Depolarizing channel operator sum representation











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In Nielsen and Chuang, it is shown that the operator sum representation of a depolarizing channel $mathcal{E}(rho) = frac{pI}{2} + (1-p)rho$ is easily seen by substituting the identity matrix with



$frac{mathbb{1}}{2} = frac{rho + Xrho X + Yrho Y +Zrho Z}{4}$.



What is the more systematic way to see this result? Particularly, for the higher dimensional analogue, I cannot see how to proceed.






quantum-information quantum-channel







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    up vote
    1
    down vote

    favorite












    In Nielsen and Chuang, it is shown that the operator sum representation of a depolarizing channel $mathcal{E}(rho) = frac{pI}{2} + (1-p)rho$ is easily seen by substituting the identity matrix with



    $frac{mathbb{1}}{2} = frac{rho + Xrho X + Yrho Y +Zrho Z}{4}$.



    What is the more systematic way to see this result? Particularly, for the higher dimensional analogue, I cannot see how to proceed.






    quantum-information quantum-channel







    share|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      In Nielsen and Chuang, it is shown that the operator sum representation of a depolarizing channel $mathcal{E}(rho) = frac{pI}{2} + (1-p)rho$ is easily seen by substituting the identity matrix with



      $frac{mathbb{1}}{2} = frac{rho + Xrho X + Yrho Y +Zrho Z}{4}$.



      What is the more systematic way to see this result? Particularly, for the higher dimensional analogue, I cannot see how to proceed.






      quantum-information quantum-channel







      share|improve this question















      In Nielsen and Chuang, it is shown that the operator sum representation of a depolarizing channel $mathcal{E}(rho) = frac{pI}{2} + (1-p)rho$ is easily seen by substituting the identity matrix with



      $frac{mathbb{1}}{2} = frac{rho + Xrho X + Yrho Y +Zrho Z}{4}$.



      What is the more systematic way to see this result? Particularly, for the higher dimensional analogue, I cannot see how to proceed.






      quantum-information quantum-channel




      quantum-information quantum-channel






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 2 hours ago









      Blue♦

      5,64011352




      5,64011352










      asked 5 hours ago









      user1936752

      2605




      2605






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          3
          down vote













          This really depends where you want to start from. For instance, you can construct the Choi state of $mathcal E$, i.e.,
          $$
          sigma = (mathcal E otimes mathbb I)(|OmegaranglelangleOmega|) ,
          $$
          with $Omega = tfrac{1}{sqrt{D}}sum_{i=1}^D |i,irangle$, and then extract the Kraus operators of $mathcal E(rho)=sum M_irho M_i^dagger$ by taking any decomposition
          $$
          sigma = sum |psi_iranglelanglepsi_i| ,tag{*}
          $$
          and writing $|psi_irangle = (M_iotimesmathbb I)|Omegarangle$ (which is always possible).



          Note that the decomposition $(*)$ is highly non-unique (any $|phi_jrangle = sum V_{ij} |psi_irangle$, with $V$ an isometry, is also a valid decomposition), which relates to the fact that the Kraus decomposition is equally non-unique. Obviously, the eigenvalue decomposition is a simple choice (which, moreover, minimizes the number of Kraus operators).




          Let's look at your example in a bit more detail. Here, $D=2$. You have that
          $$
          mathcal E(X)=pmathrm{tr}(X),frac{mathbb I}{2}+(1-p)X
          $$
          for any $X$ (due to linearity) -- the $mathrm{tr}(X)$ is required to make this trace-preserving for general $X$.



          We now have that
          begin{align}
          sigma &= (mathcal E otimes mathbb I)(|Omegaranglelangle Omega|)
          \
          & = tfrac1D sum_{ij} mathcal E(|iranglelangle j|)otimes |iranglelangle j|
          end{align}
          inserting the definition of $|Omegarangle$ and using linearity.



          This yields
          $$
          sigma = frac{p}{2D}mathbb Iotimes sum_{i}|iranglelangle i| +
          (1-p)frac1D sum_{ij}|iranglelangle j|otimes |iranglelangle j| .
          $$
          The second term is just $(1-p)|OmegaranglelangleOmega|$, and the first term is
          $frac{p}{2D}mathbb Iotimesmathbb I$.



          You can now see that one possible eigenvalue decomposition of $sigma$ is given by the four Bell states (I leave it to you to work out the weights), and it is well known and easy to check that that the four Bell states can be written as
          $$
          (sigma_kotimes mathbb I)|Omegarangle ,
          $$
          where $sigma_k$ are the three Pauli matrices or the identity.



