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 4 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
            3 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
            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











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          1 Answer
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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
            3 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
            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















          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
            3 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
            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













          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 3 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
            3 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
            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


















          • 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
            3 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
            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
















          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
          3 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
          3 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
          3 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
          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




          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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          Ellipse (mathématiques)

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          Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
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          Quarter-circle Tiles

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          up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
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          Mont Emei

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          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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