Notation for two qubit composite product state





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In my lecture notes on quantum information processing my lecturer gives an example of composite systems as $|phirangle=|0rangle |0rangle=|00rangle$. I understand that if we have two qubits then its product state will be in 2n dimensional Hilbert space and I understand the 2 qubit state $|00rangle$ to be represented in matrix representation as $begin{pmatrix} 1 & 1 \ 0 & 0 end{pmatrix}$ (if that is wrong please do correct my misunderstanding though). My question is about the notation $|0rangle|0rangle=|00rangle$, how can we calculate this with matrices on the left-hand side we have a 2 by 1 matrix multiplied by a 2 by 1 matrix which cannot be calculated. I thought perhaps it was a matter of direct products but my calculation led to an incorrect result there too.



Could anyone clarify this for me, please?



Edit: It occurred to me that I think I'm mistaken about the matrix representation of $|00rangle$, I think it would make more sense to be $begin{pmatrix} 1 \ 0\0\0 end{pmatrix}$ in which case the direct product does work and I should take the notation $|0rangle|0rangle$ to be a shorthand for the direct product not the multiplication of two matrices, is that correct?










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    In my lecture notes on quantum information processing my lecturer gives an example of composite systems as $|phirangle=|0rangle |0rangle=|00rangle$. I understand that if we have two qubits then its product state will be in 2n dimensional Hilbert space and I understand the 2 qubit state $|00rangle$ to be represented in matrix representation as $begin{pmatrix} 1 & 1 \ 0 & 0 end{pmatrix}$ (if that is wrong please do correct my misunderstanding though). My question is about the notation $|0rangle|0rangle=|00rangle$, how can we calculate this with matrices on the left-hand side we have a 2 by 1 matrix multiplied by a 2 by 1 matrix which cannot be calculated. I thought perhaps it was a matter of direct products but my calculation led to an incorrect result there too.



    Could anyone clarify this for me, please?



    Edit: It occurred to me that I think I'm mistaken about the matrix representation of $|00rangle$, I think it would make more sense to be $begin{pmatrix} 1 \ 0\0\0 end{pmatrix}$ in which case the direct product does work and I should take the notation $|0rangle|0rangle$ to be a shorthand for the direct product not the multiplication of two matrices, is that correct?










    share|improve this question









    New contributor




    can'tcauchy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      In my lecture notes on quantum information processing my lecturer gives an example of composite systems as $|phirangle=|0rangle |0rangle=|00rangle$. I understand that if we have two qubits then its product state will be in 2n dimensional Hilbert space and I understand the 2 qubit state $|00rangle$ to be represented in matrix representation as $begin{pmatrix} 1 & 1 \ 0 & 0 end{pmatrix}$ (if that is wrong please do correct my misunderstanding though). My question is about the notation $|0rangle|0rangle=|00rangle$, how can we calculate this with matrices on the left-hand side we have a 2 by 1 matrix multiplied by a 2 by 1 matrix which cannot be calculated. I thought perhaps it was a matter of direct products but my calculation led to an incorrect result there too.



      Could anyone clarify this for me, please?



      Edit: It occurred to me that I think I'm mistaken about the matrix representation of $|00rangle$, I think it would make more sense to be $begin{pmatrix} 1 \ 0\0\0 end{pmatrix}$ in which case the direct product does work and I should take the notation $|0rangle|0rangle$ to be a shorthand for the direct product not the multiplication of two matrices, is that correct?










      share|improve this question









      New contributor




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







      $endgroup$




      In my lecture notes on quantum information processing my lecturer gives an example of composite systems as $|phirangle=|0rangle |0rangle=|00rangle$. I understand that if we have two qubits then its product state will be in 2n dimensional Hilbert space and I understand the 2 qubit state $|00rangle$ to be represented in matrix representation as $begin{pmatrix} 1 & 1 \ 0 & 0 end{pmatrix}$ (if that is wrong please do correct my misunderstanding though). My question is about the notation $|0rangle|0rangle=|00rangle$, how can we calculate this with matrices on the left-hand side we have a 2 by 1 matrix multiplied by a 2 by 1 matrix which cannot be calculated. I thought perhaps it was a matter of direct products but my calculation led to an incorrect result there too.



      Could anyone clarify this for me, please?



      Edit: It occurred to me that I think I'm mistaken about the matrix representation of $|00rangle$, I think it would make more sense to be $begin{pmatrix} 1 \ 0\0\0 end{pmatrix}$ in which case the direct product does work and I should take the notation $|0rangle|0rangle$ to be a shorthand for the direct product not the multiplication of two matrices, is that correct?







      quantum-state tensor-product notation






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      edited 4 hours ago









      Sanchayan Dutta

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      can'tcauchycan'tcauchy

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          $begingroup$

          $|0rangle|0rangle$ is actually a shorthand for $|0rangle otimes |0rangle$ or $begin{bmatrix} 1 \ 0 end{bmatrix} otimes begin{bmatrix} 1 \ 0end{bmatrix} $ where $otimes$ stands for the tensor product or essentially the Kronecker product. To quote Wikipedia:




          In mathematics, the Kronecker product, denoted by $otimes$, is an operation
          on two matrices of arbitrary size resulting in a block matrix. It is a
          generalization of the outer product (which is denoted by the same
          symbol) from vectors to matrices, and gives the matrix of the tensor
          product with respect to a standard choice of basis
          . The Kronecker
          product should not be confused with the usual matrix multiplication,
          which is an entirely different operation
          .




