# How do the Type I and Type II fusions work?

I am going through the original paper on Type I and Type II fusions. But I cannot understand how this is actually implmented. They say they start with

$$\frac{1}{2} |H,H\rangle+|V,V\rangle \otimes |H,H\rangle+|V,V\rangle$$

Then they perform a PBS and then apply a detector after rotating the reflected polarization by 45 degrees, followed by a measurement. But how is this actually applied? I am not really familiar with this formalism, and since the polarizations of the photons end up reflected into different modes, I am not sure how the pre and post measured state is meant to look. Whenever I try it, I end up with something similar to a GHZ state, but with other terms that don't make sense, indicating I have applied it wrong.

Moreover, for Type II fusions, they talk about using $$\sigma_{x}$$ measurements on the larger states. But this doesn't work for GHZ states, like they have in the paper. You don't end up with a logical state between the two qubits surrounding the measured one. Only a cluster state created via local operations on a GHZ state does this. I think my confusion here is I cannot actually see the generated cluster state from the type 1's.

There is this post, but honestly it doesn't really explain how the projector is arrived at, and from it, I can't really see how that would even create the cluster state, as only 2 projectors are specified there, but the product state before the gate has two terms where the central qubits would have either $$|HV\rangle$$ or $$|VH\rangle$$. So unless the implication is that these don't get measured due to the absence of a photon due to reflection, I am unsure how these disappear from the final state.

Edit: After applying the PBS, I believe we arrive at

$$\frac{1}{2} |H,H,H,H\rangle+|H,HV,0,V\rangle+|V,0,VH,H\rangle + |V,V,V,V\rangle$$

where $$|HV\rangle$$ denotes 2 photons with different polarizations in the same mode. However from here, the rotation and measurements are still confusing me, and I still am unsure about Type 2's

Edit 2:

I think after applying the 45 degrees rotation they talk about, and measuring one of the modes, it's meant to go to:

$$\frac{|H,+,H\rangle+|V,-,V\rangle}{\sqrt{2}}$$ which gives us a 3 qubit cluster state. But I'm unclear how one of the modes is discarded, as they say the measurement destroys one of the photons

NOTE:

I've seen this line of reasoning and those calculations somewhere, but I don't remember where. Here is my reconstruction of them.

Type I:

Let's have in mind the sketch of fusion type I from the original paper, Fig. 2a). Let us call the left port A, the down port B (these are the input ports), the right port C and the upper port D (these are the port where the detectors are located). In the middle there is a PBS, and before port D there is a waveplate which rotate the incoming signal by polarization. In type I, you only detect the incoming photons from port D, but you can distinguish the polarization, that it to say you can either get no photons at all, a single H or V photons, or 2H or 2V photons.

Now, how does the input change after the transformation of the PBS + rotation? Let us assume that you have a general incoming wave, which is already connected to other qubits (for example, you have two large cluster states and you want to connect them by fusing together one photon from each). So, we have from port A: $$f_{1}a_H^{\dagger} + f_{2}a_V^{\dagger}$$, where $$f_{1}$$ and $$f_{2}$$ are the other modes, $$a$$ stands for the port and $$H$$ or $$V$$ is the polarization. From port B we have the same: $$f_{3}b_H^{\dagger} + f_{4}b_V^{\dagger}$$.

The H mode from port A doesn't feel the PBS because the photon comes in parallel to it, so it continues unharmed to a horizontal mode in port C. On the contrary, the vertical mode will turn to port D and then will be rotated. The opposite happens for the photons from port B. Finally, what will happen is:

1. $$a_{H}^{\dagger} \to c_{H}^{\dagger}$$
2. $$a_{V}^{\dagger} \to \dfrac{d_{H}^{\dagger} - d_{V}^{\dagger}}{\sqrt{2}}$$
3. $$b_{V}^{\dagger} \to c_{V}^{\dagger}$$
4. $$b_{H}^{\dagger} \to \dfrac{d_{H}^{\dagger} + d_{V}^{\dagger}}{\sqrt{2}}$$

(The reason why port B doesn't get a minus sign, is because it comes with H polarization, so it go through unharmed, and than get a rotation).

Finally, one can write:

$$(f_{1}a_{H}^{\dagger} + f_{2}a_{V}^{\dagger})(f_{3}b_{H}^{\dagger}f_{4}b_{V}^{\dagger}) \to ...$$, replace $$a$$ and $$b$$ with $$c$$ and $$d$$ using relations 1-4, do some algebra, and you should get:

$$\to \dfrac{1}{\sqrt{2}}f_{1}f_{3}c_{H}^{\dagger}(d_{H}^{\dagger} + d_{V}^{\dagger}) + f_{1}f_{4}c_{H}^{\dagger}c_{V}^{\dagger} + \dfrac{1}{2}f_{2}f_{3}((d_{H}^{\dagger})^2 - (d_{V}^{\dagger})^2) + \dfrac{1}{\sqrt{2}}f_{2}f_{4}c_{V}^{\dagger}(d_{H}^{\dagger} - d_{V}^{\dagger})$$.

