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Meaning of Bloch representation

Can the Bloch sphere be generalized to two qubits?What's the difference between a pure and mixed quantum state?Why do Bloch sphere wavefunctions have half angles?Density matrices for pure states and mixed statesWhy is an entangled qubit shown at the origin of a Bloch sphere?Homeomorphism or stereographic projection corresponding to the set of mixed states within the Bloch spherePurity of mixed states as a function of radial distance from origin of Bloch ballCompute average value of two-qubit systemWhy do Bloch sphere wavefunctions have half angles?Why is an entangled qubit shown at the origin of a Bloch sphere?Is there a place online where I can catch up with all the notational syntax associated with quantum computing?What is the meaning of the state $|1rangle-|1rangle$?What's a vector in the format of the Bloch Sphere?What utility is provided by the Bloch sphere visualization?Purity of mixed states as a function of radial distance from origin of Bloch ballNotation for two qubit composite product stateWhat does it mean to express a gate in Dirac notation?

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1

$begingroup$

What is the meaning of writing a state $|psirangle$ in its Bloch representation. Would I be correct in saying it's just writing out its Bloch vector?

notation bloch-sphere

share|improve this question

edited 38 mins ago

Sanchayan Dutta♦

6,75841556

asked 1 hour ago

can'tcauchycan'tcauchy

2015

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The short answer is "yes." Read Sanchayan's answer for a complete understanding :)
  $endgroup$
  – Will
  3 mins ago
  

  
  
  
  

add a comment | 

1

$begingroup$

What is the meaning of writing a state $|psirangle$ in its Bloch representation. Would I be correct in saying it's just writing out its Bloch vector?

notation bloch-sphere

share|improve this question

edited 38 mins ago

Sanchayan Dutta♦

6,75841556

asked 1 hour ago

can'tcauchycan'tcauchy

2015

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The short answer is "yes." Read Sanchayan's answer for a complete understanding :)
  $endgroup$
  – Will
  3 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

What is the meaning of writing a state $|psirangle$ in its Bloch representation. Would I be correct in saying it's just writing out its Bloch vector?

notation bloch-sphere

share|improve this question

edited 38 mins ago

Sanchayan Dutta♦

6,75841556

asked 1 hour ago

can'tcauchycan'tcauchy

2015

$endgroup$

share|improve this question

edited 38 mins ago

Sanchayan Dutta♦

6,75841556

asked 1 hour ago

can'tcauchycan'tcauchy

2015

share|improve this question

edited 38 mins ago

Sanchayan Dutta♦

6,75841556

asked 1 hour ago

can'tcauchycan'tcauchy

2015

edited 38 mins ago

Sanchayan Dutta♦

6,75841556

edited 38 mins ago

Sanchayan Dutta♦

6,75841556

asked 1 hour ago

can'tcauchycan'tcauchy

2015

asked 1 hour ago

can'tcauchycan'tcauchy

2015

1 Answer
1

active

oldest

votes

2

$begingroup$

The Bloch sphere formalism is used for geometrically representing pure and mixed states of two-dimensional quantum systems a.k.a qubits. Any pure state $|Psirangle$ of a qubit can be written in the form:

$$|Psirangle = cosfrac{theta}{2}|0rangle + e^{iphi}sinfrac{theta}{2}|1rangle$$ where $0leq thetaleq pi$ and $0leq phileq 2pi$. This $|Psirangle$ can be represented on the Bloch sphere as:

The Bloch vector $vec{a}in Bbb R^3$ is basically $(sintheta cosphi, sinthetasinphi, cos theta) = (a_1,a_2,a_3)$.

