Ryan Keleti

2020-07-06 › math

Last Friday I finished by first research experience for undergraduates (REU)! I learned about quantum transfer methods on graphs and quantum computation. The final draft for the project can be found here (if that link is gone, please email me for a copy). I had the great pleasure to work with Whitney Drazen (our advisor) and Noble Mushtak on this project! Read more for a bit of a rushed summary.

I'm a little burned out right now, so I might write about this more later. I'll give a brief summary here, so please forgive any errors.

In classical computer/information science, data can be represented using bits, which are either $0$ or $1$. We can consider a quantum analogue to this binary system, called a quantum bit, or qubit. Like a classical bit, a qubit can have the states $0$ or $1$, usually denoted by the ket notations $| 0 \rangle$ and $| 1 \rangle$. Unlike a classical bit, a qubit $q$ can exist in a superposition of $| 0 \rangle$ and $| 1 \rangle$, that is, a general state $| \psi \rangle$ of $q$ is linear combination $\alpha | 0 \rangle + \beta | 1 \rangle$, where are $\alpha$ and $\beta$ are complex numbers such that $|\alpha|^2 + |\beta|^2 = 1$. This gives that $| \psi \rangle$ is a unit vector is a two-dimensional $\mathbb{C}$-vector space $V$, called the state space of $q$. Another caveat: while the state of a qubit is a superposition of $| 0 \rangle$ and $| 1 \rangle$, when we "observe" it, the state collapses to either $| 0 \rangle$ and $| 1 \rangle$, and the state is lost (the probability of collapsing to $| 0 \rangle$ is $|\alpha|^2$ and the probability of collapsing to $| 1 \rangle$ is $|\beta|^2$).

Now suppose we have $n$ qubits $q_1, \dots, q_n$, with state spaces $V_1, \dots, V_n$. A naive approach to an $n$-qubit system might be to take the product of the $V_i$. But due to quantum entanglement, this is not the right notion. Instead, we take the tensor product $V = V_1 \otimes \cdots \otimes V_n$. Then entanglement is expressed as the fact that an arbitrary element of $V$ is not an elementary tensor $u \otimes v$, but instead a linear combination of the basis vectors. These basis vectors ($2^n$ of them) are precisely

$$ | b_1 \cdots b_n \rangle := | b_1 \rangle \otimes \cdots \otimes | b_n \rangle, $$

where $b_i \in \{0, 1\}$ (this is a bit of an abuse of notation).

Now one might like to ask the question: can we copy a qubit's state? In the classical setting, this is as easy as inspecting the state of the bit and copying the value. Well, not so in the quantum case. This is due to the no-cloning theorem, which gives that there does not exist an operator which can clone an arbitrary quantum state.

What's the next thing we can try? There is a concept called perfect state transfer. We might want to consider the problem of moving a qubit's state in some kind of network (usually something which is called a quantum spin network). Suppose this network has $n$ qubits. We can model the network as an undirected graph $G$ with $n$ vertices. The state of $G$ at a given time $t$ can be expressed as a vector $| \Psi(t) \rangle$ with $n$ components, one for each vertex in $G$. The axioms of quantum mechanics give that the state $| \Psi(t) \rangle$ of $G$ evolves over time according to Schrödinger's equation,

$$i\hbar\frac{d}{dt} | \Psi(t) \rangle = \mathcal{H} | \Psi(t) \rangle,$$

where $\hbar$ is the reduced Planck constant (in mathematical fashion, we can take $\hbar = 1$), and $\mathcal{H}$ is the Hamiltonian of the system. The Hamiltonian can be expressed in different ways depending on the context, but it involves the adjacency matrix $A$ of $G$. Solving Schrödinger's equation gives

$$ | \Psi(t) \rangle = e^{-it\mathcal{H}/\hbar} | \Psi(0) \rangle.$$

Now we are interested in the case where one qubit is "excited" and the others are $| 0 \rangle$. Through some math involving the Hamiltonian, we get that the evolution of this system is given by

$$ U(t) := e^{-itA}.$$ This operator is called the quantum walk operator.

Now say we have two qubits $q_i$ and $q_j$ in $G$. We say that perfect state transfer occurs at time $t$ from $q_i$ to $q_j$ if $| U(t)_{ij} | = 1$, where $U(t)_{ij}$ is the $q_i,q_j$-entry of $U(t)$ (we're indexing the vertices by the qubits). This represents the idea that at time $t$, the quantum state at $q_i$ was transfered to $q_j$ with probability $1$.

We can use various methods from spectral graph theory and algebraic graph theory to find out properties of the graph $G$ via the eigenvalues and eigenvectors of its adjacency matrix $A$, and more.

Our project examined conditions for a more general version of perfect state transfer, where instead of transfering between two qubits, we consider a transfer to a superposition over a subset of ther vertex set. This is called fractional revival, and comes up in entanglement generation (although I know little about this).

We found examples of graphs which exhibit a certain property called fractional cospectrality, which can be used sometimes to rule out fractional revival.

We might also try to figure out which graphs have pretty good fractional revival, which is an $\epsilon$-close version of fractional revival which is a bit less strict.

Okay I think that's enough for right now, I might talk about this later.

Thanks for reading! :)


If you want to discuss, please email me or post on ~ryankeleti/pub.

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