. $\endgroup$
    – DaftWullie
    Commented Sep 23, 2019 at 7:59
  • $\begingroup$ @DaftWullie Thanks for your help that I'm able to recall the exponential of a diagonal matrix and spot a mistake. $\endgroup$
    – C.C.
    Commented Sep 23, 2019 at 12:04
  • $\begingroup$ I think that the Hamiltonian should not have a minus sign. $\endgroup$
    – C.C.
    Commented Sep 23, 2019 at 12:35

In the first summation for $U$: $$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}(it\Omega \frac{H_{2}}{\Omega})^{2n}=\cos(\Omega t)\begin{pmatrix} 0&0&0\\0&1&0\\0&0&1\end{pmatrix}(\text{This is wrong})$$

$$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}(it\Omega \frac{H_{2}}{\Omega})^{2n}=\begin{pmatrix} 1&0&0\\0&0&0\\0&0&0\end{pmatrix}+\cos(\Omega t)\begin{pmatrix} 0&0&0\\0&1&0\\0&0&1\end{pmatrix}(\text{This is correct})$$

I mentioned that $m$ starts summing from $1$ so I will force it to sum from $0$, which in this case indeed I have to subtract an Identity Matrix. This will result in an exponential of $H_{1}$ subtract an Identity whereby the steps will look like:

$$\sum_{m=1}^{\infty} \frac{(-1)^m}{(2m)!}(it H_{1})^{2m}+\sum_{m=0}^{\infty}\frac{(-1)^m}{(2m+1)!}(it H_{1} )^{2m+1}\\=\sum_{m=0}^{\infty} \frac{(-1)^m}{(2m)!}(it H_{1})^{2m}+\sum_{m=0}^{\infty}\frac{(-1)^m}{(2m+1)!}(it H_{1} )^{2m+1}-\begin{pmatrix} 1&0&0\\0&1&0\\0&0&1\end{pmatrix}\\=\begin{pmatrix} e^{i\delta t}&0&0\\0&1&0\\0&0&1\end{pmatrix}-\begin{pmatrix} 1&0&0\\0&1&0\\0&0&1\end{pmatrix}$$

Adding everything up including the $sin(\Omega t)$ term will give the correct expression except for maybe some sign issues.


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