To represent mixed states you need to consider the corresponding density operator $rho$. the set of states of a single qubit can be described in terms of $2times 2$ density matrices and as ${I,X,Y,Z}$ forms a basis for the vector space of $2times 2$ Hermitian matrices, you can write the density operator as $$rho = a_0I+a_1X+a_2Y+a_3Z = frac{1}{2}begin{pmatrix}1+a_3 & a_1-ia_2 \ a_1+ia_2 & 1-a_3end{pmatrix}.$$ As density matrices always have trace $1$, and here $mathrm{tr}(rho)=2a_0$, so $a_0$ is necessarily $frac{1}{2}$. So from here you can find out the Bloch coordinates of the any mixed state i.e. $(a_1,a_2,a_3)$ after performing the Pauli decompostion of the density matrix. If you're wondering what ensures that $|vec{a}|leq 1$, it's the positive semidefiniteness! The two eigenvalues of $rho$ are $frac{1}{2}(1+|vec{a}|)$ and $frac{1}{2}(1-|vec{a}|)$. Thus, to ensure that the second eigenvalue is non-negative, $|vec{a}|leq 1$. The three properties of density matrices which you should drill into your brain are: self-adjointness, positive-semidefiniteness and unit trace.

Once you determine the values $a_1,a_2$ and $a_3$ from the density operator, you can easily find the location of the qubit state $(sintheta cosphi, sinthetasinphi, cos theta)$ inside the Bloch sphere. Let me emphasize on this point: pure states lie on the Bloch sphere (i.e. $|vec{a}|=1$) whereas mixed states lie inside the Bloch sphere (i.e. $|vec{a}|<1$); prove this as an exercise. If you're mathematically inclined, you'll also love to think about the Bloch sphere in terms of stereographic projections; it's excellently summarized in this Physics SE answer. I'll attach the image from there, which is originally from this blogpost (the article is in French, sorry :).

Here are a few "further readings" for you:

• Density matrices for pure states and mixed states


• Why do Bloch sphere wavefunctions have half angles?



• Homeomorphism or stereographic projection corresponding to the set of mixed states within the Bloch sphere




• Purity of mixed states as a function of radial distance from origin of Bloch ball




• Can the Bloch sphere be generalized to two qubits?




• Why is an entangled qubit shown at the origin of a Bloch sphere?

Essentially, go through the bloch-sphere tag; you'll find several interesting questions and answers, which should clarify most of your beginner confusions about the Bloch sphere formalism.

share|improve this answer

edited 30 mins ago

answered 44 mins ago

Sanchayan Dutta♦Sanchayan Dutta

6,75841556

$endgroup$

add a comment | 

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2

$begingroup$

The Bloch sphere formalism is used for geometrically representing pure and mixed states of two-dimensional quantum systems a.k.a qubits. Any pure state $|Psirangle$ of a qubit can be written in the form:

$$|Psirangle = cosfrac{theta}{2}|0rangle + e^{iphi}sinfrac{theta}{2}|1rangle$$ where $0leq thetaleq pi$ and $0leq phileq 2pi$. This $|Psirangle$ can be represented on the Bloch sphere as:

The Bloch vector $vec{a}in Bbb R^3$ is basically $(sintheta cosphi, sinthetasinphi, cos theta) = (a_1,a_2,a_3)$.

To represent mixed states you need to consider the corresponding density operator $rho$. the set of states of a single qubit can be described in terms of $2times 2$ density matrices and as ${I,X,Y,Z}$ forms a basis for the vector space of $2times 2$ Hermitian matrices, you can write the density operator as $$rho = a_0I+a_1X+a_2Y+a_3Z = frac{1}{2}begin{pmatrix}1+a_3 & a_1-ia_2 \ a_1+ia_2 & 1-a_3end{pmatrix}.$$ As density matrices always have trace $1$, and here $mathrm{tr}(rho)=2a_0$, so $a_0$ is necessarily $frac{1}{2}$. So from here you can find out the Bloch coordinates of the any mixed state i.e. $(a_1,a_2,a_3)$ after performing the Pauli decompostion of the density matrix. If you're wondering what ensures that $|vec{a}|leq 1$, it's the positive semidefiniteness! The two eigenvalues of $rho$ are $frac{1}{2}(1+|vec{a}|)$ and $frac{1}{2}(1-|vec{a}|)$. Thus, to ensure that the second eigenvalue is non-negative, $|vec{a}|leq 1$. The three properties of density matrices which you should drill into your brain are: self-adjointness, positive-semidefiniteness and unit trace.

