10/22/2024
Entanglement Background
Quantum entanglement is something I think most people hear in science fiction or in the news – some fancy physics jargon not well understood. There’s so much research going on regarding local and nonlocal faster-than-light entanglement (a Nobel prize was awarded for this work in 2022) that show we aren’t even close to fully understanding it.
Quantum mechanics (QM) as a whole even – Max Planck just thought of considering light as a rattling gun of photons (small packets of energy) to resolve the Ultraviolet Catastrophe – and ultimately, with work from Einstein and deBroglie, out came one of the founding principles of QM, wave-particle duality.
QM is truly beautiful in that it works, allowing us to probe and understand the mind-boggingly weird nature of elementary particles. However, we don’t truly understand it. We don’t know why non-commuting observables like position and momentum have an uncertainty principle; we don’t know why all electrons are identical; we don’t know why, fundamentally, entanglement is even a thing. Of course, we know these properties are true, and we can use math to show they are true.
But is it true that God or Nature made everything this way, with a calculator in one hand continuously computing the states of every atom in the universe. Possibly?. Mother Nature may just be some supercomputing entity controlling the time evolution of every particle in the universe. Then perhaps we don’t have any free will. I shouldn’t get into philosophy here though. The point is at the base of QM lies many assumptions. QM works! Mathematical Proofs are there! But the principle axioms of QM are just Nature, I guess.
The fact is, all electrons are utterly identical, in a way that no two classical objects can ever be. It’s not just that we don’t know which electron is which; God doesn’t know which is which. – Griffiths, Introduction to QM. 3rd ed.
Entanglement is a crazy phenomena that is a consequence of QM – physically being able to alter the state of one qubit from measuring another.
Here’s an example.
Consider the following one-qubit state,
$$ |\psi\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle).$$
The above equation tells us that if we measure \( |\psi\rangle \), we will yield the state \( |0\rangle \) with \( \frac{1}{2} \) probability and likewise \( |1\rangle \) with \( \frac{1}{2} \) probability. But now let’s consider a similar setup, but with a two-qubit system, denoted \( |\Phi_+\rangle \).
$$
|\Phi_+\rangle = \frac{1}{\sqrt{2}}(|0\rangle \otimes |0\rangle + |1\rangle \otimes
|1\rangle),
$$
$$
\hspace{-35px} = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle) .
$$
Let’s say we measure the first qubit (the one on the left). On a side note, IBM Qiskit has a really weird notation where the furthest qubit on the right is actually the first. But anyways, measuring the leftest qubit yields, just like the one-qubit state $|\psi\rangle$,
$$ |0\rangle \quad \text{probability } \frac{1}{2}, $$ $$ |1\rangle \quad \text{probability } \frac{1}{2}. $$
But if you look back again at the two-qubit system, if the first qubit collapses to \( |0\rangle \) post measurement, the two-qubit state becomes \( |00\rangle \) with certainty – the second qubit collapes to \( |0\rangle \) automatically! And likewise if the first qubit collapses to \( |1\rangle \), so too must the second qubit. What this means is measuring \( |\Phi_+\rangle \) yields the two qubit states:
$$ |00\rangle \quad \text{probability } \frac{1}{2}, $$ $$ |11\rangle \quad \text{probability } \frac{1}{2}. $$
Here, the first and second qubits, immediately post-measurement, must always be equivalent. Think about this carefully. Just by measuring the state of the first qubit – we automatically force the second qubit into a state, in this case the same state.
The Partial Trace
In the above case I used an intuitive relation between the two-qubit $|\Phi_+\rangle$ system and the single-qubit $|\psi\rangle$ system to showcase entanglement. Specifically, I said measuring the leftest qubit yields, just like the one-qubit state $|\psi\rangle$, … Or in other words, I treated the first qubit as being in the same state as $|\psi\rangle$.
Well, from Quantum Information Theory, we know that entangled states don’t have a tensor product decomposition. There exist no states $|\xi_1\rangle$ and $|\xi_2\rangle$ such that
$$ |\Phi_+\rangle = |\xi_1\rangle \otimes |\xi_2\rangle.$$
Therefore, there isn’t a purely identifiable “leftest-qubit” state. Both qubits, being entangled, are intimately connected. We can’t isolate one from the ohter. But then again, it does seem like for the $|\Phi_+\rangle$ state, the leftest-qubit can be defined by $|\psi\rangle$. Or more accurately, $|\psi\rangle$ defines the left-part of $|\Phi_+\rangle$.
