Are projective measurement bases always orthonormal?












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Are projective measurement bases always orthonormal?






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    Related: What is the difference between general measurement and projective measurement? & Wikipedia: POVM
    – Blue♦
    Dec 3 '18 at 7:39


















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Are projective measurement bases always orthonormal?






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  • 2




    Related: What is the difference between general measurement and projective measurement? & Wikipedia: POVM
    – Blue♦
    Dec 3 '18 at 7:39
















3












3








3


1





Are projective measurement bases always orthonormal?






measurement







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Are projective measurement bases always orthonormal?






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edited Dec 3 '18 at 7:21









Blue♦

5,67021354




5,67021354










asked Dec 3 '18 at 6:58









ahelwer

1,270112




1,270112








  • 2




    Related: What is the difference between general measurement and projective measurement? & Wikipedia: POVM
    – Blue♦
    Dec 3 '18 at 7:39
















  • 2




    Related: What is the difference between general measurement and projective measurement? & Wikipedia: POVM
    – Blue♦
    Dec 3 '18 at 7:39










2




2




Related: What is the difference between general measurement and projective measurement? & Wikipedia: POVM
– Blue♦
Dec 3 '18 at 7:39






Related: What is the difference between general measurement and projective measurement? & Wikipedia: POVM
– Blue♦
Dec 3 '18 at 7:39












2 Answers
2






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7














Yes.



Remember that you require several properties of a projective measurement including $P_i^2=P_i$ for each projector, and
$$
sum_iP_i=mathbb{I}.
$$
The first of these show you that the $P_i$ have eigenvalues 0 and 1. Now take a $|phirangle$ that is an eigenvector of eigenvalue 1 of a particular projector $P_i$. Use this in the identity relation:
$$
left(sum_jP_jright)|phirangle=mathbb{I}|phirangle
$$
Clearly, this simplifies to
$$
|phirangle+sum_{jneq i}P_j|phirangle=|phirangle.
$$
Hence,
$$
sum_{jneq i}P_j|phirangle=0.
$$
The $P_j$ are all non-negative, so the only way that this can be 0 is if $P_j|phirangle=0$ for all $jneq i$. (To expand upon this, assume there's a $P_k$ such that $P_k|phirangle=|psirangleneq 0$. This means that
$$
sum_{jneq i,k}langlepsi|P_j|phirangle=-langlepsi|P_k|phirangle,
$$
so some terms must be negative, which is impossible if the eigenvalues are all 0 and 1.)






share|improve this answer































    3














    Here is another way to see this.



    A projection $P$ is an operator such that $P^2=P$.



    This directly implies that we can attach to each projector $P$ a set of orthonormal states that represent it, by choosing any orthonormal base for its range. More precisely, if $P_i$ has trace $operatorname{tr}(P_i)=n$, then we can represent $P_i$ as a set of orthonormal states ${lvertpsi_{i,j}rangle}_{j=1}^n$.
    Note in particular that if $operatorname{tr}(P_i)=1$ then this choice is unique, meaning that there is always a bijection between trace-1 projections and states.



    The projector $P_i$ and the corresponding states are connected through
    $$P_i=sum_{j=1}^n lvertpsi_{ij}rangle!langle psi_{ij}rvert.$$
    In the simpler case of $operatorname{tr}(P_i)=1$ this reads $P_i=lvertpsi_irangle!langlepsi_irvert$.



    Now, if you are asking for a projective measurement basis, then you require a set of operators which describes every possible outcome of your state.
    This condition is expressed mathematically by requiring $$sum_i P_i=I,$$
    which in terms of the associated ket states reads
    $$sum_{ij}lvertpsi_{ij}rangle!langlepsi_{ij}rvert=I,$$
    which is the completeness relation for the vectors ${lvertpsi_{ij}rangle}_{ij}$.
    This immediately implies that this is also an orthonormal set (to see it, take for example the sandwich of this expression with any $lvertpsi_{ij}rangle$).



