# I Questions about an optical photon quantum computer

#### Haorong Wu

Summary
Questions about optical photon quantum computer
Hi. I'm learning the optical photon quantum computer from Nielsen's QCQI. Since I'm not familiar with quantum optics, I got some questions about it.

Q1. In page 288, the book reads: A laser outputs a state known as a coherent state $\left | \alpha \right > = e^{- \left | \alpha \right | ^2 /2 } \sum_{n=0}^{\infty} \frac {\alpha ^ n} {\sqrt {n !}} \left | n \right >$, where $\left | n \right >$ is an $n$-photon energy eigenstate. Then the mean energy is $\left < \alpha \right | n \left | \alpha \right > = \left | \alpha \right | ^2$.

But, if I remember correctly, the mean energy should be $\left < \alpha \right | H \left | \alpha \right >$, the result is the same. Am I right?

Q2. Still about the coherent state. The book reads: For example, for $\alpha = \sqrt {0.1}$, we obtain the state $\sqrt {0.90} \left | 0 \right > + \sqrt {0.09} \left | 1 \right > + \sqrt {0.002} \left | 2 \right > + \cdots$. Thus if light ever makes it through the attenuator, one know it is a single photon with probability better than 95%; the failure probability is thus 5%. After calculation, the state above is $0.6154\left | 0 \right > +0.3 \left | 1 \right > +0.04472 \left | 2 \right > + \cdots$.

However, by applying the definition of the coherent state, I got $\left | \alpha = \sqrt {0.1} \right > =0.9512 \left | 0 \right > +0.3008 \left | 1 \right > +0.06726 \left | 2 \right > + \cdots$, which doesn't match the above expression. Where am I wrong?

Also, I have no idea how the two probabilities come out.

Q3. About the phase shifter, the book reads: A phase shifter $P$ acts just like normal time evolution, but at a different rate, and localized to only the modes going through it. ... . it takes $\Delta = \left ( n-n_0 \right ) L / C_0$ more time to propagate a distance $L$ in a medium with index of refraction $n$ than in vacuum. For example, the action of $P$ on the vacuum state is to do nothing: $P \left | 0 \right > =\left | 0 \right >$, but on a single photon state, one obtains $P \left | 1 \right > = e^{i \Delta} \left | 1 \right >$.

But I think, the time evolution should be $P \left | 1 \right > =e^{-i \omega \Delta} \left | 1 \right >$, so where did the $\omega$ go?

Q4. Still about the phase shifter, the books reads: $P$ performs a transforms ... nothing more than a rotation $R_z \left ( \Delta \right )= e^{-iZ \Delta /2}$. One can thus think of $P$ as resulting from time evolution under the Hamiltonian $H=\left ( n_0 -n \right ) Z$, where $P=exp \left ( -i HL/c_0 \right )$.

Then I tried to expand the expression of $P$ using the Hamiltonian: $P=exp \left ( -i HL/c_0 \right )=exp \left ( -i \left ( n_0 -n \right ) Z L/c_0 \right )=e^{iZ \Delta}$. Then how can it be related to $R_z \left ( \Delta \right )= e^{-iZ \Delta /2}$?

Q5. About the Beamsplitter, the book reads: The Hamiltonian is $H_{bs}=i \theta \left( a b^\dagger - a ^ \dagger b \right )$, and the beamsplitter performs the unitary operation $B=exp \left [ \theta \left ( a^\dagger b - a b ^\dagger \right ) \right ]$.

Still, I think the operation should be $exp \left ( -i H t / \hbar \right )=exp \left [ \theta \left ( a^\dagger b - a b ^\dagger \right ) t/ \hbar \right ]$. Everything is fine, except where is the $t / \hbar$? So in the beamsplitter, $t / \hbar =1$?

Ok, I have post too many questions. Some of them may be silly. I'm looking forward to any helps. Thanks!

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#### vanhees71

Gold Member
ad Q1. I don't know the book, but if the author uses standard notation, then $|n \rangle$ is a state with $n$ photons occupying one specific photon mode (single-photon energy eigenmode). Then $|\alpha|^2$ is rather the mean photon number rather than the mean energy, which is given by $|\hbar \omega |\alpha|^2$ (where I have employed normal ordering, i.e., subtracted the socalled "vacuum energy" $\hbar \omega/2$ such that the vacuum is the energy eigenstate to energy eigenvalue 0, i.e., the ground-state energy is defined to be 0).

ad Q2. I can confirm your values (using Mathematica ;-)). Now the probability that at least one photon is present is given by
$$P(n \geq 1)=\sum_{n=1}^{\infty} \exp(-|\alpha|^2) \frac{|\alpha|^{2n}}{n!} =\exp(-|\alpha|^2) [\exp(|\alpha|^2-1]=1-\exp(-|\alpha|^2) \simeq 0.0951.$$
The probability that one has exactly one photon is
$$P(n=1)=|\langle 1|\alpha \rangle|^2=\exp(-|\alpha|^2)|\alpha|^2 \simeq 0.090.$$
Thus the probability for having had precisely one photon provided the (ideal) detector clicked is
$$P(n=1|n \neq 0)=\frac{P(n=1)}{P(n \geq 1)} \simeq 0.951.$$

ad Q3. You are right. The expression $\exp(\mathrm{i} \Delta)$ is non-sensical, because the argument of the exp function is of dimension [time]=s and you cannot calculate the exponential of a dimensionful quantity.

I don't know what to make of Q4 and Q5. I'd need more context, but perhaps somebody else who has access to the book can help.

#### Haorong Wu

ad Q1. I don't know the book, but if the author uses standard notation, then $|n \rangle$ is a state with $n$ photons occupying one specific photon mode (single-photon energy eigenmode). Then $|\alpha|^2$ is rather the mean photon number rather than the mean energy, which is given by $|\hbar \omega |\alpha|^2$ (where I have employed normal ordering, i.e., subtracted the socalled "vacuum energy" $\hbar \omega/2$ such that the vacuum is the energy eigenstate to energy eigenvalue 0, i.e., the ground-state energy is defined to be 0).

ad Q2. I can confirm your values (using Mathematica ;-)). Now the probability that at least one photon is present is given by
$$P(n \geq 1)=\sum_{n=1}^{\infty} \exp(-|\alpha|^2) \frac{|\alpha|^{2n}}{n!} =\exp(-|\alpha|^2) [\exp(|\alpha|^2-1]=1-\exp(-|\alpha|^2) \simeq 0.0951.$$
The probability that one has exactly one photon is
$$P(n=1)=|\langle 1|\alpha \rangle|^2=\exp(-|\alpha|^2)|\alpha|^2 \simeq 0.090.$$
Thus the probability for having had precisely one photon provided the (ideal) detector clicked is
$$P(n=1|n \neq 0)=\frac{P(n=1)}{P(n \geq 1)} \simeq 0.951.$$

ad Q3. You are right. The expression $\exp(\mathrm{i} \Delta)$ is non-sensical, because the argument of the exp function is of dimension [time]=s and you cannot calculate the exponential of a dimensionful quantity.

I don't know what to make of Q4 and Q5. I'd need more context, but perhaps somebody else who has access to the book can help.
Thanks, vanhees71. You have helped me reveal some of the mysteries, especially the probabilities. I would not figure them out alone. Many thanks!

"Questions about an optical photon quantum computer"

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