Here I read some papers about the applications of matrix product states (MPS) on quantum computation and quantum circuits. Works about this field are not so many, so they are easy to go through.

## References

[1] A. S. Bhatia and M. K. Saggi. * Implementing Entangled States on a Quantum Computer*. arXiv:1811.09833v2 (2019).

[2] D. S. Wang, D. T. Stephen, R. Raussendorf. **Qudit quantum computation on matrix product states with global symmetry.***Physical Review A* 95, 032312 (2017).

[3] C. Schön, E. Solano, F. Verstraete, J. I. Cirac, M. M. Wolf. * Sequential generation of entangled multiqubit states*.

*Physical Review Letters*. 11, 110503 (2005).

### Summary of Ref. [1]

This paper is easy to go through. They used the main idea of MPS to create a quantum circuit to generate GHZ and $W$ states. Without input, the circuits are static, it only presents a logical setup instead of any quantum state.

Firstly, they constructed quantum circuits to generate GHZ states with 2,3,4 qubits, according to the figures, they are easy to achieve. We note here that the implementation of the 4-qubit version of GHZ is quite different from the cases with fewer qubits, as the paper proposed [1],

However, the left circuit is really complicated for GHZ states, because we can easily use the extension of the right one to achieve this goal, so why did they use such a complicated circuit? What are the advantages?

Hadamard gate can be replaced by the rotation operation $R_y(\theta)$ with $\theta=\pi/2$, hence the circuit for generating GHZ can be implemented by rotation gates and CNOT gates.

Next, let us consider the situations of $W$ states. Here is the circuit for 4-qubit $W$ state generation, this circuit is interesting [1],

If we divide the whole circuit into two parts, the first part consists of $R_y, CZ$ array and the second part is the CNOT array. Here, again, $X$ gate can be replaced by $\theta=\pi$, then a controlled-$Z$gate performed, when the control qubit is $|1\rangle$, $CZ$ acts only on the $|1\rangle$ of the target qubit, hence $CZ$ acts only when both qubits have $|1\rangle$ and has symmetry between its inputs. Finally, $CZ$ gates mainly control the relative phase of eigenstate $|11\rangle$ in an entangled state:

$$\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle) \stackrel{CZ}{\longrightarrow} \frac{1}{\sqrt{2}}(|00\rangle-|11\rangle).$$

Let us consider the effect of the combination of $R_y$ and $CZ$ gates. However, this is worth thinking carefully. The effect of the second column is clear, however, the first column is needed to be analyzed.

The quantum state evolution can be presented by,

$$\frac{1}{2}(|00\rangle+|01\rangle+|10\rangle+|11\rangle),$$

$$\stackrel{CZ}{\longrightarrow} \frac{1}{2}(|00\rangle+|01\rangle+|10\rangle-|11\rangle)=\frac{1}{2}(|0\rangle+|1\rangle)|0\rangle+\frac{1}{2}(|0\rangle-|1\rangle)|1\rangle,$$

$$\stackrel{R_y}{\longrightarrow} \frac{1}{2}(|0\rangle+|1\rangle)(|0\rangle+|1\rangle)+\frac{1}{2}(|0\rangle-|1\rangle)(-|0\rangle+|1\rangle),$$

$$= \frac{1}{2}(|00\rangle+|01\rangle+|10\rangle+|11\rangle)+\frac{1}{2}(-|00\rangle+|01\rangle+|10\rangle-|11\rangle),$$

$$= \frac{1}{\sqrt{2}}(|01\rangle+|10\rangle).$$

Note that $R_y$ is not equal to $H$ gate when the input is $|1\rangle$, however $H$ equals to $(R_y+R_z)/\sqrt{2}$. We can see the change of relative phase leads to different EPR pairs, the equvalence is,

So, according to this equvalence, can we utilize this relationship to simplify the whole procedure?

One more question is that, how does it reflect the feature of MPS? How does the relationship between the neighboring qubits look like?

In the end, thanks to this manuscript, I review the equivalence of some quantum gates, see the Equivalence relationship of the quantum logic gates.

## Summary of Ref. [2,3]

This paper is not easy to follow. Some concepts are very strange to me.

### Notations

In the Ref.[2,3], some notations are different, here we use a uniform notation system.

- $A$ indicates the sub-system of ancilla qubits, and $S$ indicates the sub-system of the qubits which form the MPS.
- The sequential unitary operations carried on each $S$-$A$-qubit pair is written as $U_{[n]}^{i_n}$, where $[n]$ indicates the step, $i_n$ indicates the physical index of this tensor, in a $2$-d system, $i_n \in {|0\rangle, |1\rangle}$. $U_{[n]}$ indicates a summation of operation $U_{[n]}^{i_n=|0\rangle}$ and $U_{[n]}^{i_n=|1\rangle}$.
- In the beginning, qubits of $S$ are initialized as $|0\cdots 0\rangle$, and qubits of $A$ are initialized as $|\varphi\rangle$. $|\Psi\rangle$ indicates the whole system including $A$ and $S$ after the sequential operations. After the sub-system, $A$ decouples from the global system $|\Psi\rangle$, the qubits from $A$ are written as $|\varphi_F\rangle$, and the qubits from $S$ are written as $|\psi\rangle$.
- $A_{[n]}$ is the tensor we all know about it, it consists of $A_{[n]}^{i_n}$ which is corresponding to different inputs of physical indices. We obtain it by applying $U_{[n]}$ on a single-photon from the sub-system $S$.

