Genady

a. O = ∑i λiPi . For some i=j, Pj|ψ〉 = |φ〉. When the same measurement applied to |φ〉, Pj|φ〉=|φ〉, while for all i≠j, Pi|φ〉=0. b. Operators P but not O act on state vectors.

I am bad in this in any language. Perhaps, has something to do with my APD.

P1+P2-P3-P4

(|1><1|+|2><2|+|4><4|)+2(|3><3|+|5><5|+|6><6|)+3|7><7| The number of 1-bits is the eigenvalue.

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

Yes, I know how simple and inexpensive is notarizing in the States and how common notaries public are there (I lived in NYC 15 years.) Notarizing in the US Consulate is equally simple. Also, the Consul comes to Bonaire couple times a year to provide services to US citizens. I will avoid the Dutch notary and hope to never need them. I'm already avoiding them by NOT transferring my late father's real estate to my name. I asked what's involved and that was enough. The only limitation of not transferring is that I cannot sell it. But I don't plan to do so anyway. Otherwise, I can use it as I wish. BTW, I took the ferries from Cape Cod to Nantucket and to Martha's Vineyard several times. My wife used to live in Boston before eventually moving to NYC and marrying me.

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

On this exact point: (A|x〉)+ = (A|x〉)*T = (A*|x〉*)T = |x〉*TA*T= 〈x|A+

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>

They can and they do.

No, there is no ferry. I heard that there was one many years ago, but the sea between the islands is very rough, it took too long and was very unpleasant. It was discontinued eventually. OTOH, there are about dozen flights each way daily. It's like taking a bus. Many people go back and forth for work regularly. Takes about an hour one way*. Not terribly expensive. *The actual flight time is about 25 minutes.

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.

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.

Right. And I rather do this than pay to the pompous Dutch notary.

To notarize my signature on a US document, it is easier for me to fly to Curacao and to do it in the US consulate there, then to do it with a Dutch notary here on Bonaire.

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?

(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?

Not surprisingly the form and the contents of their post above fit ChatGPT response well.

I wonder if the OP "Alej" and "Alessandro" are the same person.

For the reference, the Bell basis: a. = 1/√2(|Φ+>+|Φ->) b. = 1/2(|0>+|1>)(|0>-|1>) = 1/2(|00>-|11>+|10>-|01>) = 1/√2(|Φ->-|Ψ->) c. = 1/√3(1/√2(|Φ+>+|Φ->)+√2|Ψ+>)

I will ask, with respect to what decomposition?

|0>|+> + |1>|- > = |0>(|0>+|1>) + |1>(|0>-|1>) = |00>+|01>+|10>-|11> =(assume) (|0>+x|1>)(|0>+y|1>) = |00>+y|01>+x|10>+xy|11> This requires x=1, y=1, xy=-1, which is impossible. Thus, the state is entangled.

Let's assume it is not entangled with respect to such decomposition. Then it can be factorized (I skip the normalization factors for simplicity): |Wn> = (a|0>+b|1>)...(y|0>+z|1>) Then, this state would have a term, a...y|0...0>. But there is no such term. It means that a...y=0. Let's say, a=0. Then, |Wn> = b|1>(...)...(y|0>+z|1>). Then, every term would have |1> for the first qubit, which is not so. Contradiction. Thus, such factorization is impossible, and the state is entangled with respect to such decomposition.

Eastern District of New York | Congressman George Santos Charged With Conspiracy, Wire Fraud, False Statements, Falsification of Records, Aggravated Identity Theft, and Credit Card Fraud | United States Department of Justice

1/2*(|00>+|11>) + 1/2*(|00>-|11>) = |00>