What does the minus sign in the four bell states represent?

Question

I am in grade 11, so answers as simple as possible. I understand that in quantum teleportation, the bell measurement must be made on the teleportee and the sender, and I understand that yields one of four possible bell states:

I have yet to understand the actual REASONING behind some of them. For example, the first one shown says 'Both Alice and Bob's qubits are the same.' I got that. The second one says 'Either Alice or Bob's qubit is 1.' I got that. But what are the other two? I don't understand analogies for them, or if the subtraction is significant.

Any help appreciated, preferably in the form of an analogy!

Answer 1

Quantum states are defined by more than the outcome probabilities in a fixed measurement setting.

As you noticed, $|\Phi^+\rangle$ and $|\Phi^-\rangle$ correspond to the same identical output probabilities when measured in a fixed basis (here the computational basis). However, they react in different ways, and result in different outcomes, when measured in different ways.

You can think of the sign between the two terms as a way to describe concisely how exactly these states behave when measured in different ways

Answer 2

If we measure states $|\Psi^-\rangle$ and $|\Psi^+\rangle$ in computational basis, both look identical: if one qubit is measured $|0\rangle$, the other is measured $|1\rangle$; if one qubit is measured $|1\rangle$, the other is measured $|0\rangle$.

Similarly, if we measure states $|\Phi^-\rangle$ and $|\Phi^+\rangle$ in computational basis, both look identical: if one qubit is measured $|0\rangle$, the other is measured $|0\rangle$; if one qubit is measured $|1\rangle$, the other is measured $|1\rangle$.

But let us change to Hadamard basis:

\begin{align} |+\rangle=\frac{|0\rangle+|1\rangle}{\sqrt{2}}\\ |-\rangle=\frac{|0\rangle-|1\rangle}{\sqrt{2}} \end{align}

The inverse transformation is

\begin{align} |0\rangle=\frac{|+\rangle+|-\rangle}{\sqrt{2}}\\ |1\rangle=\frac{|+\rangle-|-\rangle}{\sqrt{2}} \end{align}

Now

\begin{align} |\Psi^-\rangle&=\frac{1}{\sqrt{2}}\left(\frac{|+\rangle+|-\rangle}{\sqrt{2}}\otimes \frac{|+\rangle-|-\rangle}{\sqrt{2}} - \frac{|+\rangle-|-\rangle}{\sqrt{2}}\otimes \frac{|+\rangle+|-\rangle}{\sqrt{2}}\right)\\ &=\frac{1}{\sqrt{2}}\left(-|+\rangle|-\rangle+|-\rangle|+\rangle \right) \end{align}

\begin{align} |\Psi^+\rangle&=\frac{1}{\sqrt{2}}\left(\frac{|+\rangle+|-\rangle}{\sqrt{2}}\otimes \frac{|+\rangle-|-\rangle}{\sqrt{2}} + \frac{|+\rangle-|-\rangle}{\sqrt{2}}\otimes \frac{|+\rangle+|-\rangle}{\sqrt{2}}\right)\\ &=\frac{1}{\sqrt{2}}\left(|+\rangle|+\rangle-|-\rangle|-\rangle \right) \end{align}

We can see that in Hadamard basis $|\Psi^-\rangle$ and $|\Psi^+\rangle$ states behave differently:

• with $|\Psi^-\rangle$ state, if one qubit is measured $|+\rangle$, the other is measured $|-\rangle$; if one qubit is measured $|-\rangle$, the other is measured $|+\rangle$.

• with $|\Psi^+\rangle$ state, if one qubit is measured $|+\rangle$, the other is measured $|+\rangle$; if one qubit is measured $|-\rangle$, the other is measured $|-\rangle$.

You can check that $|\Phi^-\rangle$ and $|\Phi^+\rangle$ states also behave differently in Hadamard basis.

Answer 3

So the Bell states are states of entanglement this means that the state of qubit one is now correlated to the state of qubit two, if you measure one you know information about the other.

If you have a two qubit system, for which both qubits are in indepent superpostion, i.e $|q_1\rangle = |q_2\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle$, so both qubits are in state $|1\rangle$ and $|0\rangle$ at the same time! Then the combined state of the system you get a state of all four combinations of $|1\rangle$ and $|0\rangle$ for both qubits:

$|\psi\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) \otimes \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) = \frac{1}{2}(|00\rangle + |01\rangle + |10\rangle + |11\rangle)$

Now if we say that $|q_1\rangle$ and $|q_2\rangle$, are correlated, that means that if $|q_1\rangle$ is in a state $|1\rangle$ or $|0\rangle$ we can start off by determining that there are only two possible combinations of correlation; either $q_1$ and $q_2$ are in the same state so $|11\rangle$ and $|00\rangle$, or that $q_1$ and $q_2$ are in opposite states $|01\rangle$ and $|10\rangle$, so if they are always opposite from each other we can't possibly have the qubits in states where they are the same, and vice versa! This gives you the foundation of the Bell states you see.

Now we won't get bogged down into the math, but the different signs appear because of depending on the start state of the qubits (before we put them into superposition). If a qubit starts in state $|0\rangle$ when you put this qubit into superposition(via a $H$ gate) $H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$, so $|0\rangle$ and $|1\rangle$, however if you put $|1\rangle$ into superposition you get $H|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$, notice the minus sign. So the Bell States with the minus sign appear when you start with $q_1$ and $q_2$ in opposite states, one is in $|0\rangle$ and the other in $|1\rangle$ (if you do some multiplication of the two superposed states you can see how the minus sign appears), and the Bell states with the + sign occur when both $q_1$ and $q_2$ start in the same state!

So what is the implication of the minus sign, when measuring the probability of a state we can ignore the signs, but this is the only we can ignore them) because the probability is taking the square! However the minus signs are important as they determine unique states of the system, the Bell States form their own orthogonal basis, i.e. it is not possible to make another Bell State from a combination of adding or subtracting the others. Doing computation on these states with different signs may lead us to different results, so we can't ignore them!