Expanding the Square of a Linear Operator - Linear Algebra Question for Quantum Mechanics

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I have taken linear algebra and have some experience with basic application to quantum mechanics. I am reading a paper in quantum mechanics and am having trouble following some of the math. Here is a link to the paper. Here I refer to equations (5) and (6) on page 412.



I am given a vector $Psi$ and operators $A, A', B,$ and $B'$. A fifth operator is defined:



$ F = A otimes B + A otimes B' + A' otimes B - A' otimes B' $.



It is shown that $Psi$ is an eigenvector of $F$ with eigenvalue $2 sqrt2$.



Then $ F^2 Psi = 8 Psi implies i(AA' - A'A) i(BB' - B'B) Psi = 4 Psi $.



The first equality is trivial, but I do not understand why it implies the second. I attempted to expand $F^2$ but don't see a relationship with $i(AA' - A'A) i(BB' - B'B)$. Can someone help me to understand this?



I have tried asking others and searching online, but I have failed to find adequate search terms.



I should note that $i(AA' - A'A)$ and $i(BB' - B'B)$ are referred to in the paper as Hermitian combinations.




linear-algebra quantum-mechanics




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    up vote
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    I have taken linear algebra and have some experience with basic application to quantum mechanics. I am reading a paper in quantum mechanics and am having trouble following some of the math. Here is a link to the paper. Here I refer to equations (5) and (6) on page 412.



    I am given a vector $Psi$ and operators $A, A', B,$ and $B'$. A fifth operator is defined:



    $ F = A otimes B + A otimes B' + A' otimes B - A' otimes B' $.



    It is shown that $Psi$ is an eigenvector of $F$ with eigenvalue $2 sqrt2$.



    Then $ F^2 Psi = 8 Psi implies i(AA' - A'A) i(BB' - B'B) Psi = 4 Psi $.



    The first equality is trivial, but I do not understand why it implies the second. I attempted to expand $F^2$ but don't see a relationship with $i(AA' - A'A) i(BB' - B'B)$. Can someone help me to understand this?



    I have tried asking others and searching online, but I have failed to find adequate search terms.



    I should note that $i(AA' - A'A)$ and $i(BB' - B'B)$ are referred to in the paper as Hermitian combinations.




    linear-algebra quantum-mechanics




    share|cite|improve this question





















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      I have taken linear algebra and have some experience with basic application to quantum mechanics. I am reading a paper in quantum mechanics and am having trouble following some of the math. Here is a link to the paper. Here I refer to equations (5) and (6) on page 412.



      I am given a vector $Psi$ and operators $A, A', B,$ and $B'$. A fifth operator is defined:



      $ F = A otimes B + A otimes B' + A' otimes B - A' otimes B' $.



      It is shown that $Psi$ is an eigenvector of $F$ with eigenvalue $2 sqrt2$.



      Then $ F^2 Psi = 8 Psi implies i(AA' - A'A) i(BB' - B'B) Psi = 4 Psi $.



      The first equality is trivial, but I do not understand why it implies the second. I attempted to expand $F^2$ but don't see a relationship with $i(AA' - A'A) i(BB' - B'B)$. Can someone help me to understand this?



      I have tried asking others and searching online, but I have failed to find adequate search terms.



      I should note that $i(AA' - A'A)$ and $i(BB' - B'B)$ are referred to in the paper as Hermitian combinations.




      linear-algebra quantum-mechanics




      share|cite|improve this question











      I have taken linear algebra and have some experience with basic application to quantum mechanics. I am reading a paper in quantum mechanics and am having trouble following some of the math. Here is a link to the paper. Here I refer to equations (5) and (6) on page 412.



      I am given a vector $Psi$ and operators $A, A', B,$ and $B'$. A fifth operator is defined:



      $ F = A otimes B + A otimes B' + A' otimes B - A' otimes B' $.



      It is shown that $Psi$ is an eigenvector of $F$ with eigenvalue $2 sqrt2$.



      Then $ F^2 Psi = 8 Psi implies i(AA' - A'A) i(BB' - B'B) Psi = 4 Psi $.



      The first equality is trivial, but I do not understand why it implies the second. I attempted to expand $F^2$ but don't see a relationship with $i(AA' - A'A) i(BB' - B'B)$. Can someone help me to understand this?



      I have tried asking others and searching online, but I have failed to find adequate search terms.



      I should note that $i(AA' - A'A)$ and $i(BB' - B'B)$ are referred to in the paper as Hermitian combinations.





      linear-algebra quantum-mechanics





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      asked Jul 28 at 3:24









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          The trick is in an additional assumption between (4) and (5):




          At the beginning, it is convenient to restrict the eigenvalues of the operators $A$, $A'$, $B$ and $B'$ to $-1$ and $1$ only.




