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How are witness operators physically implemented?

How are quantum gates implemented in reality?Is the Kraus representation of a quantum channel equivalent to a unitary evolution in an enlarged space?Image of a sum of positive operators contains the images of each individual operator?How many Kraus operators are required to characterise a channel with different start and end dimensions?Why is $P(1,2)_textsame = frac14$ and not $frac12$ in Preskill's Bell experiment?Construction of optimal ensemble to show quantum steerabilityPartial Transpose and Positive OperatorsApplication of improved compatibilityWhat's the difference between Kraus operators and measurement operators?Are entanglement witnesses of this form optimal?

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4

$begingroup$

Let's take an example of an entanglement witness of the form $W = | phi rangle langle phi | ^T_2$ where $ | phi rangle $ is some pure entangled state.

If I wanted to test some state $rho$, I would have to perform $mathrmTr(W rho)$. I assume this is done by measuring $rho$ multiple times in the eigenbasis of $W$ and finding the expected eigenvalue, and that would be the solution to $mathrmTr(W rho)$.

1. Is this the way it is done?


2. Therefore, specifically in the above case, it is very much possible to physically apply the witness? (Even though there is a mathematical partial transpose present.)

quantum-information entanglement physical-realization quantum-operation entanglement-witness

share|improve this question

edited May 14 at 7:37

Sanchayan Dutta♦

7,12041658

asked May 14 at 7:35

Mahathi VempatiMahathi Vempati

5938

$endgroup$

add a comment | 

4

$begingroup$

Let's take an example of an entanglement witness of the form $W = | phi rangle langle phi | ^T_2$ where $ | phi rangle $ is some pure entangled state.

If I wanted to test some state $rho$, I would have to perform $mathrmTr(W rho)$. I assume this is done by measuring $rho$ multiple times in the eigenbasis of $W$ and finding the expected eigenvalue, and that would be the solution to $mathrmTr(W rho)$.

1. Is this the way it is done?


2. Therefore, specifically in the above case, it is very much possible to physically apply the witness? (Even though there is a mathematical partial transpose present.)

quantum-information entanglement physical-realization quantum-operation entanglement-witness

share|improve this question

edited May 14 at 7:37

Sanchayan Dutta♦

7,12041658

asked May 14 at 7:35

Mahathi VempatiMahathi Vempati

5938

$endgroup$

add a comment | 

$begingroup$

Let's take an example of an entanglement witness of the form $W = | phi rangle langle phi | ^T_2$ where $ | phi rangle $ is some pure entangled state.

If I wanted to test some state $rho$, I would have to perform $mathrmTr(W rho)$. I assume this is done by measuring $rho$ multiple times in the eigenbasis of $W$ and finding the expected eigenvalue, and that would be the solution to $mathrmTr(W rho)$.

1. Is this the way it is done?


2. Therefore, specifically in the above case, it is very much possible to physically apply the witness? (Even though there is a mathematical partial transpose present.)

quantum-information entanglement physical-realization quantum-operation entanglement-witness

share|improve this question

edited May 14 at 7:37

Sanchayan Dutta♦

7,12041658

asked May 14 at 7:35

Mahathi VempatiMahathi Vempati

5938

$endgroup$

share|improve this question

edited May 14 at 7:37

Sanchayan Dutta♦

7,12041658

asked May 14 at 7:35

Mahathi VempatiMahathi Vempati

5938

share|improve this question

edited May 14 at 7:37

Sanchayan Dutta♦

7,12041658

asked May 14 at 7:35

Mahathi VempatiMahathi Vempati

5938

edited May 14 at 7:37

Sanchayan Dutta♦

7,12041658

edited May 14 at 7:37

Sanchayan Dutta♦

7,12041658

asked May 14 at 7:35

Mahathi VempatiMahathi Vempati

5938

asked May 14 at 7:35

Mahathi VempatiMahathi Vempati

5938

1 Answer
1

active

oldest

votes

2

$begingroup$

This is certainly how theorists think of this being done. I don't know if there's an experimental reality to compare this to. Whether they actually decompose it in terms of the eigenvectors, or find some other terms to decompose it as.

Just as an example of what I mean, let
$$
W=left(beginarraycccc
1 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1
endarrayright).
$$
As it happens, this is just a swap gate, but ignore this for now.
You might find the eigenvectors $|00rangle,|11rangle,(|01ranglepm|10rangle)/sqrt2$ and measure those expectation values directly. Or, you might write
$$
W=(mathbbI+Zotimes Z+Xotimes X+Yotimes Y)/2,
$$
and them you might go off and measure the 3 separate observables $ZZ$, $XX$ and $YY$, those being particularly natural, accessible, things.