    Orthogonality of $P_i$ is equivalent to orthogonality of the corresponding $lvertpsi_{ij}rangle$, thus the conclusion.






    share|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      7














      Yes.



      Remember that you require several properties of a projective measurement including $P_i^2=P_i$ for each projector, and
      $$
      sum_iP_i=mathbb{I}.
      $$
      The first of these show you that the $P_i$ have eigenvalues 0 and 1. Now take a $|phirangle$ that is an eigenvector of eigenvalue 1 of a particular projector $P_i$. Use this in the identity relation:
      $$
      left(sum_jP_jright)|phirangle=mathbb{I}|phirangle
      $$
      Clearly, this simplifies to
      $$
      |phirangle+sum_{jneq i}P_j|phirangle=|phirangle.
      $$
      Hence,
      $$
      sum_{jneq i}P_j|phirangle=0.
      $$
      The $P_j$ are all non-negative, so the only way that this can be 0 is if $P_j|phirangle=0$ for all $jneq i$. (To expand upon this, assume there's a $P_k$ such that $P_k|phirangle=|psirangleneq 0$. This means that
      $$
      sum_{jneq i,k}langlepsi|P_j|phirangle=-langlepsi|P_k|phirangle,
      $$
      so some terms must be negative, which is impossible if the eigenvalues are all 0 and 1.)






      share|improve this answer




























        7














        Yes.



        Remember that you require several properties of a projective measurement including $P_i^2=P_i$ for each projector, and
        $$
        sum_iP_i=mathbb{I}.
        $$
        The first of these show you that the $P_i$ have eigenvalues 0 and 1. Now take a $|phirangle$ that is an eigenvector of eigenvalue 1 of a particular projector $P_i$. Use this in the identity relation:
        $$
        left(sum_jP_jright)|phirangle=mathbb{I}|phirangle
        $$
        Clearly, this simplifies to
        $$
        |phirangle+sum_{jneq i}P_j|phirangle=|phirangle.
        $$
        Hence,
        $$
        sum_{jneq i}P_j|phirangle=0.
        $$
        The $P_j$ are all non-negative, so the only way that this can be 0 is if $P_j|phirangle=0$ for all $jneq i$. (To expand upon this, assume there's a $P_k$ such that $P_k|phirangle=|psirangleneq 0$. This means that
        $$
        sum_{jneq i,k}langlepsi|P_j|phirangle=-langlepsi|P_k|phirangle,
        $$
        so some terms must be negative, which is impossible if the eigenvalues are all 0 and 1.)






        share|improve this answer


























          7












          7








          7






          Yes.



          Remember that you require several properties of a projective measurement including $P_i^2=P_i$ for each projector, and
          $$
          sum_iP_i=mathbb{I}.
          $$
          The first of these show you that the $P_i$ have eigenvalues 0 and 1. Now take a $|phirangle$ that is an eigenvector of eigenvalue 1 of a particular projector $P_i$. Use this in the identity relation:
          $$
          left(sum_jP_jright)|phirangle=mathbb{I}|phirangle
          $$
          Clearly, this simplifies to
          $$
          |phirangle+sum_{jneq i}P_j|phirangle=|phirangle.
          $$
          Hence,
          $$
          sum_{jneq i}P_j|phirangle=0.
          $$
          The $P_j$ are all non-negative, so the only way that this can be 0 is if $P_j|phirangle=0$ for all $jneq i$. (To expand upon this, assume there's a $P_k$ such that $P_k|phirangle=|psirangleneq 0$. This means that
          $$
          sum_{jneq i,k}langlepsi|P_j|phirangle=-langlepsi|P_k|phirangle,
          $$
          so some terms must be negative, which is impossible if the eigenvalues are all 0 and 1.)






          share|improve this answer














          Yes.



