I know I am wrong here and trying to find out where I am making a logical mistake. I'd appreciate it if you can help me untangle.

A. We know that the eigenvalues of Unitaries are all unimodular (length = 1)

B. we know that Quantum circuits/operator has to be unitaries ( for the closed system)

C. Also, I believe one of the objectives of Hamiltonian simulation is to find out the energy minimum, which is the eigenvalue of the eigenstate.

So, we do implement the Unitary circuits to represent Hamiltonian (exponentiation of Paulis) to find this eigenvalue but we know it is "1" from the A above.

I know I am wrong but not clear where and how.

  • 1
    $\begingroup$ the eigenvalues of unitaries have modulus 1; the eigenvalues of the Hamiltonians don't have to $\endgroup$
    – glS
    Oct 27, 2021 at 13:51
  • $\begingroup$ @glS thanks I agree in the physics perspective. but we can only implement through Unitaries in the QPU so.. $\endgroup$ Oct 27, 2021 at 14:41
  • $\begingroup$ @JohnParker The idea is if $\phi$ is an eigenvalue of $H$ with respect to some eigenvector $|\psi\rangle$ then $e^{-iHt}|\psi\rangle = e^{-i\phi t} |\psi\rangle$. With this in mind, you can extract $\phi$ from something like the Quantum Phase Estimation. $\endgroup$
    – KAJ226
    Oct 27, 2021 at 15:12
  • $\begingroup$ Thank you !! @KAJ226 $\endgroup$ Oct 27, 2021 at 19:00

1 Answer 1


The Idea:

Suppose you have the eigenvalue problem

$$ H|\psi \rangle_j = \phi_j |\psi\rangle_j$$

That is, if $\{ \phi_j \}$ are the eigenvalues of $H$ then $\{e^{-i\phi_j t} \}$ are the eigenvalues of $U = e^{-iHt}$, thanks to linear algebra.

To be more more specific, consider a specific case where $$H |\psi \rangle = \phi |\psi\rangle$$ then note that $H^2 |\psi \rangle = H\big( H|\psi\rangle \big) = H\big(\phi |\psi\rangle \big) = \phi^2 |\psi\rangle$.

You can argue by induction to get $H^n |\psi \rangle = \phi^n |\psi \rangle$. Now, using the Taylor series expansion definition, we have that

$$ e^{-iHt} |\psi \rangle = \bigg( \sum_{n=0}^\infty \dfrac{(-iHt)^n}{n!} \bigg)|\psi \rangle = \sum_{n=0}^\infty \dfrac{(-it)^n}{n!} H^n|\psi\rangle = \sum_{n=0}^\infty \dfrac{(-it)^n}{n!} \phi^n|\psi\rangle = \bigg( \sum_{n=0}^\infty \dfrac{(-i\phi t)^n}{n!} \bigg)|\psi \rangle = e^{-i\phi t} |\psi \rangle$$

Thus, being able to simulate the operator $U(t) = e^{-iHt}$ on our quantum computer, we essentially embedding the eigenvalue of $H$ to the overall phase factor of the eigenstate $|\psi\rangle$ after the evolution.

Extraction Step (QPE):

Given a unitary operator $U$ and suppose that $|\psi\rangle$ is one of its associated eigenvectors. Since $U$ is unitary, its eigenvalues have modulus 1. Thus, it can be written in the form $e^{i\theta}$. Now if you have a Hamiltonian with an eigenvalue of $\hat{\phi}$ with respect to the eigenvector $|\psi \rangle$ then as discussed above, $$U(t)|\psi \rangle = e^{-iHt} |\psi \rangle = e^{-i \hat{\phi} t} |\psi \rangle $$ For the sake of convenience later, we can rescale thing and get $e^{-i \hat{\phi} t} |\psi \rangle = e^{2\pi i \hat{\phi} t} |\psi \rangle $. Now, setting $t=1$ (for convenience) have the problem

$$ U|\psi\rangle = e^{2\pi i \phi} |\psi\rangle$$

The goal of the Quantum Phase Estimation (QPE) is to extract out the value of $\phi$. Note, we would like to have $U$ to be efficient implementable on a quantum computer.

Now let's see how QPE algorithm works. QPE algorithm breaks into two steps. The first step is to apply the controlled $U^j$ operation to ancilla (additional) qubits to essentially pushing the overall phase into the relative phase of the ancilla qubits. We need to do this because only relative phase can be observed. The second step is to apply the inverse Quantum Fourier Transform (QFT)

enter image description here

Step 1 of QPE: The first step of the algorithm is to create another quantum registers to encode the ancilla qubits. This is denoted as "Register 1" in the figure above. These ancilla qubits will all start out at the state $|0\rangle$ follow by a layer of Hadamard gate. How to choose how many ancilla qubits depend on how accurate we wish to estimate $\phi$ and with what probability we wish the algorithm to be successful. At this point, we have two type of quantum registers, the first one is encode the input state $|\psi\rangle$ and this register contains as many qubit as is necessary to store $|\psi\rangle$. The second register is to encode the ancilla qubits and this register contains as many qubit as is necessary to achieve the accuracy you want when approximating $\phi$. One qubit for one decimal place.