Note that there's no problem using $W$ to define a measurement. The partial transpose is irrelevant; it's still a Hermitian matrix. The partial transpose just means it might not be a valid state, but being a valid state is irrelevant as a measurement: if we say "do a Z measurement", the Z matrix certainly has nothing to do with being a state. It's just a Hermitian matrix.

share|improve this answer

answered May 14 at 8:41

DaftWullieDaftWullie

16.5k1644

$endgroup$

add a comment | 

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2

$begingroup$

This is certainly how theorists think of this being done. I don't know if there's an experimental reality to compare this to. Whether they actually decompose it in terms of the eigenvectors, or find some other terms to decompose it as.

Just as an example of what I mean, let
$$
W=left(beginarraycccc
1 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1
endarrayright).
$$
As it happens, this is just a swap gate, but ignore this for now.
You might find the eigenvectors $|00rangle,|11rangle,(|01ranglepm|10rangle)/sqrt2$ and measure those expectation values directly. Or, you might write
$$
W=(mathbbI+Zotimes Z+Xotimes X+Yotimes Y)/2,
$$
and them you might go off and measure the 3 separate observables $ZZ$, $XX$ and $YY$, those being particularly natural, accessible, things.

Note that there's no problem using $W$ to define a measurement. The partial transpose is irrelevant; it's still a Hermitian matrix. The partial transpose just means it might not be a valid state, but being a valid state is irrelevant as a measurement: if we say "do a Z measurement", the Z matrix certainly has nothing to do with being a state. It's just a Hermitian matrix.

share|improve this answer

answered May 14 at 8:41

DaftWullieDaftWullie

16.5k1644

$endgroup$

add a comment | 

2

$begingroup$

This is certainly how theorists think of this being done. I don't know if there's an experimental reality to compare this to. Whether they actually decompose it in terms of the eigenvectors, or find some other terms to decompose it as.

Just as an example of what I mean, let
$$
W=left(beginarraycccc
1 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1
endarrayright).
$$
As it happens, this is just a swap gate, but ignore this for now.
You might find the eigenvectors $|00rangle,|11rangle,(|01ranglepm|10rangle)/sqrt2$ and measure those expectation values directly. Or, you might write
$$
W=(mathbbI+Zotimes Z+Xotimes X+Yotimes Y)/2,
$$
and them you might go off and measure the 3 separate observables $ZZ$, $XX$ and $YY$, those being particularly natural, accessible, things.

Note that there's no problem using $W$ to define a measurement. The partial transpose is irrelevant; it's still a Hermitian matrix. The partial transpose just means it might not be a valid state, but being a valid state is irrelevant as a measurement: if we say "do a Z measurement", the Z matrix certainly has nothing to do with being a state. It's just a Hermitian matrix.

share|improve this answer

answered May 14 at 8:41

DaftWullieDaftWullie

16.5k1644

$endgroup$

add a comment | 

$begingroup$

This is certainly how theorists think of this being done. I don't know if there's an experimental reality to compare this to. Whether they actually decompose it in terms of the eigenvectors, or find some other terms to decompose it as.

Just as an example of what I mean, let
$$
W=left(beginarraycccc
1 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1
endarrayright).
$$
As it happens, this is just a swap gate, but ignore this for now.
You might find the eigenvectors $|00rangle,|11rangle,(|01ranglepm|10rangle)/sqrt2$ and measure those expectation values directly. Or, you might write
$$
W=(mathbbI+Zotimes Z+Xotimes X+Yotimes Y)/2,
$$
and them you might go off and measure the 3 separate observables $ZZ$, $XX$ and $YY$, those being particularly natural, accessible, things.

Note that there's no problem using $W$ to define a measurement. The partial transpose is irrelevant; it's still a Hermitian matrix. The partial transpose just means it might not be a valid state, but being a valid state is irrelevant as a measurement: if we say "do a Z measurement", the Z matrix certainly has nothing to do with being a state. It's just a Hermitian matrix.

share|improve this answer

answered May 14 at 8:41

DaftWullieDaftWullie

16.5k1644

$endgroup$

share|improve this answer

answered May 14 at 8:41

DaftWullieDaftWullie

16.5k1644

answered May 14 at 8:41

DaftWullieDaftWullie

16.5k1644

answered May 14 at 8:41

DaftWullieDaftWullie

16.5k1644