Why can’t a qbit be both entangled and in a pure state?

Question

Not sure if I should ask this on CS or physics SE, but here goes.

I’m reading on quantum computing, and one thing that keeps confusing me is the following basic fact about QM:

Say we have a qbit A, and a qbit B, then two possible quantum states for $A$ are:

$$|+\rangle=\frac {|0\rangle + |1\rangle}{\sqrt 2}$$

$$\frac {|00\rangle + |11\rangle}{\sqrt 2}$$

In the first case, the qbit $A$ is in a quantum superposition between $0$ and $1$. In the second case, the qbits $A$ and $B$ are ALSO in a quantum superposition, but they are now entangled. I understand that there is a difference between these states, in the sense that the second is an entangled state, whereas the first one is not.

However, what I don’t understand, is why, solely from the perspective of $A$, there is a difference between these states. I read that in the second, entangled state, $A$ (taken as an individual qbit) is not in a superposition state, but in a mixed state. I know that this is a very important basic fact about QM but I still don’t really get it.

• why is it not possible for $A$ to be both in an entangled state, AND in a superposition “with itself” (i.e. a pure state, rather than a mixed state). I.e. by what principle of quantum mechanics do we know this, and from what does this principle follow?

• Is there an intuition behind this? I.e. how would quantum mechanics no longer make sense if it weren’t the case.

Answer 1

First, let's look at the density matrix for A in the two cases. When A is in the pure state $|+\rangle$, we have \begin{align} \rho_+ = |+\rangle\langle+| = \frac{1}{2}\left(|0\rangle + |1\rangle\right)\left(\langle0| + \langle1|\right) \end{align} As a matrix, we'd write this \begin{align} \rho_{+} = \frac{1}{2} \begin{pmatrix} 1 & 1\\1& 1 \end{pmatrix}. \end{align} When A is entangled with B, we can obtain the (reduced) density matrix for A by tracing over all B states: \begin{align} \rho_A &= \frac{1}{2}\text{Tr}_B\left[\left(|0\rangle\otimes|0\rangle + |1\rangle\otimes|1\rangle\right)\left(\langle0|\otimes\langle0| + \langle1|\otimes\langle1|\right)\right]\\ &= \frac{1}{2}\left(|0\rangle\langle0| + |1\rangle\langle1|\right), \end{align} or, as a matrix, \begin{align} \rho_{A} = \frac{1}{2} \begin{pmatrix} 1 & 0\\0& 1 \end{pmatrix}. \end{align} Ok, you say, the density matrices are different. Who cares? Well, we don't really care about states, because we can't measure states directly. But we can measure observables, and we can find the expectation value for those measurements using our states. Specifically, we can find the expectation value of $\mathcal{O}$ by taking $\langle\mathcal{O}\rangle = \text{Tr}\,(\rho\mathcal{O})$. Let's look at the example of the $X$ operator. We have \begin{align} \langle X\rangle_+ &= \frac{1}{2}\text{Tr}\,(\rho_+ X)\\ &= \frac{1}{2}\text{Tr}\,\left[ \begin{pmatrix}1&1\\1&1\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}\right]\\ &=\frac{1}{2}\text{Tr}\,\begin{pmatrix}1&1\\1&1\end{pmatrix} = 1. \end{align} In the state $\rho_A$, however, we have \begin{align} \langle X\rangle_A &= \frac{1}{2}\text{Tr}\,(\rho_A X)\\ &= \frac{1}{2}\text{Tr}\,\left[ \begin{pmatrix}1&0\\0&1\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}\right]\\ &=\frac{1}{2}\text{Tr}\,\begin{pmatrix}0&1\\1&0\end{pmatrix} = 0. \end{align} We obtain different experimental predictions for the outcome of measurements in these two states, and so we can distinguish between them.

Edit: I didn't really answer your question beyond showing how states in the example you gave lead to different predictions. I would say the more broad lesson to draw is that when system A is entangled with system B, measurements on B can change the outcome of measurements on A. This leads to an additional, classical uncertainty in our measurements that is not present when A is in a pure, unentangled state.

Answer 2

Let $A$ and $B$ be prepared and then put apart.

First case: $A$ is in $|+\rangle$ state and $B$ is in $|0\rangle$. Now let's apply Hadamard gate to $A$ (I'll omit normalization): $$H_A(|00\rangle+|10\rangle) = |00\rangle$$ Then if we look at $A$ we'll observe it in $|0\rangle$, which is expected for such a superposition being hadamarded.

Second case: $A$ and $B$ are entangled. Again, we apply Hadamard gate to $A$ only: $$H_A(|00\rangle+|11\rangle) = |00\rangle+|01\rangle+|10\rangle-|11\rangle$$ This means that if we look at $A$, we'll observe it either in $|0\rangle$ or $|1\rangle$ equiprobably.

The difference is caused by positive and negative amplitudes contributing to the same $|10\rangle$ outcome in the first case (giving zero probability), while in the second case they go into different $|10\rangle$ and $|11\rangle$.

Despite the fact that we didn't do anything to $B$ (not even measured it), there's detectable difference caused by the entanglement. That's what entanglement means: talking about individual qubits doesn't make sense anymore.