          Now the standard choice of basis for a two-qubit system is:



          ${|00rangle = begin{bmatrix} 1 \ 0 \ 0 \ 0 end{bmatrix}, |01rangle = begin{bmatrix} 0 \ 1 \ 0 \ 0 end{bmatrix}, |10rangle = begin{bmatrix} 0 \ 0 \ 1 \ 0 end{bmatrix}, |11rangle = begin{bmatrix} 0 \ 0 \ 0 \ 1 end{bmatrix}}$



          If you wish, you can also represent the basis as (if you strictly take $otimes$ as the outer product):



          ${|00rangle = begin{bmatrix} 1 & 0 \ 0 & 0 end{bmatrix}, |01rangle = begin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix}, |10rangle = begin{bmatrix} 0 & 0 \ 1 & 0 end{bmatrix}, |11rangle = begin{bmatrix} 0 & 0 \ 0 & 1 end{bmatrix}}$



          but then while carrying out calculations like determining the action of a quantum gate on a composite state you'd have to write the state using the vector representation (carefully read the linked Mathematics SE answer).



          The key point here is that don't be bent on thinking of these linear algebraic operations in terms of matrices, but rather think in terms of linear maps. You'll get more comfortable with these things once you learn about tensors!



          P.S: Kronecker product and outer product confusion






          share|improve this answer











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            $begingroup$

            $|0rangle|0rangle$ is actually a shorthand for $|0rangle otimes |0rangle$ or $begin{bmatrix} 1 \ 0 end{bmatrix} otimes begin{bmatrix} 1 \ 0end{bmatrix} $ where $otimes$ stands for the tensor product or essentially the Kronecker product. To quote Wikipedia:




            In mathematics, the Kronecker product, denoted by $otimes$, is an operation
            on two matrices of arbitrary size resulting in a block matrix. It is a
            generalization of the outer product (which is denoted by the same
            symbol) from vectors to matrices, and gives the matrix of the tensor
            product with respect to a standard choice of basis
            . The Kronecker
            product should not be confused with the usual matrix multiplication,
            which is an entirely different operation
            .




            Now the standard choice of basis for a two-qubit system is:



            ${|00rangle = begin{bmatrix} 1 \ 0 \ 0 \ 0 end{bmatrix}, |01rangle = begin{bmatrix} 0 \ 1 \ 0 \ 0 end{bmatrix}, |10rangle = begin{bmatrix} 0 \ 0 \ 1 \ 0 end{bmatrix}, |11rangle = begin{bmatrix} 0 \ 0 \ 0 \ 1 end{bmatrix}}$



            If you wish, you can also represent the basis as (if you strictly take $otimes$ as the outer product):



            ${|00rangle = begin{bmatrix} 1 & 0 \ 0 & 0 end{bmatrix}, |01rangle = begin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix}, |10rangle = begin{bmatrix} 0 & 0 \ 1 & 0 end{bmatrix}, |11rangle = begin{bmatrix} 0 & 0 \ 0 & 1 end{bmatrix}}$



            but then while carrying out calculations like determining the action of a quantum gate on a composite state you'd have to write the state using the vector representation (carefully read the linked Mathematics SE answer).



            The key point here is that don't be bent on thinking of these linear algebraic operations in terms of matrices, but rather think in terms of linear maps. You'll get more comfortable with these things once you learn about tensors!



            P.S: Kronecker product and outer product confusion






            share|improve this answer











            $endgroup$


















              2












              $begingroup$

              $|0rangle|0rangle$ is actually a shorthand for $|0rangle otimes |0rangle$ or $begin{bmatrix} 1 \ 0 end{bmatrix} otimes begin{bmatrix} 1 \ 0end{bmatrix} $ where $otimes$ stands for the tensor product or essentially the Kronecker product. To quote Wikipedia:




              In mathematics, the Kronecker product, denoted by $otimes$, is an operation
              on two matrices of arbitrary size resulting in a block matrix. It is a
              generalization of the outer product (which is denoted by the same
              symbol) from vectors to matrices, and gives the matrix of the tensor
              product with respect to a standard choice of basis
              . The Kronecker
              product should not be confused with the usual matrix multiplication,
              which is an entirely different operation
              .