          Thus, you get that the $M_i$ in the Kraus representation are the Paulis and the identity, with the weight given by the eigenvalue decomposition of $sigma$.






          share|improve this answer























          • Could you add to your answer exactly how this works for the specific example of the two qubit state and depolarizing channel (such that one obtains the Pauli matrices)? I am not sure how to express $mathcal{E}$ in the first equation you have written? I assume $Omega$ is the Bell state for two qubits but I'm not sure what exactly $(mathcal{E}otimesmathbb{1})(vertOmegaranglelangleOmegavert)$ looks like.
            – user1936752
            4 hours ago










          • @user1936752 Well, you have to know how you are given the channel. But in whichever form you are given the channel, you should have a way to apply it to an input state. -- Maybe could you first explain what you tried to apply this to your example?
            – Norbert Schuch
            4 hours ago










          • I see. I'm only given the effect of the channel i.e. $mathcal{E}(rho) = frac{pI}{2} + (1-p)rho$. Is this what you mean? The presentation of this in Nielsen and Chuang is that one can see that the identity operator can be expressed (as in the question) as a summation of Pauli operators. This gives immediately the Kraus operators. However, I cannot see how to get the same result through your suggestion (and also I don't know how to generalize this to higher dimensions).
            – user1936752
            3 hours ago










          • @user1936752 Have edited.
            – Norbert Schuch
            1 hour ago











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          oldest

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          active

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          up vote
          3
          down vote













          This really depends where you want to start from. For instance, you can construct the Choi state of $mathcal E$, i.e.,
          $$
          sigma = (mathcal E otimes mathbb I)(|OmegaranglelangleOmega|) ,
          $$
          with $Omega = tfrac{1}{sqrt{D}}sum_{i=1}^D |i,irangle$, and then extract the Kraus operators of $mathcal E(rho)=sum M_irho M_i^dagger$ by taking any decomposition
          $$
          sigma = sum |psi_iranglelanglepsi_i| ,tag{*}
          $$
          and writing $|psi_irangle = (M_iotimesmathbb I)|Omegarangle$ (which is always possible).



          Note that the decomposition $(*)$ is highly non-unique (any $|phi_jrangle = sum V_{ij} |psi_irangle$, with $V$ an isometry, is also a valid decomposition), which relates to the fact that the Kraus decomposition is equally non-unique. Obviously, the eigenvalue decomposition is a simple choice (which, moreover, minimizes the number of Kraus operators).




          Let's look at your example in a bit more detail. Here, $D=2$. You have that
          $$
          mathcal E(X)=pmathrm{tr}(X),frac{mathbb I}{2}+(1-p)X
          $$
          for any $X$ (due to linearity) -- the $mathrm{tr}(X)$ is required to make this trace-preserving for general $X$.



          We now have that
          begin{align}
          sigma &= (mathcal E otimes mathbb I)(|Omegaranglelangle Omega|)
          \
          & = tfrac1D sum_{ij} mathcal E(|iranglelangle j|)otimes |iranglelangle j|
          end{align}
          inserting the definition of $|Omegarangle$ and using linearity.



          This yields
          $$
          sigma = frac{p}{2D}mathbb Iotimes sum_{i}|iranglelangle i| +
          (1-p)frac1D sum_{ij}|iranglelangle j|otimes |iranglelangle j| .
          $$
          The second term is just $(1-p)|OmegaranglelangleOmega|$, and the first term is
          $frac{p}{2D}mathbb Iotimesmathbb I$.



          You can now see that one possible eigenvalue decomposition of $sigma$ is given by the four Bell states (I leave it to you to work out the weights), and it is well known and easy to check that that the four Bell states can be written as
          $$
          (sigma_kotimes mathbb I)|Omegarangle ,
          $$
          where $sigma_k$ are the three Pauli matrices or the identity.