### The circuit for MPS states

This is a brilliant way of generating MPS states. Let’s take a look.

Firstly, we must take a look at the input state, and it is divided into two parts, A is the ancilla qubits and S is a group of 2D single-qubits. Hence the input is

$$|\varphi_I\rangle |i_n,\cdots,i_1\rangle,$$

where $|\varphi_I\rangle \in H_A$ represents the initial state of ancilla qubits, $|i_n,\cdots,i_1\rangle$ represents the other $n$ single-qubits of system S which are initialized in states $|0\rangle$. In fact, the operations here are performed sequentially, and the key point is how should we express the unitary operators $U_n$. Sure $U_n$ involves the two parts, i.e., the ancilla qubits and one of the single-qubits, due to the single-qubits’ initial states $|0\rangle$, we ignore the input, then the $U_n$ is a mapping $H_A \rightarrow H_A \otimes H_B$ [3],

$$U_{[n]}=\sum_{i_n,\alpha,\beta}U^{i_n}_{\alpha,\beta}|\alpha,i_n\rangle\langle\beta|,$$

where $i$ is the physical index in the language of MPS, $\alpha$ and $\beta$ are the basis of sub-system $A$ and $S$ respectively, besides $U^{i_n}_{\alpha,\beta}$ is a scalar. And

$$|i_n\rangle \in {|0\rangle, |1\rangle},$$

$$|\alpha\rangle \in {|1\rangle, |2\rangle, \cdots, |D\rangle},$$

$$|\beta\rangle \in {|0\rangle, |1\rangle},$$

hence, $|\alpha, i\rangle$ is in a $D \times 2$ dimensional space, and $\beta$ is in a $2$ dimensional space, but what about the matrix $U_{[n]}$, what does it represent, and what size is it? Here is another difinition of the unitary operator $U_{[n]}$ [2],

$$U_{[n]}|0\rangle =\sum_{i_n}|i_n\rangle \otimes A^{i_n}_{[n]},$$

where $|0\rangle$ is the new input single-photon, however term $|i_n\rangle \otimes A_{i_n}$ is ONE situation of a tensor, while there are totally two situations here ($|i_n\rangle=|0\rangle,|i_n\rangle=|1\rangle$):

Obviously, $|i_n\rangle$ designates the physical indices, and the following $A_{i_n=|0\rangle}$ is the matrix of this tensor when the input state is $|0\rangle$. However, in this view, how can we analyse the two sub-systems $A$ and $S$? And the other question is, how can we present the $U$ operator and how can we use this scheme to generate a specified quantum state?

Finally, a MPS state (sub-system $S$) can be expressesd as

$$|\psi\rangle=\sum_{i_{1} \ldots i_{n}=0}^{1}\left\langle\varphi_{F}\left|U_{[n]}^{i_{n}} \ldots U_{[1]}^{i_{1}}\right| \varphi_{I}\right\rangle\left|i_{n}, \ldots, i_{1}\right\rangle,$$

This result is extremely important. We know that, a MPS state, in fact, indicates a certain quantum state which can be in an eigenstate or a superposition state, just like others. The difference is that the probability amplitude is presented by a series of matrix products, hence, the above equation can be understood by two parts seperately. Firstly, $\left|i_{n}, \ldots, i_{1}\right\rangle$ represents an eigenstate in this system. Secondly, $\left\langle\varphi_{F}\left|U_{[n]}^{i_{n}} \ldots U_{[1]}^{i_{1}}\right| \varphi_{I}\right\rangle$ represents its probability amplitude, and it is derived by the products of matrices. The last but most important, $U_{[n]}^{i_{n}}$ is the unitary operator carried on current $S$ and $A$ qubit(s), and it directly determines the result tensor $A_{i_n}$, although they are not equivelent due to qubits from $S$.

### Operator \(U^{i_n}_{[n]}\) and the boundary conditions

$U^{i_n}*{[n]}$ is the unitary operator for certain $S$-qubit and input of physical input, different value of $i_n$ corresponds to different matrix $A^{i_n}*{[n]}$. In the last section, we gave the

In this expression, each $A^{i_n}_{[n]}$ indicates the relationship between current tensor (2-dimensional) and the rest of the system (D-dimensional). Specifically, when applying SVD, one tensor is a bridge to connect the left-side tensor and the right-side tensor:

Of course, we does not apply SVD here, every $U_{[n]}$ as well as $A_{[n]}$ is a $D \times D$ matrix, and every $A^{i_n}*{[n]}$ is a $\frac{D}{2} \times D$ matrix, hence the structure of $U*{[k]}$ is,

The final question is, what are the $D$-dimensional vectors? It is reasonable to regard the vectors as the states of the whole system, there are $D=2^n$ states can exist in this MPS because dimension of ancilla qubits $D=2^n$, as well as the qubit number of $S$ system is $n$.

