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What happens when an operator is applied only to some bits of a mixed state?

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2

$begingroup$

What happens when an operator is applied only to some bits of a mixed state?
For instance, assume $vert xranglevert f(x)rangle$ is entangled. Then what is the result of $vert Uxranglevert f(x)rangle$ (how to compute the amplitudes) ?
What if U is Grover's diffusion? Will it still work (without uncomputing $f(x)$) ?

Update $y$ replaced with $f(x)$

quantum-state entanglement

share|improve this question

edited May 29 at 9:13

mike_dole_z3

asked May 29 at 4:38

mike_dole_z3mike_dole_z3

133

New contributor

mike_dole_z3 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

2

$begingroup$

What happens when an operator is applied only to some bits of a mixed state?
For instance, assume $vert xranglevert f(x)rangle$ is entangled. Then what is the result of $vert Uxranglevert f(x)rangle$ (how to compute the amplitudes) ?
What if U is Grover's diffusion? Will it still work (without uncomputing $f(x)$) ?

Update $y$ replaced with $f(x)$

quantum-state entanglement

share|improve this question

edited May 29 at 9:13

mike_dole_z3

asked May 29 at 4:38

mike_dole_z3mike_dole_z3

133

New contributor

mike_dole_z3 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

What happens when an operator is applied only to some bits of a mixed state?
For instance, assume $vert xranglevert f(x)rangle$ is entangled. Then what is the result of $vert Uxranglevert f(x)rangle$ (how to compute the amplitudes) ?
What if U is Grover's diffusion? Will it still work (without uncomputing $f(x)$) ?

Update $y$ replaced with $f(x)$

quantum-state entanglement

share|improve this question

edited May 29 at 9:13

mike_dole_z3

asked May 29 at 4:38

mike_dole_z3mike_dole_z3

133

New contributor

mike_dole_z3 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

edited May 29 at 9:13

mike_dole_z3

asked May 29 at 4:38

mike_dole_z3mike_dole_z3

133

New contributor

mike_dole_z3 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited May 29 at 9:13

mike_dole_z3

asked May 29 at 4:38

mike_dole_z3mike_dole_z3

133

New contributor

mike_dole_z3 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked May 29 at 4:38

mike_dole_z3mike_dole_z3

133

New contributor

mike_dole_z3 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked May 29 at 4:38

mike_dole_z3mike_dole_z3

133

2 Answers
2

active

oldest

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3

$begingroup$

$|xrangle|yrangle = |xrangleotimes |yrangle$ is the notation for disentangled state. Entangled state can't be written this way. In general, every pure state (entangled or disentangled) on a bipartite system is a linear combination of disentangled states
$$
|phirangle_AB = sum_i alpha_i |x_irangle_Aotimes|y_irangle_B
$$

Application of $U$ on the first subsystem is equivalent to application of $U otimes I$ on the whole system. The result will be
$$
(Uotimes I) |phirangle_AB = sum_i alpha_i U|x_irangle_Aotimes|y_irangle_B
$$
Mixed state is a different thing (do not confuse it with entangled state). Mixed state can be seen as probability distribution over pure states: $p_i,, p_i>0, sum_ip_i=1$. It has the corresponding density matrix $rho=sum_ip_i|phi_iranglelanglephi_i|$. Note that every $|phi_irangle$ can be entangled.

The result of application of $U$ on the first subsystem of a mixed state is the probability distribution $p_i,(Uotimes I)$, or, in terms of density matrices, $(Uotimes I) rho (U^daggerotimes I)$.

share|improve this answer

edited May 29 at 7:29

answered May 29 at 7:03

Danylo YDanylo Y

1,03516

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank Danylo. Please take another look, I've updated the question.
  $endgroup$
  – mike_dole_z3
  May 29 at 9:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $|xrangle otimes |f(x)rangle$ is also disentangled. So the result will be exactly $ (U|xrangle) otimes vert f(x)rangle$. You can compute the amplitudes $alpha_i$ of $U|xrangle = sum_i alpha_i |irangle$ and then take the product $otimes$ with $|f(x)rangle$.
  $endgroup$
  – Danylo Y
  May 29 at 9:44
  
  

  
  
  
  

add a comment | 

3

$begingroup$

$newcommandket[1]#1rangleketxkety$ is a pure state, not mixed, and is a product state, which is not entangled by definition, so your example is rather confusing.