          Thus, you get that the $M_i$ in the Kraus representation are the Paulis and the identity, with the weight given by the eigenvalue decomposition of $sigma$.






          share|improve this answer























          • Could you add to your answer exactly how this works for the specific example of the two qubit state and depolarizing channel (such that one obtains the Pauli matrices)? I am not sure how to express $mathcal{E}$ in the first equation you have written? I assume $Omega$ is the Bell state for two qubits but I'm not sure what exactly $(mathcal{E}otimesmathbb{1})(vertOmegaranglelangleOmegavert)$ looks like.
            – user1936752
            4 hours ago










          • @user1936752 Well, you have to know how you are given the channel. But in whichever form you are given the channel, you should have a way to apply it to an input state. -- Maybe could you first explain what you tried to apply this to your example?
            – Norbert Schuch
            4 hours ago










          • I see. I'm only given the effect of the channel i.e. $mathcal{E}(rho) = frac{pI}{2} + (1-p)rho$. Is this what you mean? The presentation of this in Nielsen and Chuang is that one can see that the identity operator can be expressed (as in the question) as a summation of Pauli operators. This gives immediately the Kraus operators. However, I cannot see how to get the same result through your suggestion (and also I don't know how to generalize this to higher dimensions).
            – user1936752
            3 hours ago










          • @user1936752 Have edited.
            – Norbert Schuch
            1 hour ago















          up vote
          3
          down vote













          This really depends where you want to start from. For instance, you can construct the Choi state of $mathcal E$, i.e.,
          $$
          sigma = (mathcal E otimes mathbb I)(|OmegaranglelangleOmega|) ,
          $$
          with $Omega = tfrac{1}{sqrt{D}}sum_{i=1}^D |i,irangle$, and then extract the Kraus operators of $mathcal E(rho)=sum M_irho M_i^dagger$ by taking any decomposition
          $$
          sigma = sum |psi_iranglelanglepsi_i| ,tag{*}
          $$
          and writing $|psi_irangle = (M_iotimesmathbb I)|Omegarangle$ (which is always possible).



          Note that the decomposition $(*)$ is highly non-unique (any $|phi_jrangle = sum V_{ij} |psi_irangle$, with $V$ an isometry, is also a valid decomposition), which relates to the fact that the Kraus decomposition is equally non-unique. Obviously, the eigenvalue decomposition is a simple choice (which, moreover, minimizes the number of Kraus operators).




          Let's look at your example in a bit more detail. Here, $D=2$. You have that
          $$
          mathcal E(X)=pmathrm{tr}(X),frac{mathbb I}{2}+(1-p)X
          $$
          for any $X$ (due to linearity) -- the $mathrm{tr}(X)$ is required to make this trace-preserving for general $X$.



          We now have that
          begin{align}
          sigma &= (mathcal E otimes mathbb I)(|Omegaranglelangle Omega|)
          \
          & = tfrac1D sum_{ij} mathcal E(|iranglelangle j|)otimes |iranglelangle j|
          end{align}
          inserting the definition of $|Omegarangle$ and using linearity.



          This yields
          $$
          sigma = frac{p}{2D}mathbb Iotimes sum_{i}|iranglelangle i| +
          (1-p)frac1D sum_{ij}|iranglelangle j|otimes |iranglelangle j| .
          $$
          The second term is just $(1-p)|OmegaranglelangleOmega|$, and the first term is
          $frac{p}{2D}mathbb Iotimesmathbb I$.



          You can now see that one possible eigenvalue decomposition of $sigma$ is given by the four Bell states (I leave it to you to work out the weights), and it is well known and easy to check that that the four Bell states can be written as
          $$
          (sigma_kotimes mathbb I)|Omegarangle ,
          $$
          where $sigma_k$ are the three Pauli matrices or the identity.



          Thus, you get that the $M_i$ in the Kraus representation are the Paulis and the identity, with the weight given by the eigenvalue decomposition of $sigma$.






          share|improve this answer























          • Could you add to your answer exactly how this works for the specific example of the two qubit state and depolarizing channel (such that one obtains the Pauli matrices)? I am not sure how to express $mathcal{E}$ in the first equation you have written? I assume $Omega$ is the Bell state for two qubits but I'm not sure what exactly $(mathcal{E}otimesmathbb{1})(vertOmegaranglelangleOmegavert)$ looks like.
            – user1936752
            4 hours ago










          • @user1936752 Well, you have to know how you are given the channel. But in whichever form you are given the channel, you should have a way to apply it to an input state. -- Maybe could you first explain what you tried to apply this to your example?
            – Norbert Schuch
            4 hours ago










          • I see. I'm only given the effect of the channel i.e. $mathcal{E}(rho) = frac{pI}{2} + (1-p)rho$. Is this what you mean? The presentation of this in Nielsen and Chuang is that one can see that the identity operator can be expressed (as in the question) as a summation of Pauli operators. This gives immediately the Kraus operators. However, I cannot see how to get the same result through your suggestion (and also I don't know how to generalize this to higher dimensions).
            – user1936752
            3 hours ago