Once you determine the values $a_1,a_2$ and $a_3$ from the density operator, you can easily find the location of the qubit state $(sintheta cosphi, sinthetasinphi, cos theta)$ inside the Bloch sphere. Let me emphasize on this point: pure states lie on the Bloch sphere (i.e. $|vec{a}|=1$) whereas mixed states lie inside the Bloch sphere (i.e. $|vec{a}|<1$); prove this as an exercise. If you're mathematically inclined, you'll also love to think about the Bloch sphere in terms of stereographic projections; it's excellently summarized in this Physics SE answer. I'll attach the image from there, which is originally from this blogpost (the article is in French, sorry :).

Here are a few "further readings" for you:

• Density matrices for pure states and mixed states


• Why do Bloch sphere wavefunctions have half angles?



• Homeomorphism or stereographic projection corresponding to the set of mixed states within the Bloch sphere




• Purity of mixed states as a function of radial distance from origin of Bloch ball




• Can the Bloch sphere be generalized to two qubits?




• Why is an entangled qubit shown at the origin of a Bloch sphere?

Essentially, go through the bloch-sphere tag; you'll find several interesting questions and answers, which should clarify most of your beginner confusions about the Bloch sphere formalism.

share|improve this answer

edited 30 mins ago

answered 44 mins ago

Sanchayan Dutta♦Sanchayan Dutta

6,75841556

$endgroup$

add a comment | 

2

$begingroup$

The Bloch sphere formalism is used for geometrically representing pure and mixed states of two-dimensional quantum systems a.k.a qubits. Any pure state $|Psirangle$ of a qubit can be written in the form:

$$|Psirangle = cosfrac{theta}{2}|0rangle + e^{iphi}sinfrac{theta}{2}|1rangle$$ where $0leq thetaleq pi$ and $0leq phileq 2pi$. This $|Psirangle$ can be represented on the Bloch sphere as:

The Bloch vector $vec{a}in Bbb R^3$ is basically $(sintheta cosphi, sinthetasinphi, cos theta) = (a_1,a_2,a_3)$.

To represent mixed states you need to consider the corresponding density operator $rho$. the set of states of a single qubit can be described in terms of $2times 2$ density matrices and as ${I,X,Y,Z}$ forms a basis for the vector space of $2times 2$ Hermitian matrices, you can write the density operator as $$rho = a_0I+a_1X+a_2Y+a_3Z = frac{1}{2}begin{pmatrix}1+a_3 & a_1-ia_2 \ a_1+ia_2 & 1-a_3end{pmatrix}.$$ As density matrices always have trace $1$, and here $mathrm{tr}(rho)=2a_0$, so $a_0$ is necessarily $frac{1}{2}$. So from here you can find out the Bloch coordinates of the any mixed state i.e. $(a_1,a_2,a_3)$ after performing the Pauli decompostion of the density matrix. If you're wondering what ensures that $|vec{a}|leq 1$, it's the positive semidefiniteness! The two eigenvalues of $rho$ are $frac{1}{2}(1+|vec{a}|)$ and $frac{1}{2}(1-|vec{a}|)$. Thus, to ensure that the second eigenvalue is non-negative, $|vec{a}|leq 1$. The three properties of density matrices which you should drill into your brain are: self-adjointness, positive-semidefiniteness and unit trace.

Once you determine the values $a_1,a_2$ and $a_3$ from the density operator, you can easily find the location of the qubit state $(sintheta cosphi, sinthetasinphi, cos theta)$ inside the Bloch sphere. Let me emphasize on this point: pure states lie on the Bloch sphere (i.e. $|vec{a}|=1$) whereas mixed states lie inside the Bloch sphere (i.e. $|vec{a}|<1$); prove this as an exercise. If you're mathematically inclined, you'll also love to think about the Bloch sphere in terms of stereographic projections; it's excellently summarized in this Physics SE answer. I'll attach the image from there, which is originally from this blogpost (the article is in French, sorry :).

Here are a few "further readings" for you:

• Density matrices for pure states and mixed states


• Why do Bloch sphere wavefunctions have half angles?



• Homeomorphism or stereographic projection corresponding to the set of mixed states within the Bloch sphere




• Purity of mixed states as a function of radial distance from origin of Bloch ball




• Can the Bloch sphere be generalized to two qubits?