          Remember that you require several properties of a projective measurement including $P_i^2=P_i$ for each projector, and
          $$
          sum_iP_i=mathbb{I}.
          $$
          The first of these show you that the $P_i$ have eigenvalues 0 and 1. Now take a $|phirangle$ that is an eigenvector of eigenvalue 1 of a particular projector $P_i$. Use this in the identity relation:
          $$
          left(sum_jP_jright)|phirangle=mathbb{I}|phirangle
          $$
          Clearly, this simplifies to
          $$
          |phirangle+sum_{jneq i}P_j|phirangle=|phirangle.
          $$
          Hence,
          $$
          sum_{jneq i}P_j|phirangle=0.
          $$
          The $P_j$ are all non-negative, so the only way that this can be 0 is if $P_j|phirangle=0$ for all $jneq i$. (To expand upon this, assume there's a $P_k$ such that $P_k|phirangle=|psirangleneq 0$. This means that
          $$
          sum_{jneq i,k}langlepsi|P_j|phirangle=-langlepsi|P_k|phirangle,
          $$
          so some terms must be negative, which is impossible if the eigenvalues are all 0 and 1.)







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Dec 3 '18 at 16:20

























          answered Dec 3 '18 at 7:42









          DaftWullie

          12.1k1537




          12.1k1537

























              3














              Here is another way to see this.



              A projection $P$ is an operator such that $P^2=P$.



              This directly implies that we can attach to each projector $P$ a set of orthonormal states that represent it, by choosing any orthonormal base for its range. More precisely, if $P_i$ has trace $operatorname{tr}(P_i)=n$, then we can represent $P_i$ as a set of orthonormal states ${lvertpsi_{i,j}rangle}_{j=1}^n$.
              Note in particular that if $operatorname{tr}(P_i)=1$ then this choice is unique, meaning that there is always a bijection between trace-1 projections and states.



              The projector $P_i$ and the corresponding states are connected through
              $$P_i=sum_{j=1}^n lvertpsi_{ij}rangle!langle psi_{ij}rvert.$$
              In the simpler case of $operatorname{tr}(P_i)=1$ this reads $P_i=lvertpsi_irangle!langlepsi_irvert$.



              Now, if you are asking for a projective measurement basis, then you require a set of operators which describes every possible outcome of your state.
              This condition is expressed mathematically by requiring $$sum_i P_i=I,$$
              which in terms of the associated ket states reads
              $$sum_{ij}lvertpsi_{ij}rangle!langlepsi_{ij}rvert=I,$$
              which is the completeness relation for the vectors ${lvertpsi_{ij}rangle}_{ij}$.
              This immediately implies that this is also an orthonormal set (to see it, take for example the sandwich of this expression with any $lvertpsi_{ij}rangle$).



              Orthogonality of $P_i$ is equivalent to orthogonality of the corresponding $lvertpsi_{ij}rangle$, thus the conclusion.






              share|improve this answer


























                3














                Here is another way to see this.



                A projection $P$ is an operator such that $P^2=P$.



                This directly implies that we can attach to each projector $P$ a set of orthonormal states that represent it, by choosing any orthonormal base for its range. More precisely, if $P_i$ has trace $operatorname{tr}(P_i)=n$, then we can represent $P_i$ as a set of orthonormal states ${lvertpsi_{i,j}rangle}_{j=1}^n$.
                Note in particular that if $operatorname{tr}(P_i)=1$ then this choice is unique, meaning that there is always a bijection between trace-1 projections and states.



                The projector $P_i$ and the corresponding states are connected through
                $$P_i=sum_{j=1}^n lvertpsi_{ij}rangle!langle psi_{ij}rvert.$$
                In the simpler case of $operatorname{tr}(P_i)=1$ this reads $P_i=lvertpsi_irangle!langlepsi_irvert$.



