# Why are there only perfectly anti-correlated quantum states, not perfectly correlated?

The singlet state of two qubits is anticorrelated in every basis. For example, in the Pauli bases, it can be expressed,

$\frac{1}{\sqrt{2}} ( | 01 \rangle - | 10 \rangle) = \frac{1}{\sqrt{2}} ( | +- \rangle - | -+ \rangle) = \frac{1}{\sqrt{2}} ( | \circlearrowright \circlearrowleft \rangle - | \circlearrowleft\circlearrowright \rangle)$.

However, there is no state that is correlated in every basis. The best we can get is correlation in two Pauli bases, and anti-correlation in the third. For example,

$\frac{1}{\sqrt{2}} ( | 00 \rangle + | 11 \rangle) = \frac{1}{\sqrt{2}} ( | ++ \rangle + | -- \rangle) = \frac{1}{\sqrt{2}} ( | \circlearrowright \circlearrowleft \rangle + | \circlearrowleft\circlearrowright \rangle)$.

Why is this? Would it violate some principal, like allowing for superluminal communication?

• The anti-correlated state works that way because it has zero total angular momentum. So if one particle is spin +w the other must be -w (for w being any direction). I can't think of any simple principle that forbids the symmetric case. However, if you measure the second state you mention with one basis on the first particle, and a basis for the second particle related to the first by complex conjugate then it is symmetric. This works for larger dimensional states as well. – Dan Stahlke Jul 2 '12 at 16:20

From a purely algebraic standpoint, you could never have a state which is perfectly correlated with respect to every basis, simply because it is incompatible with the probabilistic formulation of quantum mechanics. Specifically, if you accept the current linear formulation of quantum mechanics, any two-qubit state which had perfect positively correlated measurement outcomes for each basis would have a density operator whose eigendecomposition includes negative probabilities. So, you would either have to explain what negative probabilities correspond to, or you must abandon the linear structure of quantum mechanics. Either alternative is likely to raise people's hackles, and give rise to exotic phenomena such as superluminal signalling.

The proof is not difficult, if you are comfortable with Pauli (spin) operators. On two-qubit states, the fact that the singlet is anticorrelated in all bases is related to the fact that $$\rho \;:=\; |\Psi^-\rangle\langle\Psi^-| \;\;=\;\; \tfrac{1}{4}\bigl(\mathbf 1 \otimes \mathbf 1 \;-\; X \otimes X \;-\; Y \otimes Y \;-\; Z \otimes Z\bigr)$$ which indicates at least that the outcomes of X, Y, and Z observable measurements will be anticorrelated: for each observable $\mathcal O \in \{X \otimes X, \;Y \otimes Y,\; Z \otimes Z\}$, we have $\mathop{\mathrm{Tr}}(\mathcal O \rho) = -1$. This is the maximum negative value such an expectation value may have by virtue of the fact that $\rho$ has eigenvalues bounded between zero and one, and that each observable $X \otimes X$,  $Y \otimes Y$, and $Z \otimes Z$ have eigenvalues $\pm 1$. This also means that the anticorrelation is the greatest possible; that is, the outcomes of the measurements will be opposite with certainty.