To answer the question in your first sentence, applying a unitary operator $U_A$ to one subsystem of a bipartite system is equivalent to applying the operator $U_Aotimesmathbb1_B$ to the whole system, where $mathbb1_B$ is the identity for subsystem $B$, so you can treat $U_Aotimesmathbb1_B$ as one big unitary and apply it how you normally would. If you have a pure state $ketpsi = sum_i psi_i keti_A otimes keti_B$, this means that applying such unitary gives $sum_i psi_i (U_A keti_A) otimes keti_B$, and analogously for a mixed state.

share|improve this answer

answered May 29 at 6:35

BFGBFG

711

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. Please take another look, I've updated the question.
  $endgroup$
  – mike_dole_z3
  May 29 at 9:18
  

  
  
  
  

add a comment | 

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3

$begingroup$

$|xrangle|yrangle = |xrangleotimes |yrangle$ is the notation for disentangled state. Entangled state can't be written this way. In general, every pure state (entangled or disentangled) on a bipartite system is a linear combination of disentangled states
$$
|phirangle_AB = sum_i alpha_i |x_irangle_Aotimes|y_irangle_B
$$

Application of $U$ on the first subsystem is equivalent to application of $U otimes I$ on the whole system. The result will be
$$
(Uotimes I) |phirangle_AB = sum_i alpha_i U|x_irangle_Aotimes|y_irangle_B
$$
Mixed state is a different thing (do not confuse it with entangled state). Mixed state can be seen as probability distribution over pure states: $p_i,, p_i>0, sum_ip_i=1$. It has the corresponding density matrix $rho=sum_ip_i|phi_iranglelanglephi_i|$. Note that every $|phi_irangle$ can be entangled.

The result of application of $U$ on the first subsystem of a mixed state is the probability distribution $p_i,(Uotimes I)$, or, in terms of density matrices, $(Uotimes I) rho (U^daggerotimes I)$.

share|improve this answer

edited May 29 at 7:29

answered May 29 at 7:03

Danylo YDanylo Y

1,03516

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank Danylo. Please take another look, I've updated the question.
  $endgroup$
  – mike_dole_z3
  May 29 at 9:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $|xrangle otimes |f(x)rangle$ is also disentangled. So the result will be exactly $ (U|xrangle) otimes vert f(x)rangle$. You can compute the amplitudes $alpha_i$ of $U|xrangle = sum_i alpha_i |irangle$ and then take the product $otimes$ with $|f(x)rangle$.
  $endgroup$
  – Danylo Y
  May 29 at 9:44
  
  

  
  
  
  

add a comment | 

3

$begingroup$

$|xrangle|yrangle = |xrangleotimes |yrangle$ is the notation for disentangled state. Entangled state can't be written this way. In general, every pure state (entangled or disentangled) on a bipartite system is a linear combination of disentangled states
$$
|phirangle_AB = sum_i alpha_i |x_irangle_Aotimes|y_irangle_B
$$

Application of $U$ on the first subsystem is equivalent to application of $U otimes I$ on the whole system. The result will be
$$
(Uotimes I) |phirangle_AB = sum_i alpha_i U|x_irangle_Aotimes|y_irangle_B
$$
Mixed state is a different thing (do not confuse it with entangled state). Mixed state can be seen as probability distribution over pure states: $p_i,, p_i>0, sum_ip_i=1$. It has the corresponding density matrix $rho=sum_ip_i|phi_iranglelanglephi_i|$. Note that every $|phi_irangle$ can be entangled.

The result of application of $U$ on the first subsystem of a mixed state is the probability distribution $p_i,(Uotimes I)$, or, in terms of density matrices, $(Uotimes I) rho (U^daggerotimes I)$.

share|improve this answer

edited May 29 at 7:29

answered May 29 at 7:03

Danylo YDanylo Y

1,03516

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank Danylo. Please take another look, I've updated the question.
  $endgroup$
  – mike_dole_z3
  May 29 at 9:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $|xrangle otimes |f(x)rangle$ is also disentangled. So the result will be exactly $ (U|xrangle) otimes vert f(x)rangle$. You can compute the amplitudes $alpha_i$ of $U|xrangle = sum_i alpha_i |irangle$ and then take the product $otimes$ with $|f(x)rangle$.
  $endgroup$
  – Danylo Y
  May 29 at 9:44
  
  

  
  