                Now, if you are asking for a projective measurement basis, then you require a set of operators which describes every possible outcome of your state.
                This condition is expressed mathematically by requiring $$sum_i P_i=I,$$
                which in terms of the associated ket states reads
                $$sum_{ij}lvertpsi_{ij}rangle!langlepsi_{ij}rvert=I,$$
                which is the completeness relation for the vectors ${lvertpsi_{ij}rangle}_{ij}$.
                This immediately implies that this is also an orthonormal set (to see it, take for example the sandwich of this expression with any $lvertpsi_{ij}rangle$).



                Orthogonality of $P_i$ is equivalent to orthogonality of the corresponding $lvertpsi_{ij}rangle$, thus the conclusion.






                share|improve this answer
























                  3












                  3








                  3






                  Here is another way to see this.



                  A projection $P$ is an operator such that $P^2=P$.



                  This directly implies that we can attach to each projector $P$ a set of orthonormal states that represent it, by choosing any orthonormal base for its range. More precisely, if $P_i$ has trace $operatorname{tr}(P_i)=n$, then we can represent $P_i$ as a set of orthonormal states ${lvertpsi_{i,j}rangle}_{j=1}^n$.
                  Note in particular that if $operatorname{tr}(P_i)=1$ then this choice is unique, meaning that there is always a bijection between trace-1 projections and states.



                  The projector $P_i$ and the corresponding states are connected through
                  $$P_i=sum_{j=1}^n lvertpsi_{ij}rangle!langle psi_{ij}rvert.$$
                  In the simpler case of $operatorname{tr}(P_i)=1$ this reads $P_i=lvertpsi_irangle!langlepsi_irvert$.



                  Now, if you are asking for a projective measurement basis, then you require a set of operators which describes every possible outcome of your state.
                  This condition is expressed mathematically by requiring $$sum_i P_i=I,$$
                  which in terms of the associated ket states reads
                  $$sum_{ij}lvertpsi_{ij}rangle!langlepsi_{ij}rvert=I,$$
                  which is the completeness relation for the vectors ${lvertpsi_{ij}rangle}_{ij}$.
                  This immediately implies that this is also an orthonormal set (to see it, take for example the sandwich of this expression with any $lvertpsi_{ij}rangle$).



                  Orthogonality of $P_i$ is equivalent to orthogonality of the corresponding $lvertpsi_{ij}rangle$, thus the conclusion.






                  share|improve this answer












                  Here is another way to see this.



                  A projection $P$ is an operator such that $P^2=P$.



                  This directly implies that we can attach to each projector $P$ a set of orthonormal states that represent it, by choosing any orthonormal base for its range. More precisely, if $P_i$ has trace $operatorname{tr}(P_i)=n$, then we can represent $P_i$ as a set of orthonormal states ${lvertpsi_{i,j}rangle}_{j=1}^n$.
                  Note in particular that if $operatorname{tr}(P_i)=1$ then this choice is unique, meaning that there is always a bijection between trace-1 projections and states.



                  The projector $P_i$ and the corresponding states are connected through
                  $$P_i=sum_{j=1}^n lvertpsi_{ij}rangle!langle psi_{ij}rvert.$$
                  In the simpler case of $operatorname{tr}(P_i)=1$ this reads $P_i=lvertpsi_irangle!langlepsi_irvert$.



                  Now, if you are asking for a projective measurement basis, then you require a set of operators which describes every possible outcome of your state.
                  This condition is expressed mathematically by requiring $$sum_i P_i=I,$$
                  which in terms of the associated ket states reads
                  $$sum_{ij}lvertpsi_{ij}rangle!langlepsi_{ij}rvert=I,$$
                  which is the completeness relation for the vectors ${lvertpsi_{ij}rangle}_{ij}$.
                  This immediately implies that this is also an orthonormal set (to see it, take for example the sandwich of this expression with any $lvertpsi_{ij}rangle$).



                  Orthogonality of $P_i$ is equivalent to orthogonality of the corresponding $lvertpsi_{ij}rangle$, thus the conclusion.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Dec 3 '18 at 19:04









                  glS

                  3,566437




                  3,566437






























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