• Why is an entangled qubit shown at the origin of a Bloch sphere?

Essentially, go through the bloch-sphere tag; you'll find several interesting questions and answers, which should clarify most of your beginner confusions about the Bloch sphere formalism.

share|improve this answer

edited 30 mins ago

answered 44 mins ago

Sanchayan Dutta♦Sanchayan Dutta

6,75841556

$endgroup$

add a comment | 

$begingroup$

The Bloch sphere formalism is used for geometrically representing pure and mixed states of two-dimensional quantum systems a.k.a qubits. Any pure state $|Psirangle$ of a qubit can be written in the form:

$$|Psirangle = cosfrac{theta}{2}|0rangle + e^{iphi}sinfrac{theta}{2}|1rangle$$ where $0leq thetaleq pi$ and $0leq phileq 2pi$. This $|Psirangle$ can be represented on the Bloch sphere as:

The Bloch vector $vec{a}in Bbb R^3$ is basically $(sintheta cosphi, sinthetasinphi, cos theta) = (a_1,a_2,a_3)$.

To represent mixed states you need to consider the corresponding density operator $rho$. the set of states of a single qubit can be described in terms of $2times 2$ density matrices and as ${I,X,Y,Z}$ forms a basis for the vector space of $2times 2$ Hermitian matrices, you can write the density operator as $$rho = a_0I+a_1X+a_2Y+a_3Z = frac{1}{2}begin{pmatrix}1+a_3 & a_1-ia_2 \ a_1+ia_2 & 1-a_3end{pmatrix}.$$ As density matrices always have trace $1$, and here $mathrm{tr}(rho)=2a_0$, so $a_0$ is necessarily $frac{1}{2}$. So from here you can find out the Bloch coordinates of the any mixed state i.e. $(a_1,a_2,a_3)$ after performing the Pauli decompostion of the density matrix. If you're wondering what ensures that $|vec{a}|leq 1$, it's the positive semidefiniteness! The two eigenvalues of $rho$ are $frac{1}{2}(1+|vec{a}|)$ and $frac{1}{2}(1-|vec{a}|)$. Thus, to ensure that the second eigenvalue is non-negative, $|vec{a}|leq 1$. The three properties of density matrices which you should drill into your brain are: self-adjointness, positive-semidefiniteness and unit trace.

Once you determine the values $a_1,a_2$ and $a_3$ from the density operator, you can easily find the location of the qubit state $(sintheta cosphi, sinthetasinphi, cos theta)$ inside the Bloch sphere. Let me emphasize on this point: pure states lie on the Bloch sphere (i.e. $|vec{a}|=1$) whereas mixed states lie inside the Bloch sphere (i.e. $|vec{a}|<1$); prove this as an exercise. If you're mathematically inclined, you'll also love to think about the Bloch sphere in terms of stereographic projections; it's excellently summarized in this Physics SE answer. I'll attach the image from there, which is originally from this blogpost (the article is in French, sorry :).

Here are a few "further readings" for you:

• Density matrices for pure states and mixed states


• Why do Bloch sphere wavefunctions have half angles?



• Homeomorphism or stereographic projection corresponding to the set of mixed states within the Bloch sphere




• Purity of mixed states as a function of radial distance from origin of Bloch ball




• Can the Bloch sphere be generalized to two qubits?




• Why is an entangled qubit shown at the origin of a Bloch sphere?

Essentially, go through the bloch-sphere tag; you'll find several interesting questions and answers, which should clarify most of your beginner confusions about the Bloch sphere formalism.

share|improve this answer

edited 30 mins ago

answered 44 mins ago

Sanchayan Dutta♦Sanchayan Dutta

6,75841556

$endgroup$

share|improve this answer

edited 30 mins ago

answered 44 mins ago

Sanchayan Dutta♦Sanchayan Dutta

6,75841556

answered 44 mins ago

Sanchayan Dutta♦Sanchayan Dutta

6,75841556

answered 44 mins ago

Sanchayan Dutta♦Sanchayan Dutta

6,75841556