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What are the $2$ qubits of the state $a\lvert 00\rangle+b\lvert11\rangle$?

Are they $a'\vert0\rangle+b'\vert1\rangle$ and $c'\vert0\rangle+d'\vert1\rangle$?

How are they measured, and what would be the outcome on each qubit?

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  • $\begingroup$ The qubits are the leftmost qubit and the rightmost qubit. They are entangled; when you measure them in the computational (assuming your normalization factor is $1/\sqrt2$) you either get $00$ or $11$. The left qubit is $0$ or $1$, and the right qubit is the same as the left qubit (it will be measured to be the same value). $\endgroup$ – Mark S Jun 19 at 12:00
  • $\begingroup$ When you say the left qubit is 0 or 1 you mean |0⟩+|1⟩ ? measurement of the last qubit |0⟩+|1⟩ could be either 0 or 1 so it's not the same value as you stated $\endgroup$ – user1319236 Jun 19 at 12:07
  • $\begingroup$ as far as i know there's no qubit 0 or 1 but |0⟩ or |1⟩ $\endgroup$ – user1319236 Jun 19 at 12:08
  • $\begingroup$ If you want to take the left qubit and send it to Mars, and the right qubit and send it to Jupiter, they would respectively be in the (mixed) state of your edit. To me, a qubit is something physical, however. It's like saying a (classical) bit corresponds to a voltage on a wire, with a value of either $0$ or $1$, while a qubit corresponds to a spin, with a state being $\frac{1}{\sqrt 2}(\vert 0\rangle+\vert 1\rangle)$. $\endgroup$ – Mark S Jun 19 at 12:22
  • $\begingroup$ As soon as you measure the qubit, you "collapse" the value to a classical bit. $\endgroup$ – Mark S Jun 19 at 12:24
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You can't associate some pure states $|x\rangle$ and $|y\rangle$ to the subsystems of the 2-qubit system in the entangled state ($a\vert 00\rangle+b\vert11\rangle, a\ne 0, b \ne 0$ in our case). This is the definition of entanglement. You can verify that there is no vectors $|x\rangle$, $|y\rangle$ such that $|x\rangle \otimes |y\rangle = a\vert 0\rangle \otimes |0\rangle+b\vert1\rangle \otimes |1\rangle$.

But we can associate mixed states to the subsystems of this entangled state. Mixed state is a probability distribution over pure states. Both mixed states in this case coincide and equal to $\{ |0\rangle$ with probability $|a|^2$, $|1\rangle$ with probability $|b|^2\}$.

The general method of calculating (mixed) state of a subsystem is via partial trace of density matrix.

In our case $$\text{tr}_1((a|00\rangle + b|11\rangle)(\bar{a}\langle 00| + \bar{b}\langle11|)) = |a^2|\cdot |0\rangle\langle0| + |b^2| \cdot |1\rangle\langle1| $$

So the density matrix of the second qubit is $|a^2|\cdot |0\rangle\langle0| + |b^2| \cdot |1\rangle\langle1|$.
Similarly for the 1st qubit.

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