### An example of cavity QED

For understanding this section, it is better to read my another post 腔QED与量子信息过程的简单学习 first. As to a two-level hyperfine* ground states $|a\rangle$ and $|b\rangle$. Then an outer laser can control the state like this,

$$|a,0\rangle \rightarrow \cos(\theta)|a,0\rangle + \sin(\theta)e^{i\theta}|b,1\rangle=c_1|a,0\rangle+s_1|b,1\rangle$$

where $c_i=\cos(\phi_1)$ and $s_i=\sin(\phi_1)e^{i\varphi_1}$, subscript $i$ indicates the step.

- 超精细结构即原子核具有自旋导致电子能级的分裂，总得来说就是有两个能级 $|a\rangle$ 和 $|b\rangle$。

It is exactly a unitary operator. Here it is an entangled state of two degrees of freedom, that is the energy level and cavity mode, $|0\rangle$ means no photon presence and $|1\rangle$ means photon presence. Here we need to understand that, the ponton was generated by the dropping of the excited atom to the lower energy level. In the beginning, there is no photon, the cavity mode is $|0\rangle$, however, after the rotation, the energy level may drop to $|b\rangle$, one photon would be generated, the cavity mode changes to $|1\rangle$. In my opinion, the cavity mode is, in fact, a Fock state.

This is clever, dropping to the lower level indeed generate a photon, however, when we analyze the evolution using the tool of Fock state, and regard ‘no dropping’ as state $|0\rangle$ and regard ‘dropping’ as state $|1\rangle$, the operator is still unitary and no need the creation operator.

The unitary operator can be written as the combination of $R_y(\theta_y)$ and a phase gate $\exp(i\theta_z/2) R_z(\theta_z)$:

Next, we perform such unitary operator again, in the last step, we obtained $c_1|a,0\rangle+s_1|b,1\rangle$. For term $c_1|a,0\rangle$, again

$$|a,0\rangle \rightarrow c_2|a,0,0\rangle + s_2|b,1,0\rangle,$$

$$|b,1\rangle \rightarrow |a,0,1\rangle.$$

Here we give an explain, only when the atom is in $|a\rangle$ state, it can omit a photon. Hence, the result of step 2 is,

$$c_1|a,0\rangle+s_1|b,1\rangle \rightarrow c_1c_2|a,0,0\rangle+c_1s_2|b,1,0\rangle+s_1|b,0,1\rangle,$$

$$=c_1c_2|a,0,0\rangle+|b\rangle(c_1s_2|1,0\rangle+s_1|0,1\rangle).$$

After $n$ times of repeat, we can obtain,

$$c_1 \cdots c_n|a\rangle|0,\cdots,0\rangle +|b\rangle(c_1\cdots c_{n-1}s_n|1,0,\cdots,0\rangle$$

$$+c_1\cdots c_{n-2}s_{n-1}|0,1,0,\cdots,0\rangle)+s_{1}|0,\cdots,0,1\rangle).$$

After we measure the atom state and obtain $|b\rangle$, then we obtain the $W$ state. Another thing we need to note is the expression of photon-generation process, which we have given in last section,

$$U_{[n]}=\sum_{i_n,\alpha,\beta}U^{i_n}_{\alpha,\beta}|\alpha,i_n\rangle\langle\beta|,$$

could be simplified to

$$U_{[i]}=c_i|0\rangle_i|a\rangle\langle a|+s_i|1\rangle_i|b\rangle\langle a|+|1\rangle_i|b\rangle\langle b|.$$

If we give a input state of $|\varphi_I\rangle$, it is the ancilla state of the atom, then we obtain the ouput

$$|\Psi\rangle = U_{[n]}\cdots U_{[1]}|\varphi_I\rangle$$

and the output state contains omitted photon state, which consists of the target state. Next, like we did before, we need a projective operation to get a certain state $|\varphi_F\rangle$ of the atom ancilla qubit, then we obtain the target MPS state,

$$|\Psi\rangle=|\varphi_F\rangle\langle\varphi_F|U_{[n]}\cdots U_{[1]}|\varphi_I\rangle$$

In this example, $|\varphi_F\rangle=|b\rangle$.

### Is a state can be generated?

A critical problem is judging whether a state can be generated by certain resources, how can we achieve this? Using SVD!

Repeat the SVD procedure for *$n$* times, we can obtain such a state,

its state can be given as

$$|\widetilde{\psi}\rangle=U^{\prime}_{[n]}\cdots U^{\prime}_{[n]}|\varphi_I\rangle$$

what information does the result of the SVD procedure give us? Why they use SVD to prove that **the MPS representation is a general recipe for states’ generation**.

### An equivalent characterization

In this characterization, we use $D$ atoms, every atom spans a Hilbert space $\mathbb{C}^2$, hence the whole system is in Hilbert space $\mathbb{C}^2\otimes\mathbb{C}^D$.