For simplicity, and to be aligned with the paper, let us assume that you started with two bell states, such that $$f_{1}a_{H}^{\dagger} = |H, H\rangle$$, $$f_{2}a_{V}^{\dagger} = |V, V\rangle$$, and the same for port B ($$f_{3}$$ is H and $$f_{4}$$ is V). Also, the final 3-qubit state (after post-selecting the output from the detector) is written in the order $$|A, B, C\rangle$$.

Now, when do you get a GHZ? If you obtained a single photon in detector D, no matter if it is in the H or V polarization. Because, in that case, you only left with the first and last terms, which are $$\dfrac{1}{\sqrt{2}}(|H, H, H\rangle \pm |V, V, V\rangle)$$, depending if you detected (H, 0) or (0, V), respectively.

You can also check what are the results from the other options, and see that this is not the case.

Type II:

If you understand this, type II is very similar, in the sense that you just need to write the general input signal (which is exactly the same) and do the transformation for the output ports. In that case, the sketch is somewhat different (see Fig. 2b) in the paper), because each port first has a rotation. So first of all, the waveplate transforms the input as follows:

1. $$a_{H}^{\dagger} \to \dfrac{a_{H}^{\dagger} + a_{V}^{\dagger}}{\sqrt{2}}$$
2. $$a_{V}^{\dagger} \to \dfrac{a_{H}^{\dagger} - a_{V}^{\dagger}}{\sqrt{2}}$$
3. $$b_{H}^{\dagger} \to \dfrac{b_{H}^{\dagger} + b_{V}^{\dagger}}{\sqrt{2}}$$
4. $$b_{V}^{\dagger} \to \dfrac{b_{H}^{\dagger} - b_{V}^{\dagger}}{\sqrt{2}}$$

Then there is a PBS and another rotation, so each individual polarization goes to:

1. $$a_{H}^{\dagger} \to \dfrac{c_{H}^{\dagger} + c_{V}^{\dagger}}{\sqrt{2}}$$
2. $$a_{V}^{\dagger} \to \dfrac{d_{H}^{\dagger} - d_{V}^{\dagger}}{\sqrt{2}}$$
3. $$b_{H}^{\dagger} \to \dfrac{d_{H}^{\dagger} + d_{V}^{\dagger}}{\sqrt{2}}$$
4. $$b_{V}^{\dagger} \to \dfrac{c_{H}^{\dagger} - c_{V}^{\dagger}}{\sqrt{2}}$$

Now what is left is to do the same algebra, to obtain that when one gets a single detection in each detector it acts like a parity measurement that projects the "f"'s into the even/odd parity of the input, depends if one gets the same or different polarization.

Notice that, in fact, in type II fusion the two qubits are gone. It seems wired, but because (when you have a "successful" fusion, in the aforementioned sense) you know exactly to which subspace you projected the state, you can do a computation, similar to a single-qubit measurement in measurement-based quantum computation.

• Thank you for replying David. I have never used this sort of notation, so beyond this being some sort of creation/ annihilation operator evolution, it's really not clear to me what's going on, especially on what is or is not being transmitted reflected. Forgetting the rotation operator on mode d. After the action of the PBS, could you tell me what $$\frac{1}{2} |H,H,H,H\rangle+|H,H,V,V\rangle+|V,V,H,H\rangle + |V,V,V,V\rangle$$ goes to? I get $$\frac{1}{2} |H,H,H,H\rangle+|H,HV,0,V\rangle+|V,V,H,H\rangle + |V,V,V,V\rangle$$ But I think this is wrong Commented Jul 16 at 12:01
• I meant to type $$\frac{1}{2} |H,H,H,H\rangle+|H,HV,0,V\rangle+|V,0,VH,H\rangle + |V,V,V,V\rangle$$ Commented Jul 16 at 12:07
• Yes, You are correct. But, you see, for me it is a little bit confusing to write it like this, because I want to keep track on which is which. Commented Jul 16 at 13:42
• Forget about the fancy operator - $a_{H}^{\dagger}$ just tells you that there is a horizontal mode in A, so you can write it also as $|H\rangle$ instead. But if you have several qubits, and several ports, one can get easily confused. Instead I write the first Bell-pair as (up to normalization) $f_{1}a_{H} + f_{2}a_{V}$ which is exactly $|HH_{a}\rangle + |VV_{a}\rangle$, such that when I know that H from A goes to H in C and V from A goes to V in D, I can easily keep track. Commented Jul 16 at 13:52
• Yes I am beginning to see why the use of my notation may not be altogether the best approach here. Commented Jul 16 at 13:53