Let's try to simulate the time evolution generated by more general Hamiltonians. This begs the question: how do we exponentiate more complicated matrices? A naive approach comes from diagonalizing the Hamiltonian, so let's see if this is useful. We have seen that the Pauli $Z$ operator, when exponentiated, acts simply on eigenvectors of $Z$, for instance, $$ e^{i\alpha t Z} \vert 0\rangle = e^{i\alpha t} \vert 0\rangle, $$ We can perform the same trick with any Hamiltonian! Recall that the Hamiltonian $\hat{H}$ measures energy. It is an observable, and it has some associated eigenvectors $\vert E\rangle$ with eigenvalues $E$: $$ \hat{H}\vert E \rangle = E\vert E \rangle. $$ The eigenstates $\vert E \rangle$ are called the **energy eigenstates** of the system, and $E$ the **energy levels**. In the basis $(\vert E_1 \rangle, \vert E_2 \rangle, \ldots, \vert E_N \rangle)$ comprised of $N$ eigenstates, the Hamiltonian takes a particularly simple diagonal form: $$ \hat{H} = \begin{bmatrix} E_1 & & & \\ &E_2 & & \\ & & \ddots & \\ & & & E_N \end{bmatrix}. $$ To evolve the system for a time $t$ in this basis, we exponentiate as usual. But the exponential of a diagonal matrix is simply the exponential of the diagonal entries, as we can immediately show from the Taylor series: $$ \begin{align*} e^{-it\hat{H}/\hbar} & = \sum_{k=0}^\infty \frac{(-it/\hbar)^k}{k!}\hat{H}^k \\ & = \sum_{k=0}^\infty \frac{(-it/\hbar)^k}{k!}\begin{bmatrix} E_1^k & & & \\ &E_2^k & & \\ & & \ddots & \\ & & & E_N^k \end{bmatrix} \\ & = \begin{bmatrix} \sum_{k=0}^\infty\frac{(-it/\hbar)^k}{k!}E_1^k & & & \\ & \sum_{k=0}^\infty\frac{(-it/\hbar)^k}{k!}E_2^k & & \\ & & \ddots & \\ & & &\sum_{k=0}^\infty\frac{(-it/\hbar)^k}{k!} E_N^k \end{bmatrix} \\ & = \begin{bmatrix} e^{-itE_1/\hbar} & & & \\ &e^{-itE_2/\hbar} & & \\ & & \ddots & \\ & & & e^{-itE_N/\hbar} \end{bmatrix}. \end{align*} $$ It seems like all we need to do is figure out the energy levels and associated eigenstates, and the simulation becomes trivial! And you probably learned how to find eigenvalues in your linear algebra class, so it seems like a straightforward problem. So does diagonalizing indeed simplify everything? As a simple example, suppose our Hamiltonian is $$ \hat{H} = \alpha Z \otimes Z. $$ Our goal is to find the eigenvectors and eigenvalues. Since $Z|x\rangle = (-1)^x |x\rangle$, it follows that there are four eigenvectors, $\vert x y\rangle$ for $x, y \in \{0, 1\}$, with associated eigenvalues $(-1)^{x+y}$, since $$ (Z \otimes Z) \vert x y \rangle = (-1)^{x+y}\vert x y \rangle. $$ Time evolution just puts these eigenvalues in the exponential: $$ U(t) \vert xy \rangle = \exp\left[-\frac{i\alpha t}{\hbar} (-1)^{x+y}\right]\vert xy \rangle. \tag{1} $$ Since we know the eigenvalues and eigenstates, it should be easy to simulate right? Well, here we run into our first snag: we are trying to do this on a quantum computer! We have a **limited palette of gates available.** In this case it turns out to be doable, since we can "mock up" the right eigenvalue $(-1)^{x+y}$ by acting with a $Z$ rotation on a *single qubit* state $\vert x + y\rangle$. More precisely, $$ e^{-(i\alpha t/\hbar) Z}\vert x + y\rangle = \exp\left[-\frac{i\alpha t}{\hbar} (-1)^{x+y}\right]\vert x + y\rangle. $$ We now need to think about how to produce the single-qubit state $\vert x + y\rangle$ starting with the two-qubit state $\vert xy\rangle$. This turns out to be straightforward: we simply apply the CNOT gate to $\vert x y\rangle$, since $$ \text{CNOT} \vert xy\rangle = \vert x\rangle \otimes \vert x + y\rangle, $$ so the state we want lives on the second qubit. We perform a $Z$ rotation on this qubit, and restore our original state $\vert xy\rangle$ using another CNOT. As a circuit:

where $\theta = 2\alpha t/\hbar$. --- ***Exercise H.4.1.*** Write a circuit for simulating a system with Hamiltonian $\hat{H} = \alpha X \otimes X$.

Hint.

Use $HZH = X$, where $H$ is the Hadamard gate.

Solution.

Since $HZH = X$, we can convert the $X \otimes X$ Hamiltonian into the $Z \otimes Z$ Hamiltonian we just solved: $$ \hat{H} = \alpha (H \otimes H)(Z \otimes Z)(H \otimes H). $$ Since $(H \otimes H)^2 = I$, any number of powers of $\hat{H}$ can be written $$ \hat{H}^k = [\alpha (H \otimes H)(Z \otimes Z)(H \otimes H)]^k = (H \otimes H)[\alpha (Z \otimes Z)]^k(H \otimes H). $$ Plugging this into exponential gives $$ e^{-it\hat{H}/\hbar} = (H\otimes H)e^{-it\alpha Z \otimes Z/\hbar}(H\otimes H). $$ Thus, we simply bookend our circuit for exponentiating $\alpha Z \otimes Z$ with Hadamard gates, giving a new circuit:

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--- We can see that, even when we know the energy levels for a simple Hamiltonian, implementing it on a quantum computer takes work. But even discovering the energy levels is in general very hard! The reason is simply that the matrices we are dealing with are huge. For an $N \times N$ matrix, general-purpose numerical algorithms for finding the eigenvalues and eigenvectors take a polynomial number of steps in $N$. If we have a Hamiltonian acting on $n$ qubits, then $N = 2^n$, and hence the number of steps is exponential! This is the sort of exponential cost Feynman was thinking about when he proposed computers built using quantum-mechanical principles.

Implementing time evolution by diagonalizing and working with energy levels sounds good on paper, but we have run into two issues: 1. Implementing the exponential for general Hamiltonians is difficult due to the types of gates we can use. We need to be creative to find the most efficient way to put gates together that emulate the action of the exponential of the Hamiltonian. 2. Diagonalizing is a very costly procedure, since it scales exponentially with the number of qubits. Even if it heps us find the necessary gates "by hand," it is not an approach worth pursuing.

Codercise H.4.1a - Evolution of two qubits

Codercise H.4.1b - An Ising gate

Codercise H.4.1c - Built-in evolution

Codercise H.4.2a - Building the Graph Hamiltonian

Codercise H.4.2b - Guessing the ground state

Energy levels and time evolution

The ground state