Measurement of Multiple-Qubit States

[Genady]

(From Rieffel, Eleanor G.; Polak, Wolfgang H.. Quantum Computing: A Gentle Introduction.)

a. 

1	0
0	0

b. = (|0>+|1>)<0| - i(|0>-|1>)<1| = |0><0| + |1><0| - i|0><1| +i|1><1|

1	-i
1	i

c. 

1	0	0	0
0	1	0	0
0	0	0	0
0	0	0	0

d.  

1	0	0	0
0	1	0	0
0	0	0	1
0	1	0	0

e. = 1/2(|00>+|11>)(<00|+<11|) = 1/2(|00><00|+|00><11|+|11><00|+|11><11|)

1/2	0	0	1/2
0	0	0	0
0	0	0	0
1/2	0	0	1/2

 

All right?

Edited October 12, 2023 by Genady

[Genady]

a. 1/√2(|0><0|+|0><1|+|1><0|-|1><1|)

b. |0><1|+|1><0|

c. |0><1|-|1><0|

d. |0><0|-|1><1|

e. 23|00><00|-5|01><01|+9|11><11|

f. (|0><1|+|1><0|)⊗(|0><1|+|1><0|) = |00><11|+|01><10|+|10><01|+|11><00|

g. (|0><1|+|1><0|)⊗(|0><0|-|1><1|) = |00><10|-|01><11|+|10><00|-|11><01|

h. 1/2(|00><00|+|00><01|+|01><00|-|01><01|+|00><10|+|00><11|+|01><10|-|01><11|+

           |10><00|+|10><01|+|11><00|-|11><01|-|10><10|-|10><11|-|11><10|+|11><11|

i. P₁ = |++><++| + |-- >< --|

   P₂ = |+- ><+-| + |-+>< -+|

 

Correct?

[Genady]

Let's consider a direct sum decomposition of the space V = S ⊕ S^⊥, where S and S^⊥ are orthogonal, and let's consider a projection operator P of V onto S.

Take two vectors (|v〉+|v^⊥〉) and (|w〉+|w^⊥〉) in V, where |v〉 and |w〉 are in S and |v^⊥〉 and |w^⊥〉 are in S^⊥. By definition, P(|v〉+|v^⊥〉)=|v〉.

(〈w|+〈w^⊥|)(P(|v〉+|v^⊥〉)) = (〈w|+〈w^⊥|)(|v〉) = 〈w|v〉.

OTOH, by definition, P(|w〉+|w^⊥〉)=|w〉.

((〈w|+〈w^⊥|)P)(|v〉+|v^⊥〉) = (〈w|)(|v〉+|v^⊥〉) = 〈w|v〉.

Thus,

(〈w|+〈w^⊥|)(P(|v〉+|v^⊥〉)) = ((〈w|+〈w^⊥|)P)(|v〉+|v^⊥〉),

i.e., P is its own adjoint.

Edited October 12, 2023 by Genady

[joigus]

10 hours ago, Genady said:

b. = (|0>+|1>)<0| - i(|0>-|1>)<1| = |0><0| + |1><0| - i|0><1| +i|1><1|

1	-i
1	i

Give me some time to check conventions. They differ from QC to "atoms-and-molecules" QM...

But arent you missing 1/sqrt(2) factors there? Plus and minus kets should be normalised.

Didn't get the time to check everything else yet. One of the most common mistakes is when taking the adjoint of |0> + i |1> which should be <0|-i<1|

but I'm guessing you didn't make that mistake.

[Genady]

4 minutes ago, joigus said:

But arent you missing 1/sqrt(2) factors there?

Yes, I am. Thank you.

Thank you for checking. The book does not give answers of the exercises, so I don't have a way to check myself.

[joigus]

55 minutes ago, Genady said:

Yes, I am. Thank you.

Thank you for checking. The book does not give answers of the exercises, so I don't have a way to check myself.

You're most welcome.

The rest of the expansions seem right to me.

As to proof 4.3, I see what you're doing there, and it's correct too, AFAICS. The only glitch is for these kind of proofs is that it's perhaps best to drop Dirac's notation, because it kind of stands in the way of distinguishing the vector, the operator, and the action of the inner product more clearly, so you wouldn't have to use the --somewhat awkward, IMO-- double parenthesis on your last line.