          • @user1936752 Have edited.
            – Norbert Schuch
            1 hour ago













          up vote
          3
          down vote










          up vote
          3
          down vote









          This really depends where you want to start from. For instance, you can construct the Choi state of $mathcal E$, i.e.,
          $$
          sigma = (mathcal E otimes mathbb I)(|OmegaranglelangleOmega|) ,
          $$
          with $Omega = tfrac{1}{sqrt{D}}sum_{i=1}^D |i,irangle$, and then extract the Kraus operators of $mathcal E(rho)=sum M_irho M_i^dagger$ by taking any decomposition
          $$
          sigma = sum |psi_iranglelanglepsi_i| ,tag{*}
          $$
          and writing $|psi_irangle = (M_iotimesmathbb I)|Omegarangle$ (which is always possible).



          Note that the decomposition $(*)$ is highly non-unique (any $|phi_jrangle = sum V_{ij} |psi_irangle$, with $V$ an isometry, is also a valid decomposition), which relates to the fact that the Kraus decomposition is equally non-unique. Obviously, the eigenvalue decomposition is a simple choice (which, moreover, minimizes the number of Kraus operators).




          Let's look at your example in a bit more detail. Here, $D=2$. You have that
          $$
          mathcal E(X)=pmathrm{tr}(X),frac{mathbb I}{2}+(1-p)X
          $$
          for any $X$ (due to linearity) -- the $mathrm{tr}(X)$ is required to make this trace-preserving for general $X$.



          We now have that
          begin{align}
          sigma &= (mathcal E otimes mathbb I)(|Omegaranglelangle Omega|)
          \
          & = tfrac1D sum_{ij} mathcal E(|iranglelangle j|)otimes |iranglelangle j|
          end{align}
          inserting the definition of $|Omegarangle$ and using linearity.



          This yields
          $$
          sigma = frac{p}{2D}mathbb Iotimes sum_{i}|iranglelangle i| +
          (1-p)frac1D sum_{ij}|iranglelangle j|otimes |iranglelangle j| .
          $$
          The second term is just $(1-p)|OmegaranglelangleOmega|$, and the first term is
          $frac{p}{2D}mathbb Iotimesmathbb I$.



          You can now see that one possible eigenvalue decomposition of $sigma$ is given by the four Bell states (I leave it to you to work out the weights), and it is well known and easy to check that that the four Bell states can be written as
          $$
          (sigma_kotimes mathbb I)|Omegarangle ,
          $$
          where $sigma_k$ are the three Pauli matrices or the identity.



          Thus, you get that the $M_i$ in the Kraus representation are the Paulis and the identity, with the weight given by the eigenvalue decomposition of $sigma$.






          share|improve this answer














          This really depends where you want to start from. For instance, you can construct the Choi state of $mathcal E$, i.e.,
          $$
          sigma = (mathcal E otimes mathbb I)(|OmegaranglelangleOmega|) ,
          $$
          with $Omega = tfrac{1}{sqrt{D}}sum_{i=1}^D |i,irangle$, and then extract the Kraus operators of $mathcal E(rho)=sum M_irho M_i^dagger$ by taking any decomposition
          $$
          sigma = sum |psi_iranglelanglepsi_i| ,tag{*}
          $$
          and writing $|psi_irangle = (M_iotimesmathbb I)|Omegarangle$ (which is always possible).



          Note that the decomposition $(*)$ is highly non-unique (any $|phi_jrangle = sum V_{ij} |psi_irangle$, with $V$ an isometry, is also a valid decomposition), which relates to the fact that the Kraus decomposition is equally non-unique. Obviously, the eigenvalue decomposition is a simple choice (which, moreover, minimizes the number of Kraus operators).




          Let's look at your example in a bit more detail. Here, $D=2$. You have that
          $$
          mathcal E(X)=pmathrm{tr}(X),frac{mathbb I}{2}+(1-p)X
          $$
          for any $X$ (due to linearity) -- the $mathrm{tr}(X)$ is required to make this trace-preserving for general $X$.



          We now have that
          begin{align}
          sigma &= (mathcal E otimes mathbb I)(|Omegaranglelangle Omega|)
          \
          & = tfrac1D sum_{ij} mathcal E(|iranglelangle j|)otimes |iranglelangle j|
          end{align}
          inserting the definition of $|Omegarangle$ and using linearity.