              Now the standard choice of basis for a two-qubit system is:



              ${|00rangle = begin{bmatrix} 1 \ 0 \ 0 \ 0 end{bmatrix}, |01rangle = begin{bmatrix} 0 \ 1 \ 0 \ 0 end{bmatrix}, |10rangle = begin{bmatrix} 0 \ 0 \ 1 \ 0 end{bmatrix}, |11rangle = begin{bmatrix} 0 \ 0 \ 0 \ 1 end{bmatrix}}$



              If you wish, you can also represent the basis as (if you strictly take $otimes$ as the outer product):



              ${|00rangle = begin{bmatrix} 1 & 0 \ 0 & 0 end{bmatrix}, |01rangle = begin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix}, |10rangle = begin{bmatrix} 0 & 0 \ 1 & 0 end{bmatrix}, |11rangle = begin{bmatrix} 0 & 0 \ 0 & 1 end{bmatrix}}$



              but then while carrying out calculations like determining the action of a quantum gate on a composite state you'd have to write the state using the vector representation (carefully read the linked Mathematics SE answer).



              The key point here is that don't be bent on thinking of these linear algebraic operations in terms of matrices, but rather think in terms of linear maps. You'll get more comfortable with these things once you learn about tensors!



              P.S: Kronecker product and outer product confusion






              share|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                $|0rangle|0rangle$ is actually a shorthand for $|0rangle otimes |0rangle$ or $begin{bmatrix} 1 \ 0 end{bmatrix} otimes begin{bmatrix} 1 \ 0end{bmatrix} $ where $otimes$ stands for the tensor product or essentially the Kronecker product. To quote Wikipedia:




                In mathematics, the Kronecker product, denoted by $otimes$, is an operation
                on two matrices of arbitrary size resulting in a block matrix. It is a
                generalization of the outer product (which is denoted by the same
                symbol) from vectors to matrices, and gives the matrix of the tensor
                product with respect to a standard choice of basis
                . The Kronecker
                product should not be confused with the usual matrix multiplication,
                which is an entirely different operation
                .




                Now the standard choice of basis for a two-qubit system is:



                ${|00rangle = begin{bmatrix} 1 \ 0 \ 0 \ 0 end{bmatrix}, |01rangle = begin{bmatrix} 0 \ 1 \ 0 \ 0 end{bmatrix}, |10rangle = begin{bmatrix} 0 \ 0 \ 1 \ 0 end{bmatrix}, |11rangle = begin{bmatrix} 0 \ 0 \ 0 \ 1 end{bmatrix}}$



                If you wish, you can also represent the basis as (if you strictly take $otimes$ as the outer product):



                ${|00rangle = begin{bmatrix} 1 & 0 \ 0 & 0 end{bmatrix}, |01rangle = begin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix}, |10rangle = begin{bmatrix} 0 & 0 \ 1 & 0 end{bmatrix}, |11rangle = begin{bmatrix} 0 & 0 \ 0 & 1 end{bmatrix}}$



                but then while carrying out calculations like determining the action of a quantum gate on a composite state you'd have to write the state using the vector representation (carefully read the linked Mathematics SE answer).



                The key point here is that don't be bent on thinking of these linear algebraic operations in terms of matrices, but rather think in terms of linear maps. You'll get more comfortable with these things once you learn about tensors!



                P.S: Kronecker product and outer product confusion






                share|improve this answer











                $endgroup$



                $|0rangle|0rangle$ is actually a shorthand for $|0rangle otimes |0rangle$ or $begin{bmatrix} 1 \ 0 end{bmatrix} otimes begin{bmatrix} 1 \ 0end{bmatrix} $ where $otimes$ stands for the tensor product or essentially the Kronecker product. To quote Wikipedia:




                In mathematics, the Kronecker product, denoted by $otimes$, is an operation
                on two matrices of arbitrary size resulting in a block matrix. It is a
                generalization of the outer product (which is denoted by the same
                symbol) from vectors to matrices, and gives the matrix of the tensor
                product with respect to a standard choice of basis
                . The Kronecker
                product should not be confused with the usual matrix multiplication,
                which is an entirely different operation
                .




                Now the standard choice of basis for a two-qubit system is:



                ${|00rangle = begin{bmatrix} 1 \ 0 \ 0 \ 0 end{bmatrix}, |01rangle = begin{bmatrix} 0 \ 1 \ 0 \ 0 end{bmatrix}, |10rangle = begin{bmatrix} 0 \ 0 \ 1 \ 0 end{bmatrix}, |11rangle = begin{bmatrix} 0 \ 0 \ 0 \ 1 end{bmatrix}}$



                If you wish, you can also represent the basis as (if you strictly take $otimes$ as the outer product):



                ${|00rangle = begin{bmatrix} 1 & 0 \ 0 & 0 end{bmatrix}, |01rangle = begin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix}, |10rangle = begin{bmatrix} 0 & 0 \ 1 & 0 end{bmatrix}, |11rangle = begin{bmatrix} 0 & 0 \ 0 & 1 end{bmatrix}}$



                but then while carrying out calculations like determining the action of a quantum gate on a composite state you'd have to write the state using the vector representation (carefully read the linked Mathematics SE answer).



                The key point here is that don't be bent on thinking of these linear algebraic operations in terms of matrices, but rather think in terms of linear maps. You'll get more comfortable with these things once you learn about tensors!



                P.S: Kronecker product and outer product confusion







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 5 hours ago

























                answered 5 hours ago









                Sanchayan DuttaSanchayan Dutta

                6,67641556




                6,67641556






















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