So, for example, I would write something like,

for every \( w \), \( v \) in \( \mathscr{H} \) (the Hilbert space of states),

\[ \left( w, P\right) = \left( P^{\dagger}w,v \right) \]

(That is just a definition of \( P^{\dagger} \), of course)

I would also write,

\[ v=v_{perp}+v_{parallel} \]

etc, with,

\[ Pv_{perp}=0 \]

\[ Pv_{parallel}=v_{parallel} \]

 

for an arbitrary vector \( v \).

And then, as you say,

\[ \left( w, Pv \right) = \left( w,w_{\parallel} \right) = \left( w_{\parallel}, v_{\parallel} \right) = \left( w_{\parallel}, v \right) = \left( Pw, v \right) \]

 

So indeed \( P^{\dagger}=P.

To me, it's a bit more transparent with this notation, but I understood what you meant, and if you think about it we're saying the same thing.

The devil is in the details, as they say. In infinite dimension one would have to be much more careful than this, but I don't have the chops for it. 😊

 

Ok. Something got messed up in the LaTeX rendering, I'm afraid... I hope you can see what it is.

I always have to be very careful that my text is not interpreted as rich text at some point (eg, when the editor refreshes) so the rendering is messed up.

It might have to do with the software at my end. I dunno.

[Genady]

1 hour ago, joigus said:

it's perhaps best to drop Dirac's notation, because it kind of stands in the way of distinguishing the vector, the operator, and the action of the inner product more clearly, so you wouldn't have to use the --somewhat awkward, IMO-- double parenthesis on your last line.

Yes, I think we do the same thing in different notations.

I use the Dirac's notation because it seems to be the standard in this book and maybe in QC. The nested parenthesis might be easier to read if I give more spaces, e.g.,

(〈w|+〈w^⊥|) (P (|v〉+|v^⊥〉) ) = ( (〈w|+〈w^⊥|) P ) (|v〉+|v^⊥〉).

Perhaps, I should've made the vectors |q> = |v〉+|v^⊥〉 and |r> = |w〉+|w^⊥〉, and the last line would be then

<r| (P|q>) = (<r|P) |q>

Edited October 12, 2023 by Genady

[joigus]

Yes. On the one hand stating more clearly that what you're saying is that every \( \left| q \right\rangle) \) in the space can be written as \( \left| v \right\rangle + \left| v_{\perp} \right\rangle \) (what I've written as \( v=v_{\parallel}+v_{\perp} \) ), and on the other hand, noticing that taking the action of \( P \) from the second to the first factor in the scalar product is not simply "looking at it as acting on its left", but also complex conjugating.

In matrix notation: q⁺Pr=(P⁺q)⁺r

where "+" as a superindex means complex conjugate and transpose. It's only when P⁺=P that you can write, q⁺P=(P⁺q)⁺=(Pq)⁺

Otherwise, you can just say that q⁺P=(P⁺q)⁺, which is always true, because it's a definition.

This you cannot say clearly in Dirac notation, as Dirac notation automatically assumes that any operator "sandwiched" between both factors is Hermitian. Otherwise the notation is ambiguous. That's why someone as careful as Weinberg sometimes drops it in his proofs. Weinberg proved most everything he said, so he was very careful about these questions.

The point of the exercise being that the eigenvalues of any projector are always real. Something that is not true for any linear operator.

Another way of seeing it is by investigating the eigenvalue equation. As P²=P, the eigenvalues must satisfy p²=p, so either p=0 or 1.

[Genady]

1 hour ago, joigus said:

Otherwise, you can just say that q⁺P=(P⁺q)⁺, which is always true, because it's a definition.

On this exact point:

(A|x〉)⁺ = (A|x〉)^*T = (A^*|x〉^*)^T = |x〉^*TA^*T= 〈x|A⁺

Edited October 12, 2023 by Genady

[joigus]

Exactly. As you see, Dirac's notation can handle it by using extra round braces. But it kind of becomes more awkward --or perhaps just uglier-- for those cases.

And it's a huge inconvenience when you want to talk about time inversion operation, because it's not even linear, but antilinear, so it has a complex conjugation in its guts that "interferes" with the complex conjugation involved in passing from ket to bra.

[Genady]

|0><0|+|7><7|-(|1><1|+|2><2|+...+|6><6|)

[Genady]

|0><0|+|3><3|+|5><5|+|6><6|-(the other four: 1, 2, 4, 7)

[Genady]

(|1><1|+|2><2|+|4><4|)+2(|3><3|+|5><5|+|6><6|)+3|7><7|

The number of 1-bits is the eigenvalue.