          This yields
          $$
          sigma = frac{p}{2D}mathbb Iotimes sum_{i}|iranglelangle i| +
          (1-p)frac1D sum_{ij}|iranglelangle j|otimes |iranglelangle j| .
          $$
          The second term is just $(1-p)|OmegaranglelangleOmega|$, and the first term is
          $frac{p}{2D}mathbb Iotimesmathbb I$.



          You can now see that one possible eigenvalue decomposition of $sigma$ is given by the four Bell states (I leave it to you to work out the weights), and it is well known and easy to check that that the four Bell states can be written as
          $$
          (sigma_kotimes mathbb I)|Omegarangle ,
          $$
          where $sigma_k$ are the three Pauli matrices or the identity.



          Thus, you get that the $M_i$ in the Kraus representation are the Paulis and the identity, with the weight given by the eigenvalue decomposition of $sigma$.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 1 hour ago

























          answered 4 hours ago









          Norbert Schuch

          1,188211




          1,188211












          • Could you add to your answer exactly how this works for the specific example of the two qubit state and depolarizing channel (such that one obtains the Pauli matrices)? I am not sure how to express $mathcal{E}$ in the first equation you have written? I assume $Omega$ is the Bell state for two qubits but I'm not sure what exactly $(mathcal{E}otimesmathbb{1})(vertOmegaranglelangleOmegavert)$ looks like.
            – user1936752
            4 hours ago










          • @user1936752 Well, you have to know how you are given the channel. But in whichever form you are given the channel, you should have a way to apply it to an input state. -- Maybe could you first explain what you tried to apply this to your example?
            – Norbert Schuch
            4 hours ago










          • I see. I'm only given the effect of the channel i.e. $mathcal{E}(rho) = frac{pI}{2} + (1-p)rho$. Is this what you mean? The presentation of this in Nielsen and Chuang is that one can see that the identity operator can be expressed (as in the question) as a summation of Pauli operators. This gives immediately the Kraus operators. However, I cannot see how to get the same result through your suggestion (and also I don't know how to generalize this to higher dimensions).
            – user1936752
            3 hours ago










          • @user1936752 Have edited.
            – Norbert Schuch
            1 hour ago


















          • Could you add to your answer exactly how this works for the specific example of the two qubit state and depolarizing channel (such that one obtains the Pauli matrices)? I am not sure how to express $mathcal{E}$ in the first equation you have written? I assume $Omega$ is the Bell state for two qubits but I'm not sure what exactly $(mathcal{E}otimesmathbb{1})(vertOmegaranglelangleOmegavert)$ looks like.
            – user1936752
            4 hours ago










          • @user1936752 Well, you have to know how you are given the channel. But in whichever form you are given the channel, you should have a way to apply it to an input state. -- Maybe could you first explain what you tried to apply this to your example?
            – Norbert Schuch
            4 hours ago










          • I see. I'm only given the effect of the channel i.e. $mathcal{E}(rho) = frac{pI}{2} + (1-p)rho$. Is this what you mean? The presentation of this in Nielsen and Chuang is that one can see that the identity operator can be expressed (as in the question) as a summation of Pauli operators. This gives immediately the Kraus operators. However, I cannot see how to get the same result through your suggestion (and also I don't know how to generalize this to higher dimensions).
            – user1936752
            3 hours ago










          • @user1936752 Have edited.
            – Norbert Schuch
            1 hour ago
















          Could you add to your answer exactly how this works for the specific example of the two qubit state and depolarizing channel (such that one obtains the Pauli matrices)? I am not sure how to express $mathcal{E}$ in the first equation you have written? I assume $Omega$ is the Bell state for two qubits but I'm not sure what exactly $(mathcal{E}otimesmathbb{1})(vertOmegaranglelangleOmegavert)$ looks like.
          – user1936752
          4 hours ago




          Could you add to your answer exactly how this works for the specific example of the two qubit state and depolarizing channel (such that one obtains the Pauli matrices)? I am not sure how to express $mathcal{E}$ in the first equation you have written? I assume $Omega$ is the Bell state for two qubits but I'm not sure what exactly $(mathcal{E}otimesmathbb{1})(vertOmegaranglelangleOmegavert)$ looks like.
          – user1936752
          4 hours ago












          @user1936752 Well, you have to know how you are given the channel. But in whichever form you are given the channel, you should have a way to apply it to an input state. -- Maybe could you first explain what you tried to apply this to your example?
          – Norbert Schuch
          4 hours ago




