Intuition for why $\frac{|00\rangle+|11\rangle}{\sqrt{2}}$ can be written as $\frac{|++\rangle+|--\rangle}{\sqrt{2}}$

Question

In analyzing measurement of $\frac{|00\rangle+|11\rangle}{\sqrt{2}}$ in the local $|+\rangle$, $|−\rangle$ basis, through algebra manipulation, the initial state is first written as $\frac{|++\rangle+|--\rangle}{\sqrt{2}}$.

I understand how to get to this result algebraically. However, I'm surprised by how similar this state is compared to the state of $\frac{|00\rangle+|11\rangle}{\sqrt{2}}$ - you literally substitude the 0 with +, 1 with -. Thus, I am wondering if there's any intuition for why the two states look so similar?

Answer 1

Notice that it is somewhat a coincidence of that particular Bell state and choice of basis. The states $|0\rangle$ and $|1\rangle$ are in the $z$ axis of the Bloch sphere and $|+\rangle$,$|-\rangle$ are in the $x$-axis.

The state you chose is a sum of products of single states that are the same, and it turns out that the same is true when you convert it to the basis of the $x$ axis, as $$H\otimes H(|00\rangle+|11\rangle)/\sqrt{2}=(|++\rangle+|--\rangle)/\sqrt2$$ but if you had started with a different Bell state you may get something different, for example $$H\otimes H(|00\rangle-|11\rangle)/\sqrt2=(|+-\rangle+|-+\rangle)/\sqrt2$$ which looks a bit different in the $x$-basis as now the state is a linear combination of tensor products of two orthogonal states. We can also take your initial state and convert it to the $y$-axis with states $|\circlearrowleft\rangle$ and $|\circlearrowright\rangle$ using $HS^\dagger$: $$(H\otimes H)(S^\dagger\otimes S^\dagger)(|00\rangle+|11\rangle)/\sqrt{2}=(|\circlearrowright\circlearrowleft\rangle+|\circlearrowleft\circlearrowright\rangle)/\sqrt{2}$$ which again returns a state of sums of products of states in opposite directions.

What can be said is then that any mutual unitary transformation $U\otimes U$ for some single unitary gate $U$ is going to transform a Bell state into another Bell state in another basis (which may or may not have the same form).

Is there a Bell state that is more robust to this kind of transformations? Yes, the Bell state $|\psi^-\rangle=(|01\rangle-|10\rangle)/\sqrt2$ as $$H\otimes H|\psi^-\rangle=(|+-\rangle-|-+\rangle)/\sqrt2$$ $$(H\otimes H)(S^\dagger\otimes S^\dagger)|\psi^-\rangle=(|\circlearrowright\circlearrowleft\rangle-|\circlearrowleft\circlearrowright\rangle)/\sqrt{2}$$ which is always a sum of products of opposite states no matter the basis (it is called a Werner state). In this case this is possible because unitary transformations are generated by angular momentum operators, and $|\psi^-\rangle$ is a singlet which has $0$ angular momentum.

Answer 2

Yet another derivation

Applying a local unitary $U^A$ on the first subsystem of a bipartite maximally entangled state $|\psi^{AB}\rangle$ is equivalent to applying a possibly different unitary $V^B$ on the second subsystem

$$ (U^A\otimes I)|\psi^{AB}\rangle = (I\otimes V^B)|\psi^{AB}\rangle\tag1. $$

In the specific case of the Bell state $(|00\rangle+|11\rangle)/\sqrt{2}$, one transformation is the transpose of the other

$$ (I\otimes U) \frac{|00\rangle+|11\rangle}{\sqrt2} = (U^T \otimes I) \frac{|00\rangle+|11\rangle}{\sqrt2}\tag2 $$

as is easy to check by direct calculation. This suggests that in order to express $(|00\rangle+|11\rangle)/\sqrt{2}$ in the $|+\rangle$, $|-\rangle$ basis on each qubit, we can write

$$ \begin{align} \frac{|00\rangle+|11\rangle}{\sqrt2} &= (I \otimes H H^T) \frac{|00\rangle+|11\rangle}{\sqrt2} \\ &= (H \otimes H) \frac{|00\rangle+|11\rangle}{\sqrt2} \\ &= \frac{|{++}\rangle+|{--}\rangle}{\sqrt2} \end{align}\tag3 $$

where the first equality uses $HH^\dagger = HH^T = I$ and the second uses $(2)$.

Intuitive interpretation

This derivation of the identity has an intuitive interpretation. First, note that it is the hallmark of maximally entangled states that every local unitary on one subsystem corresponds to a local unitary on the other subsystem. It is thus possible for a pair of local basis change transformations, such as the two Hadamards above, to jointly correspond to identity. This happens when the two transformations cancel out as in $(3)$ above. Whenever this is the case, the state can be written down as many equivalent expressions that differ by the local basis change transformations, but look otherwise the same, i.e. have the same number of terms and the same amplitudes.

One intuitive way to think about how it is possible for the action of $U$ on one subsystem in equation $(2)$ to be equivalent to the action of $U^T$ on the other potentially distant subsystem is provided by the ER=EPR conjecture which views the Bell state $(|00\rangle+|11\rangle)/\sqrt{2}$ as a microscopic wormhole.

Answer 3

One strong element of the intuition is related to the fact that it is maximally entangled. One definition of a pure state $|\psi\rangle$ being maximally entangled is that the individual systems have density matrices $$ \rho_A=\text{Tr}_B(|\psi\rangle\langle\psi|)=\frac{I}{d} $$ where $d$ is the dimension of $A$'s Hilbert space.

Now, one thing that the reduced density matrix tells you is the basis of the Schmidt decomposition of $|\psi\rangle$. But the cool thing about the identity is that you can use any orthonormal basis you want thanks to the relation $I=\sum_k|u_k\rangle\langle u_k|$ for all orthonormal bases. So, in fact, $|\psi\rangle$ can have a very similar decomposition using any orthonormal basis (but there is a slight variation in what states of the second system the basis states are correlated with)

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Another way of looking at it is that the state $|\psi\rangle$ is a $+1$ eigenstate of the two operators $X\otimes X$ and $Z\otimes Z$ (indeed, this fact uniquely defines the state). So, there's a symmetry between the $X$ and $Z$ bases, which is exactly what you're seeing.