# A "false" superimposed qubit vs a "real one"

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1. Aug 15, 2015

### Julian Blair

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A "false" (equally superimposed qubit) is created by mechanically firing with 50/50 probability a resonance photon at a Hydrogen atom qubit in the ground state. This qubit is sent to Alice and it now has 50/50 probability of being in state |0> or state |1>, but it is not a quantum mechanical supoerimposed state! Alice is also sent a "real" 50/50 superimposed qubit created by a Rabi process.
QUESTION: Is there any way that Alice can perform gates and measurements on the two qubits to determine which is which?

Last edited by a moderator: Aug 15, 2015
2. Aug 15, 2015

### Staff: Mentor

If the phase between the two states in the second case is preserved and nothing else disturbs the measurement, she should be able to use Rabi oscillations again. The superposition case should allow to get 100% ground state (or excited state) atoms while the probabilistic case should not.

3. Aug 15, 2015

### Julian Blair

Sorry, I don't quite understand your argument. Alice doesn't know if the "probabilistic case" is in |0> or in |1>, but in either case the qubit is in a 100% state, one or the other. Can you please give me some more details as to how you would use the Rabi oscillations on the two different qubits?
Thanks.

4. Aug 16, 2015

### Staff: Mentor

If that is the case, then where is the difference?
Also, why two different qbits? What does the second one do?

5. Aug 16, 2015

### Strilanc

You're asking about distinguishing pure states from mixed states.

A qubit in the pure state $\frac{1}{\sqrt{2}} (\left| 0 \right\rangle + \left| 1 \right\rangle)$ can be probabilistically distinguished from a qubit in the mixed state $50\% \left| 0 \right\rangle + 50\% \left| 1 \right\rangle$ by applying a Hadamard operation to the qubit before measuring it.

The Hadamard operation, equivalent to a 180 degree rotation around the diagonal X+Z axis of the Bloch sphere, turns $\left| 0 \right\rangle$ into $\frac{1}{\sqrt{2}} (\left| 0 \right\rangle + \left| 1 \right\rangle)$ but turns $\left| 1 \right\rangle$ into $\frac{1}{\sqrt{2}} (\left| 0 \right\rangle - \left| 1 \right\rangle)$. It's also its own inverse.

In your case a Hadamard operation will turn the pure state you're asking about into the state $\left| 0 \right\rangle$ while turning the mixed state into $50\% \frac{1}{\sqrt{2}} (\left| 0 \right\rangle + \left| 1 \right\rangle) + 50\% \frac{1}{\sqrt{2}} (\left| 0 \right\rangle - \left| 1 \right\rangle)$.

When you measure the Hadamard'ed pure state, you will always measure 0. When you measure the Hadamard'ed mixed state, you will measure 0 half of the time and 1 half of the time. So if someone says they are sending you a stream of qubits in the 0+1 pure state, you can gain more and more confidence that they are telling the truth as you apply Hadamard operations and measure each qubit in turn while seeing more and more 0s without seeing any 1s. You never become infinitely certain, but you do accumulate evidence.

(Note that other operations, like a square root of not, can be used in place of the Hadamard operation. Also note that the relative phase of the 0 and 1 parts of the pure state matters; $H(\frac{1}{\sqrt{2}} (\left| 0 \right\rangle + \left| 1 \right\rangle)) = \left| 0 \right\rangle$ but $H(\frac{1}{\sqrt{2}} (\left| 0 \right\rangle - \left| 1 \right\rangle)) = \left| 1 \right\rangle$.)