          @user1936752 Well, you have to know how you are given the channel. But in whichever form you are given the channel, you should have a way to apply it to an input state. -- Maybe could you first explain what you tried to apply this to your example?
          – Norbert Schuch
          4 hours ago












          I see. I'm only given the effect of the channel i.e. $mathcal{E}(rho) = frac{pI}{2} + (1-p)rho$. Is this what you mean? The presentation of this in Nielsen and Chuang is that one can see that the identity operator can be expressed (as in the question) as a summation of Pauli operators. This gives immediately the Kraus operators. However, I cannot see how to get the same result through your suggestion (and also I don't know how to generalize this to higher dimensions).
          – user1936752
          3 hours ago




          I see. I'm only given the effect of the channel i.e. $mathcal{E}(rho) = frac{pI}{2} + (1-p)rho$. Is this what you mean? The presentation of this in Nielsen and Chuang is that one can see that the identity operator can be expressed (as in the question) as a summation of Pauli operators. This gives immediately the Kraus operators. However, I cannot see how to get the same result through your suggestion (and also I don't know how to generalize this to higher dimensions).
          – user1936752
          3 hours ago












          @user1936752 Have edited.
          – Norbert Schuch
          1 hour ago




          @user1936752 Have edited.
          – Norbert Schuch
          1 hour ago


















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          flock() on closed filehandle LOCK_FILE at /usr/bin/apt-mirror

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          0 I recently upgraded an Ubuntu 16.04 virtual machine to 18.04. I regularly use this VM to run apt-mirror at school on their high-speed internet, then rsync these files to my rack-mount server at home for updating multiple computers at home without wasting the tiny amount of bandwidth I have there. Since updating I can't get apt-mirror to run on my VM. It keeps giving the error: flock() on closed filehandle LOCK_FILE at /usr/bin/apt-mirror line 214 immediately followed by: apt-mirror is already running, exiting at /usr/bin/apt-mirror line 217 apt-mirror is NOT running, and does not have a cron job. I've tried rebooting, and even looked for the old-style lockfile to delete, but I guess that method is no longer used. Is there some way to force unlock this so I can run apt-mirror? UPDATE: So... I f...
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          Mangá

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            Nota:  Não confundir com Manga. Mangás e Animes Mangá Mangaká • Tankōbon Film comic La nouvelle manga Mangá original em inglês Anime História Animação influenciada por animes • Filler • OVA • Seiyū Públicos Kodomo • Shōjo • Shōnen Josei • Seinen Gêneros Mahō shōjo • Mecha • Harém Ecchi • Hentai ( Yaoi • Yuri Futanari • Shotacon • Lolicon Toddlercon ) Termos Bishōjo • Chibi • Fan service Gaiden • Kawaii • Kaitō • Light novel Mahō shōnen • Moe (antropomorfismo • Kemonomimi Nekomimi ) • Super deformed Omake • Tsundere • Yonkoma Listas Mangás • Animes • Ecchi • Mangaka • Estúdios • Revistas Fãs ( otaku ) Anime Music Video Convenções • Cosplay Dōjinshi • Fansub • Fandub • Fujoshi OS-tan • Scanlation Portal de Anime e mangá v d e O mangá (português brasileiro) ou manga (português europeu) [ 1 ] (em japonês: 漫画 , manga ? , lit. “história em quadrinhos”) , é a ...
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          Eduardo VII do Reino Unido

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          Eduardo VII Rei do Reino Unido da Grã-Bretanha e Irlanda e dos Domínios Britânicos de Além-mar Imperador da Índia Rei do Reino Unido e Imperador da Índia Reinado 22 de janeiro de 1901 a 6 de maio de 1910 Coroação 9 de agosto de 1902 Delhi Durbar 1 de janeiro de 1903 Predecessora Vitória Sucessor Jorge V   Esposa Alexandra da Dinamarca Descendência Alberto Vitor, Duque de Clarence e Avondale Jorge V do Reino Unido Luísa, Princesa Real Vitória de Gales Maud de Gales Alexandre João de Gales Casa Saxe-Coburgo-Gota Nome completo Alberto Eduardo Nascimento 9 de novembro de 1841   Palácio de Buckingham, Londres, Reino Unido Morte 6 de maio de 1910 (68 anos)   Palácio de Buckingham, Londres, Reino Unido Enterro Capela de São Jorge, Windsor, Berkshire, Reino Unido 20 de maio de 1910 Pai Alberto de Saxe-Coburgo-Gota Mãe Vitória do Reino Unido Assinatura ...
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