[Genady]

P₁+P₂-P₃-P₄

[Genady]

a. O = ∑_i λ_iP_i . For some i=j, P_j|ψ〉 = |φ〉. When the same measurement applied to |φ〉, P_j|φ〉=|φ〉, while for all i≠j, P_i|φ〉=0.

b. Operators P but not O act on state vectors.

[Genady]

For the reference:

a. |01〉 = 1/√2 (|Φ⁺〉-|Φ^-〉); 1/2 each

    |10〉 = 1/√2 (|Ψ⁺〉+|Ψ^-〉); 1/2 each

    |11〉 = 1/√2 (|Ψ⁺〉-|Ψ^-〉); 1/2 each

b. = 1/√2 (a₀₀(|Φ⁺〉+|Φ^-〉)+a₀₁(|Φ⁺〉-|Φ^-〉)+a₁₀(|Ψ⁺〉+|Ψ^-〉)+a₁₁(|Ψ⁺〉-|Ψ^-〉))

    = 1/√2 ((a₀₀+a₀₁)|Φ⁺〉+(a₀₀-a₀₁)|Φ^-〉+(a₁₀+a₁₁)|Ψ⁺〉+(a₁₀-a₁₁)|Ψ^-〉)

    Outcomes: |Φ⁺〉, |a₀₀+a₀₁|²/2; |Φ^-〉, |a_00-a₀₁|²/2; |Ψ⁺〉, |a₁₀+a₁₁|²/2; |Ψ^-〉, |a₁₀-a₁₁|²/2.

[Genady]

For the reference:

a. B = (|0〉〈0|+π|1〉〈1|)⊗(|0〉〈0|+|1〉〈1|)

b. Subspace decomposition of Q is V=S₀⊕S₁ and subspace decomposition of I is V=S. Subspace decomposition of Q⊗I is V⊗V=S₀⊗S⊕S₁⊕S.

c. Subspace decomposition of Q is V=S₀⊕S₁ and subspace decomposition of I is V=S. Subspace decomposition of I⊗Q is V⊗V=S⊗S₀⊕S⊕S₁. It measures the second qubit.

[Genady]

a. Rows of U^† are conjugate transposed columns of U. Multiplying them by other columns results in 0, thus they are orthogonal. Multiplying them by the same column results in 1, thus they are normalized.

b. (UOU^-1)^† = (U^-1)^†O^†U^†=UOU^-1

c. The eigenvalue equation, (A-λI)v=0, constitutes n equations with n+1 unknowns. Hence it has at least one nontrivial solution.

d. Use induction. Any 1x1 matrix is upper triangular. Assume it's true for any nxn matrix and take A to be a (n+1)x(n+1) matrix.

Take the eigenvector v of A and choose any n orthonormal vectors orthogonal to v. Make matrix B, which has (λ,0,...0)^T as its first column and the other chosen orthonormal vectors as its other columns. B is unitary. Make matrix C = B^-1AB.

C(1,0,...0)^T = B^-1Av = B^-1λv= λB^-1v = (λ,0,...0)^T. Thus, (λ,0,...0)^T is the C matrix' first column.

The nxn matrix to the lower right of this column can be made upper triangular by the induction assumption. This completes the proof.

[Genady]

(Finishing exercise 4.16 above)

e. Let's make an upper triangular matrix T=UAU^-1. If A is Hermitian, then

T^† = (U^-1)^†A^†U^† = UAU^-1 = T

Thus, T is upper triangular and T^† is upper triangular, i.e., T is diagonal. It has n orthonormal eigenvectors (1,0,...0), ... (0,...0,1) with the same eigenvalues as A. Then the columns of U^-1 are eigenvectors of A. They are orthogonal. Thus, the direct sum of their corresponding eigenspaces gives the whole space.

[joigus]

On 10/13/2023 at 12:30 AM, Genady said:

|0><0|+|7><7|-(|1><1|+|2><2|+...+|6><6|)

I understand they're asking for a projector that embodies this measurement. You should not use a difference of projectors, as a difference of projectors is not a projector, even though the addition of two mutuallly orthogonal projectors is:

\[ \left(P+P^{\perp}\right)^{2}=P^{2}+\left(P^{\perp}\right)^{2}+PP^{\perp}+P^{\perp}P=P+P^{\perp} \]

But,

\[ \left(P-P^{\perp}\right)^{2}=P^{2}+\left(P^{\perp}\right)^{2}+PP^{\perp}+P^{\perp}P=P+P^{\perp} \]

But the operator you've written does commute with both \( P \) and \( P^{\perp} \), so it could implement such measurement, IMO.

