Measurement of Multiple-Qubit States

[Genady]

a. Initial state |0〉. Measure in {|+〉, |-〉}. Let's say, the resulting state is |+〉. Now measure in {|0〉, |1〉}. The resulting state can be |1〉.

This effect cannot be achieved by a single measurement. Assume such measurement exists: |1〉 = (a|0〉 + b|1〉) (a^*〈0| + b*〈1|) |0〉 =

|a|²|0〉 + a^*b|1〉. Thus, |a|² = 0, a^* = 0, |1〉 = 0.

 

[Genady]

Continuing exercise 4.21 above:

a (continued). This property is generally true because product of projectors is not generally a projector: (P₂P₁)² = P₂P₁P₂P₁ ≠ P₂P₁, generally.

 

b. If S₂ is a subspace of S₁ then P₂P₁ = P₂.

[Genady]

Continuing exercise 4.21 above:

c. Generally, if a measurement of the first qubit is H₁ =

a	b
c	d

and a measurement of the second qubit is H₂ =

x	y
z	v

then the measurement of the two-qubit system would be H = H₁ ⊗ H₂ =

ax	ay	bx	by
az	av	bz	bv
cx	cy	dx	dy
cz	cv	dz	dv

 

The reference:

The measurement A cannot be achieved by two single qubit measurements, because for ax = 0 either av = 0 or dx = 0, but in A, av = 1 and dx = 2.

 

The reference:

H₁ = 

1	0
0	π

H₂ = I

 

The reference:

The measurement C cannot be achieved by two single qubit measurements, because for ax = 2, av = 3, dx = 3, dv = 2 the ratio a/d = 2/3 and a/d = 3/2.

 

The reference:

H₁ = H₂ =

0	0
0	1

 

Edited October 15, 2023 by Genady

[joigus]

23 hours ago, Genady said:

They are asking for an observable that embodies this measurement. That is the Hermitian operator built out of the combination of the projectors.

Ok. The only thing I'd say is that it's not unique.

As to the sequence of 0s & 1s actually representing the expansion of the ordering in base 2, I had never noticed. I suppose that's why people in quantum computing chose that convention for the product. It's nice to know.

And... wow! You're on a QC binge!