After applying a layer of Hadamard gate to put all the ancilla qubit in superposition, we will operate sequence of controlled-$U^{2^j}$ gate in a cascading fashion as shown in the figure above. Note that $$U^{2j}|\psi \rangle = U^{2j -1}\bigg(U|\psi\rangle\bigg) = U^{2j -1}\bigg( e^{2\pi i \phi} |\psi\rangle\bigg) = \cdots = e^{2\pi i 2^j \phi} |\psi \rangle$$ And the controlled-$U^{2^j}$ gate, $CU^{2^j}$, will applies the $U^{2^j}$ operator to the ancilla qubit, when the control qubit (ancilla qubit) is $|1\rangle$. That is, it turns the ancilla qubit (after going through Hadamard gate) from the state $\dfrac{|0\rangle + |1 \rangle}{\sqrt{2}}$ to the state $\dfrac{ |0\rangle + e^{2\pi i 2^j \phi}|1 \rangle}{\sqrt{2}}$. Even more mathematical precise,

$$ CU^{2^j}: \bigg( \dfrac{1}{\sqrt{2}} \big( |0\rangle + |1\rangle \big) \bigg)|\psi\rangle \to \dfrac{1}{\sqrt{2}} \bigg( |0\rangle |\psi \rangle + |1\rangle e^{2\pi i 2^j \phi} |\psi\rangle \bigg) = \dfrac{1}{\sqrt{2}} \bigg( |0\rangle + e^{2\pi i 2^j \phi} |1\rangle \bigg)|\psi\rangle $$

Thus, the overall process of the step 1 in the algorithm maps the ancilla qubits state $|00 \cdots 0 \rangle $ (first register) to the state
$$ \dfrac{1}{2^{n/2}} \bigg( |0 \rangle + e^{2\pi i 2^{n-1} \phi }|1 \rangle \bigg) \bigg( |0 \rangle + e^{2\pi i 2^{n-2} \phi }|1 \rangle \bigg) \cdots \bigg( |0 \rangle + e^{2\pi i 2^{0}\phi }|1 \rangle \bigg) = \dfrac{1}{2^{n/2}} \sum_{k=0}^{2^n-1} e^{2\pi i \phi k}|k\rangle $$

Note that the state $|\psi\rangle$ doesn't change throughout this process. In fact, we don't even need to apply any measurement to these qubits as it only waste our quantum resources. The key point is we have succesfully transfer the phase information (to some degree of accuracy) resulted from applying $U$ to the state $|\psi\rangle$ to the relative phase of the ancilla qubits. And this, this is what we want. This is because eventhough $-|\psi \rangle$ and $|\psi\rangle$ are essentially indistinguishable from one another, no matter which measurement basis we are using, however $|\phi \rangle - |\psi\rangle$ and $|\phi \rangle + |\psi\rangle$ can be distinguish from one another from choosing the right measurement basis.

Step 2 of QPE: Now we are at the step of retrieving the phase $\phi$, which is now encoded as the relative phase of the ancilla qubits. How can we do this? It's simple, really. We just need to perform the inverse Quantum Fourier Transform. Recall during our QFT discussion in chapter 1 about how QFT makes the following transformation

\begin{align} QFT\big( |\phi\rangle \big) &= QFT \big( |\phi_1 \phi_2 \cdots \phi_n \rangle \big) = \dfrac{1}{2^{n/2}} \bigotimes_{k=1}^n \bigg( |0\rangle + e^{2 \pi i 2^{-k} j} |1\rangle \bigg)= \dfrac{1}{2^{n/2}}\bigotimes_{k=1}^n \bigg( |0\rangle + e^{2 \pi i [0.\phi_{n-k+1} \phi_{n-k+2} \cdots \phi_n ]} |1\rangle \bigg) \\ &= \dfrac{\bigg(|0\rangle + e^{2\pi i 0.\phi_n} |1\rangle \bigg) \bigg(|0\rangle + e^{2\pi i 0.\phi_{n-1}\phi_n}|1\rangle\bigg) \bigg(|0\rangle + e^{2\pi i 0.\phi_{n-2}\phi_{n-1}\phi_n}|1\rangle\bigg) \cdots \bigg(|0\rangle + e^{2\pi i 0.\phi_1 \phi_{2}\cdots \phi_{n-1} \phi_n}|1\rangle\bigg) }{2^{n/2}} \\ \end{align}

After the first step of the QPE algorithm, we are essentially at the final equation above. Thus, by doing an inverse QFT, we will get back to the state $|\phi \rangle$, which is a good estimator for $\phi$ when measured.

** Please refer to Chapter 5.2 (pg.221) in Mike and Ike for more details analysis.


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