  
  

add a comment | 

$begingroup$

$|xrangle|yrangle = |xrangleotimes |yrangle$ is the notation for disentangled state. Entangled state can't be written this way. In general, every pure state (entangled or disentangled) on a bipartite system is a linear combination of disentangled states
$$
|phirangle_AB = sum_i alpha_i |x_irangle_Aotimes|y_irangle_B
$$

Application of $U$ on the first subsystem is equivalent to application of $U otimes I$ on the whole system. The result will be
$$
(Uotimes I) |phirangle_AB = sum_i alpha_i U|x_irangle_Aotimes|y_irangle_B
$$
Mixed state is a different thing (do not confuse it with entangled state). Mixed state can be seen as probability distribution over pure states: $p_i,, p_i>0, sum_ip_i=1$. It has the corresponding density matrix $rho=sum_ip_i|phi_iranglelanglephi_i|$. Note that every $|phi_irangle$ can be entangled.

The result of application of $U$ on the first subsystem of a mixed state is the probability distribution $p_i,(Uotimes I)$, or, in terms of density matrices, $(Uotimes I) rho (U^daggerotimes I)$.

share|improve this answer

edited May 29 at 7:29

answered May 29 at 7:03

Danylo YDanylo Y

1,03516

$endgroup$

share|improve this answer

edited May 29 at 7:29

answered May 29 at 7:03

Danylo YDanylo Y

1,03516

answered May 29 at 7:03

Danylo YDanylo Y

1,03516

answered May 29 at 7:03

Danylo YDanylo Y

1,03516

3

$begingroup$

$newcommandket[1]#1rangleketxkety$ is a pure state, not mixed, and is a product state, which is not entangled by definition, so your example is rather confusing.

To answer the question in your first sentence, applying a unitary operator $U_A$ to one subsystem of a bipartite system is equivalent to applying the operator $U_Aotimesmathbb1_B$ to the whole system, where $mathbb1_B$ is the identity for subsystem $B$, so you can treat $U_Aotimesmathbb1_B$ as one big unitary and apply it how you normally would. If you have a pure state $ketpsi = sum_i psi_i keti_A otimes keti_B$, this means that applying such unitary gives $sum_i psi_i (U_A keti_A) otimes keti_B$, and analogously for a mixed state.

share|improve this answer

answered May 29 at 6:35

BFGBFG

711

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. Please take another look, I've updated the question.
  $endgroup$
  – mike_dole_z3
  May 29 at 9:18
  

  
  
  
  

add a comment | 

3

$begingroup$

$newcommandket[1]#1rangleketxkety$ is a pure state, not mixed, and is a product state, which is not entangled by definition, so your example is rather confusing.

To answer the question in your first sentence, applying a unitary operator $U_A$ to one subsystem of a bipartite system is equivalent to applying the operator $U_Aotimesmathbb1_B$ to the whole system, where $mathbb1_B$ is the identity for subsystem $B$, so you can treat $U_Aotimesmathbb1_B$ as one big unitary and apply it how you normally would. If you have a pure state $ketpsi = sum_i psi_i keti_A otimes keti_B$, this means that applying such unitary gives $sum_i psi_i (U_A keti_A) otimes keti_B$, and analogously for a mixed state.

share|improve this answer

answered May 29 at 6:35

BFGBFG

711

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. Please take another look, I've updated the question.
  $endgroup$
  – mike_dole_z3
  May 29 at 9:18
  

  
  
  
  

add a comment | 

$begingroup$

$newcommandket[1]#1rangleketxkety$ is a pure state, not mixed, and is a product state, which is not entangled by definition, so your example is rather confusing.

To answer the question in your first sentence, applying a unitary operator $U_A$ to one subsystem of a bipartite system is equivalent to applying the operator $U_Aotimesmathbb1_B$ to the whole system, where $mathbb1_B$ is the identity for subsystem $B$, so you can treat $U_Aotimesmathbb1_B$ as one big unitary and apply it how you normally would. If you have a pure state $ketpsi = sum_i psi_i keti_A otimes keti_B$, this means that applying such unitary gives $sum_i psi_i (U_A keti_A) otimes keti_B$, and analogously for a mixed state.

share|improve this answer

answered May 29 at 6:35

BFGBFG

711

$endgroup$

share|improve this answer

answered May 29 at 6:35

BFGBFG

711

answered May 29 at 6:35

BFGBFG

711

answered May 29 at 6:35

BFGBFG

711