Measuring Qubits: Basics for Quantum Computing Beginners

[Ananthan9470]

I am newly learning quantum computing and am confused about some concepts. Suppose your qbit is the electron of a hydrogen atom and its in the state α|0> + β|1> . As far as I can understand, this means that if you measure the qbit in |0>, |1> basis, you will get a ground state electron with the probability |α|² and an exited one with the probability |β|². So what does it mean when you say you are measuring the state in some other basis? say |+>, |-> basis. How do you physically do this? Thanks.

 

[naima]

It is easier to see that things when you measure a spin along a direction with a Stern and Gerlach. You get 0> or 1> You will get another orthogonal basis by rotating the SG apparatus.

 

[Ananthan9470]

It is easier to see that things when you measure a spin along a direction with a Stern and Gerlach. You get 0> or 1> You will get another orthogonal basis by rotating the SG apparatus.

I kind of understood the thing about spin and how it is measured but I am trying to do understand the same thing for the H atom system. There has to be some similar measuring in a different basis thing for this as well right?

 

[Strilanc]

I am newly learning quantum computing and am confused about some concepts. Suppose your qbit is the electron of a hydrogen atom and its in the state α|0> + β|1> . As far as I can understand, this means that if you measure the qbit in |0>, |1> basis, you will get a ground state electron with the probability |α|² and an exited one with the probability |β|². So what does it mean when you say you are measuring the state in some other basis? say |+>, |-> basis. How do you physically do this? Thanks.

Generally what you do is apply some operations that do a basis transform so the basis you want gets mapped to the computational basis, then measure, then undo the basis transform.

For example, suppose you want to measure the X observable (instead of the Z observable that the computational basis corresponds to). The Hadamard operation happens to switch between those two basises, so you can use this circuit:

    |ψ> ──[H]──[Measure]──[H]──

or this:

    ψ ──H──•──H── ψ [collapsed]
           │
    0 ─────X───── result

If you don't want to figure out the basis transform, but have a gate whose operation is the observable's matrix U, and are able to do a controlled-U, then you can always use this circuit:

    ψ ─────U───── ψ [collapsed]
           │
    0 ──H──•──H── result

(The X observable is unique in that *both* solutions use the Hadamard gate so you can reverse which gates go on which wire and it would still work.)

Try measuring in the Y basis with this simulator.

 

[Ananthan9470]

Generally what you do is apply some operations that do a basis transform so the basis you want gets mapped to the computational basis, then measure, then undo the basis transform.

For example, suppose you want to measure the X observable (instead of the Z observable that the computational basis corresponds to). The Hadamard operation happens to switch between those two basises, so you can use this circuit:

    |ψ> ──[H]──[Measure]──[H]──

or this:

    ψ ──H──•──H── ψ [collapsed]
           │
    0 ─────X───── result

If you don't want to figure out the basis transform, but have a gate whose operation is the observable's matrix U, and are able to do a controlled-U, then you can always use this circuit:

    ψ ─────U───── ψ [collapsed]
           │
    0 ──H──•──H── result

(The X observable is unique in that *both* solutions use the Hadamard gate so you can reverse which gates go on which wire and it would still work.)

Try measuring in the Y basis with this simulator.

Thank you so much. It makes much more sense now. So basically, you apply a gate to the hydrogen atom and then you measure it in the standard basis. After the gate, the electron will be in a corresponding superposition and then when you measure, you will get outcomes depending on the new state and the new probability amplitudes(which depends on the gate you used). Did I get everything right? Thanks again.

 

[naima]

here u> and d> are spin of a particle along z.
If you couple your atom with this particle, the state is g> d> + e> u>
= g> (|+> + |->) + e> (|+> - |->) = (|g> + |e>)|+> + (|g> - |e>) |->
So if you rotate your SG along x and measure its spin, your atom is projected on one of your Schrodinger cat's state.