People who work in QC are very old-school though, so I assume, for them, nothing but a projector will do.

Also, I don't remember the ordering of the computational basis. Clearly,

\[ \left|0\right\rangle :=\left|000\right\rangle \]

and,

\[ \left|7\right\rangle :=\left|111\right\rangle \]

But the rest I forget.

Sorry, there should be a couple of minus signs in the last equality, but it doesn't matter, as they are mutually orthogonal. That's why I was sloppy.

[Genady]

47 minutes ago, joigus said:

I understand they're asking for a projector that embodies this measurement. You should not use a difference of projectors, as a difference of projectors is not a projector, even though the addition of two mutuallly orthogonal projectors is:

 

(P+P⊥)2=P2+(P⊥)2+PP⊥+P⊥P=P+P⊥

 

But,

 

(P−P⊥)2=P2+(P⊥)2+PP⊥+P⊥P=P+P⊥

 

But the operator you've written does commute with both P and P⊥ , so it could implement such measurement, IMO.

People who work in QC are very old-school though, so I assume, for them, nothing but a projector will do.

Also, I don't remember the ordering of the computational basis. Clearly,

 

|0〉:=|000〉

 

and,

 

|7〉:=|111〉

 

But the rest I forget.

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

The other numbers are just decimal representations of the binary sequences, i.e. 1=001, 2=010, 3=011, ...

The observable

|0><0|+|7><7|-(|1><1|+|2><2|+...+|6><6|)

has an eigenspace with eigenvalue 1 of states with equal bit values, and another eigenspace with eigenvalue -1 of states with unequal bit values.

Edited October 14, 2023 by Genady

[Genady]

a. An unentangled state can be factorized by single-qubit states as v⊗w. A single-qubit measurement can be represented by an observable Q⊗I. The measurement with Q would change v into v'. The measurement with I would leave w unchanged. The result would be v'⊗w, which is unentangled.

b. Yes, e.g., a measurement with Bell states would produce an entangled state.

c. Yes, e.g., such a measuring of any Bell state.

Edited October 14, 2023 by Genady

[Genady]

Let's assume that the two qubits are unentangled. Then their state can be factorized, v⊗w. If the first qubit is measured by projecting its state onto the subspace spanned by v, its result will be certain contrary to the assumption that there is no such measurement. Thus, they are entangled.

[Genady]

The reference:

|v〉〈v|-|v^⊥〉〈v^⊥| = cos²θ|0〉〈0| + sin²θ|1〉〈1| + cosθsinθ|0〉〈1| + cosθsinθ|1〉〈0| - (sin²θ|0〉〈0| + cos²θ|1〉〈1| - cosθsinθ|0〉〈1| - cosθsinθ|1〉〈0|) =

cos2θ|0〉〈0| - cos2θ|1〉〈1| + sin2θ|0〉〈1| + sin2θ|1〉〈0|

 

cos2θ	sin2θ
sin2θ	-cos2θ

 

 

 

Edited October 14, 2023 by Genady

[Genady]

The reference:

 

a. Projector onto the subspace of agreeing measurements is |v₁v₂〉〈v₁v₂|+|v₁^⊥v₂^⊥〉〈v₁^⊥v₂^⊥|.

|v₁v₂〉 = (cos θ₁ |0〉 + sin θ₁ |1〉)⊗(cos θ₂ |0〉 + sin θ₂ |1〉)

When it acts on |ψ〉 only the terms |00〉 and |11〉 survive, i.e.,

〈v₁v₂|ψ〉 = 1/√2 (cos θ₁ cos θ₂ + sin θ₁ sin θ₂) = 1/√2 cos (θ₁ - θ₂)

Similarly,

〈v₁^⊥v₂^⊥|ψ〉 = 1/√2 (cos θ₁ cos θ₂ + sin θ₁ sin θ₂) = 1/√2 cos (θ₁ - θ₂).

Thus, the projection results in the state

1/c (1/√2 cos (θ₁ - θ₂) |v₁v₂〉 + 1/√2 cos (θ₁ - θ₂) |v₁^⊥v₂^⊥〉)

with the probability c² = cos² (θ₁ - θ₂).

 

b. θ₁ = θ₂

c. θ₁ - θ₂= π/2

d. θ₁ - θ₂= π/4

e. For these pairs, the θ₁ - θ₂ is either 60⁰ or 120⁰ with cos² (θ₁ - θ₂) = 1/4.

Edited October 15, 2023 by Genady