Suppose instead that you wanted a Hermitian operator $\varrho$, whose eigenvalues have magnitude at most one and which sum to exactly one, representing a state which was perfectly positively correlated in each basis. In particular, such an operator would have to satisfy $\mathop{\mathrm{Tr}}(\mathcal O \varrho) = +1$ for each $\mathcal O \in \{X \otimes X, \;Y \otimes Y,\; Z \otimes Z\}$, so we could decompose it as $$\varrho \;\;=\;\; \tfrac{1}{4}\bigl(\mathbf 1 \otimes \mathbf 1 \;+\; X \otimes X \;+\; Y \otimes Y \;+\; Z \otimes Z \;+\; R\bigr),$$ where $R$ is some operator consisting of a linear combination of distinct Pauli operators $P \otimes Q$ for $P \ne Q$ (taking advantage of the fact that we may use the Pauli operators as a basis with real coefficients, for Hermitian operators on two qubit state vectors). In particular, the operators $(\mathbf 1 \otimes \mathbf 1)R$, $(X \otimes X)R$, $(Y \otimes Y)R$, and $(Z \otimes Z)R$ have zero trace. Because $\varrho^2$ has trace bounded above by one, and as $\mathbf 1 \otimes \mathbf 1$ is the only two-qubit Pauli operator with non-zero trace, we may expand $\varrho^2$ to obtain \begin{align*} 1 \;& \geqslant\; \mathop{\mathrm{Tr}}(\varrho^2) \;=\; \cdots \\ &= \tfrac{1}{16}\Bigl(\mathop{\mathrm{Tr}}\bigl((\mathbf 1 \otimes \mathbf 1)^2\bigr) + \mathop{\mathrm{Tr}}\bigl((X \otimes X)^2\bigr) + \mathop{\mathrm{Tr}}\bigl((Y \otimes Y)^2\bigr) + \mathop{\mathrm{Tr}}\bigl((Z \otimes Z)^2\bigr) + \mathop{\mathrm{Tr}}\bigl(R^2\bigr)\Bigr) \\ &= \tfrac{1}{16}\Bigl(4 + 4 + 4 + 4 + \mathop{\mathrm{Tr}}(R^2)\Bigr). \end{align*} Then $R^2$ must be traceless; but as the square of a Hermitian operator, it can only be traceless if all of the eigenvalues of $R$ were zero, that is if $R = 0$. Thus it follows that $$\varrho \;\;=\;\; \tfrac{1}{4}\bigl(\mathbf 1 \otimes \mathbf 1 \;+\; X \otimes X \;+\; Y \otimes Y \;+\; Z \otimes Z\bigr) \;=\; \tfrac{1}{2}(\mathbf 1 \otimes \mathbf 1) - \rho,$$ where $\rho$ again is the density operator for the singlet state which we wrote above. Because the Bell states form an eigenbasis for both $\rho$ and $\mathbf 1 \otimes \mathbf 1$, they form an eigenbasis for $\varrho$ as well, yielding the following eigen-decomposition of $\varrho$: $$\varrho \;=\; \tfrac{1}{2}|\Phi^+\rangle\langle\Phi^+| \;+\; \tfrac{1}{2}|\Phi^-\rangle\langle\Phi^-| \;+\; \tfrac{1}{2}|\Psi^+\rangle\langle\Psi^+| \;-\; \tfrac{1}{2}|\Psi^-\rangle\langle\Psi^-|.$$ This is not a positive operator, and thus not a density operator. In particular, if you performed a Bell measurement, there isn't any well-defined meaning to the probability distribution you would obtain, if you supposed that you could have a system in a state described by $\varrho$.

• As a side note, the (non-positive) operator $\varrho$ would indeed have perfect correlations for each measurement basis. This can be seen from the fact that the three triplet states have the same eigenvalues, and are an orthonormal basis for the symmetric subspace. So measuring with respect to any basis $A$ on both qubits should give the same correlation as for some fixed basis, e.g. the standard basis. By construction, the $Z Z$ correlator for $\varrho$ is $+1$, so that a measurement with respect to any basis would indeed be perfectly correlated --- if we could prepare $\varrho$, that is. – Niel de Beaudrap Jul 2 '12 at 16:44

$\newcommand{\ket}{\left|#1\right>}$ Niel de Beaudrap's answer led me to the following answer, based on the stabilizer code formalism.

Basically, a state $\ket\psi$ is said to be stabilized by the operator $S$ if and only if $S\ket{\psi}=\ket\psi$. Obviously, if $\ket\psi$ is stabilized by $S_1$ and $S_2$, it is stabilized by $S_1\cdot S_2$. In other words, the set of the stabilizers of a state is a group.

Let $\ket\psi$ be a bipartite state, and $O$ be a binary ($\pm1$) local observable. The following equivalences hold :

• $\ket\psi$ is correlated in $O$ $\Leftrightarrow O\otimes O\ket\psi=\ket\psi \Leftrightarrow$ $\ket\psi$ is stabilized by $O\otimes O$
• $\ket\psi$ is anticorrelated in $O$ $\Leftrightarrow O\otimes O\ket\psi=-\ket\psi \Leftrightarrow$ $\ket\psi$ is stabilized by $-O\otimes O$

The singlet state can be defined by the state anticorrelated in the Pauli operators $X$ and $Y$, i.e. the state stabilized by $-X\otimes X$ and $-Y\otimes Y$. It is therefore stabilized by $$(-X\otimes X)(-Y\otimes Y)=(-1)^2XY\otimes XY= iZ\otimes iZ=-Z\otimes Z,$$ where we have used the $XY=iZ$ relation. $\ket\psi$ is thus stabilized by $-Z\otimes Z$, or anticorrelated in Z.

In the same way, let's look at a state correlated in $X$ and $Y$ (the same reasoning holds for $\{Y,Z\}$ and $\{Z,X\}$. This state would be stabilized by $$(X\otimes X)(Y\otimes Y)=-Z\otimes Z,$$ and is then anticorrelated in $Z$.