          Thus as $A$,$A'$,$B$ and $B'$ are Hermitian operators we can deduce that:
          $$A^2=A'^2=B'^2=B^2=I$$
          So:
          beginequation
          (AB+AB'+A'B-A'B')^2=((A+A') B +(A-A')B')^2=(A+A')^2B^2+(A-A')^2 B'^2+(A+A')(A-A')BB'+(A-A')(A+A')B'B=(AA'+A'A+2 I)+(2I -AA'-A'A)+(A'A-AA')BB'+(AA'-A'A)B'B
          endequation
          thus:
          $$F^2=4 I -(AA'-A'A)(BB'-B'B)$$
          And:
          $$F^2 psi= 8 psi=4psi+i(AA'-A'A)i(BB'-B'B) psi$$






          share|cite|improve this answer























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            up vote
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            down vote



            accepted










            The trick is in an additional assumption between (4) and (5):




            At the beginning, it is convenient to restrict the eigenvalues of the operators $A$, $A'$, $B$ and $B'$ to $-1$ and $1$ only.




            Thus as $A$,$A'$,$B$ and $B'$ are Hermitian operators we can deduce that:
            $$A^2=A'^2=B'^2=B^2=I$$
            So:
            beginequation
            (AB+AB'+A'B-A'B')^2=((A+A') B +(A-A')B')^2=(A+A')^2B^2+(A-A')^2 B'^2+(A+A')(A-A')BB'+(A-A')(A+A')B'B=(AA'+A'A+2 I)+(2I -AA'-A'A)+(A'A-AA')BB'+(AA'-A'A)B'B
            endequation
            thus:
            $$F^2=4 I -(AA'-A'A)(BB'-B'B)$$
            And:
            $$F^2 psi= 8 psi=4psi+i(AA'-A'A)i(BB'-B'B) psi$$






            share|cite|improve this answer



























              up vote
              3
              down vote



              accepted










              The trick is in an additional assumption between (4) and (5):




              At the beginning, it is convenient to restrict the eigenvalues of the operators $A$, $A'$, $B$ and $B'$ to $-1$ and $1$ only.




              Thus as $A$,$A'$,$B$ and $B'$ are Hermitian operators we can deduce that:
              $$A^2=A'^2=B'^2=B^2=I$$
              So:
              beginequation
              (AB+AB'+A'B-A'B')^2=((A+A') B +(A-A')B')^2=(A+A')^2B^2+(A-A')^2 B'^2+(A+A')(A-A')BB'+(A-A')(A+A')B'B=(AA'+A'A+2 I)+(2I -AA'-A'A)+(A'A-AA')BB'+(AA'-A'A)B'B
              endequation
              thus:
              $$F^2=4 I -(AA'-A'A)(BB'-B'B)$$
              And:
              $$F^2 psi= 8 psi=4psi+i(AA'-A'A)i(BB'-B'B) psi$$






              share|cite|improve this answer

























                up vote
                3
                down vote



                accepted







                up vote
                3
                down vote



                accepted






                The trick is in an additional assumption between (4) and (5):




                At the beginning, it is convenient to restrict the eigenvalues of the operators $A$, $A'$, $B$ and $B'$ to $-1$ and $1$ only.




                Thus as $A$,$A'$,$B$ and $B'$ are Hermitian operators we can deduce that:
                $$A^2=A'^2=B'^2=B^2=I$$
                So:
                beginequation
                (AB+AB'+A'B-A'B')^2=((A+A') B +(A-A')B')^2=(A+A')^2B^2+(A-A')^2 B'^2+(A+A')(A-A')BB'+(A-A')(A+A')B'B=(AA'+A'A+2 I)+(2I -AA'-A'A)+(A'A-AA')BB'+(AA'-A'A)B'B
                endequation
                thus:
                $$F^2=4 I -(AA'-A'A)(BB'-B'B)$$
                And:
                $$F^2 psi= 8 psi=4psi+i(AA'-A'A)i(BB'-B'B) psi$$






                share|cite|improve this answer















                The trick is in an additional assumption between (4) and (5):




                At the beginning, it is convenient to restrict the eigenvalues of the operators $A$, $A'$, $B$ and $B'$ to $-1$ and $1$ only.




                Thus as $A$,$A'$,$B$ and $B'$ are Hermitian operators we can deduce that:
                $$A^2=A'^2=B'^2=B^2=I$$
                So:
                beginequation
                (AB+AB'+A'B-A'B')^2=((A+A') B +(A-A')B')^2=(A+A')^2B^2+(A-A')^2 B'^2+(A+A')(A-A')BB'+(A-A')(A+A')B'B=(AA'+A'A+2 I)+(2I -AA'-A'A)+(A'A-AA')BB'+(AA'-A'A)B'B
                endequation
                thus:
                $$F^2=4 I -(AA'-A'A)(BB'-B'B)$$
                And:
                $$F^2 psi= 8 psi=4psi+i(AA'-A'A)i(BB'-B'B) psi$$







                share|cite|improve this answer















                share|cite|improve this answer



                share|cite|improve this answer








                edited Jul 31 at 16:01









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                answered Jul 28 at 6:49









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