It turns out, there is actually a way, to trace-out, or extract, the single qubit component of a multi-qubit system, even for a system that can not be decomposed into tensor products of states. The partial-trace is kind of like the inverse of a tensor product. Rather than building states and combining them, it throws out states we don’t care about and restricts our view to the rest.
For a density operator \( \rho = \xi_1 \otimes \xi_2 \) defining the system \( \mathcal{H}_A \otimes \mathcal{H}_B \) with \( \xi_1 \) corresponding to the \( \mathcal{H}_A \) part and \( \xi_2 \) corresponding to the \( \mathcal{H}_B \) part, we define tracing away \( \xi_1 \) and \( \xi_2 \) as
$$ \text{tracing away } |\xi_1\rangle: \quad tr_{\xi_1} \xi_1 \otimes \xi_2 = (tr \xi_1) \cdot \xi_2, $$
$$ \text{tracing away } |\xi_2\rangle: \quad tr_{\xi_2} \xi_1 \otimes \xi_2 = \xi_1 \cdot (tr \xi_2). $$
It quite literally is taking the trace of the part you want to throw out.
Let’s work with \( |\Phi_0\rangle \)
$$ |\Phi_0\rangle = \frac{1}{\sqrt{2}} (|00\rangle + |11\rangle) .$$
I’ll call the left-most (first) qubit \( A \) and the second qubit \( B \). Let’s try to extract the “$A$-part” of the $|\Phi_+\rangle$ state. I’ll denote the density matrix of \( |\Phi_+\rangle \) as \( \rho_+ = |\Phi_+\rangle \langle \Phi_+|\). I’ll begin by calculating $\rho_+$.
$$ \rho_+ = |\Phi_+\rangle \langle \Phi_+\rangle $$ $$ = \frac{1}{2}[ |00\rangle \langle 00| + |00\rangle \langle 11| + |11\rangle \langle 00| + |11\rangle\langle 11|]. $$
Unfortunately, like I said before, this matrix can not be represented in a nice \( \xi_1 \otimes \xi_2 \) form. Luckily, the trace operator is linear: \( tr_a (u + v) = tr_a u + tr_a v \). And since intrinsically
$$|00\rangle \langle 00 | = (|0\rangle \otimes |0\rangle)(\langle 0 | \otimes \langle 0 |) = |0\rangle \langle 0| \otimes |0\rangle \langle 0),$$
we can just apply the linear trace operator to each summand separately. We trace out \( B \) to extract \( A \) as follows:
$$ tr_B \rho_+ = \frac{1}{2} (|0\rangle \langle 0| \cdot tr |0\rangle \langle 0 | + |0\rangle \langle 1| \cdot tr |0\rangle \langle 1| $$ $$ \hspace{62px} + \hspace{30px}| 1\rangle \langle 0 | \cdot tr|1\rangle \langle 0| + |1\rangle \langle 1| \cdot tr |1\rangle \langle 1|)$$
And since for two basis vectors \( i,j \) the matrix \( |i\rangle \langle j | \) just has a nonzero element in the \( (i,j) \) position, the only nonzero \( tr |i\rangle \langle j| \) is when \( i = j \). Therefore we can simplify to get
$$ tr_B \rho_+ = \frac{1}{2} (|0\rangle \langle 0 | + |1\rangle \langle 1|) = \frac{1}{2} \mathbb{1} .$$
Going back to the single-qubit $|\psi\rangle$ state… Considering that its post-measurement state is either
$$ |0\rangle \text{ probability } \frac{1}{2}, $$ $$ |1\rangle \text{ probability } \frac{1}{2}. $$
This means the post-measurement states’ density matrix is just the sum of all the pure state density matrices, scaled by their corresponding probabilities. i.e.,
$$ \rho_\psi = \frac{1}{2} |0\rangle \langle 0 | + \frac{1}{2}|1\rangle \langle1| = \frac{1}{2}\mathbb{1}. $$
Which is the exact same as \( tr_B \hspace{3px} \rho_+ \). We started with an entangled pair \( |\Phi_+\rangle \). By tracing out the second qubit, we effectively measured and threw out the result of the one that’s left.
It turns out we can actually use partial traces with von Neumann entropy to quantify how entangled a state is. This is extremely cool. We can actually quantify how intimate and correlated particles are. In my next post, hopefully I’ll present some code than can act as a quantum entanglement detector with some cool visualizations.
$$ \hspace{3px} $$ $$ \hspace{3px} $$