# B Bell's state measurement in the Bell basis

1. Jul 26, 2016

### limarodessa

Hi all

My question:

One of the four Bell's state are measuring in the Bell basis. Whether the result of measurement of this one of the four Bell's state will be the same Bell's state (just that Bell's state which are measuring) ? The each of four Bell's state is a quantum superposition of the two entangled qubits. Whether the result of measurement in the case of Bell basis will be superposition as well ? Whether there will not be collapse of the superposition ?

2. Jul 26, 2016

### StevieTNZ

So you are wondering if entangled qubits are in a certain bell-state, if you perform a bell-state measurement the entanglement between the qubits doesn't change?

3. Jul 26, 2016

### limarodessa

Yes. In my mind the result of any quantum measurement is the collapse of superposition

4. Jul 26, 2016

### Strilanc

Measurement only destroys superpositions when all the measurement results correspond to states that aren't in superposition. If the measurement results correspond to states in superposition, measurement will create superpositions.

So you have it exactly backwards. Doing a Bell state measurement won't just keep a system in a Bell state, it will force the system into a Bell state

Here is an example circuit in Quirk which demonstrates this fact:

At the start, the two top qubits are Off. Then they are each conditionally NOT-ed into the two ancilla qubits at the bottom, using Z-axis and then X-axis controls. This copies the Z-parity and X-parity of the two top qubits into the bottom qubits, where they can be measured. This is a Bell basis measurement. Then the circuit focuses on the measurement result where both parities were Off (the same basic thing happens in the other three cases). Notice that the two top qubits are no longer Off. In fact they are now entangled into the state $|00\rangle + |11\rangle$.

You can also perform the measurement inline by transforming out of of the Bell basis, performing a measurement, then transforming back into the Bell basis. This makes it very obvious that the output is in superposition. (It's so obvious that I felt like it didn't address your question, which is why I did the more complicated measurement using extra qubits instead.) Here's the inline circuit: