Use of component tensor product in quantum mechanics?

[jk22]

suppose we consider the measurement operator A=diag(1,-1).

Then the tensor product of A by itself is in components : A\otimes A=a_{ij}a_{kl}=c_{ijkl}

giving c_{1111}=c_{2222}=1, c_{1122}=c_{2211}=-1 and all other component 0.

to diagonalize a tensor of order 4, we write :

c_{ijkl}b_{kl}=\lambda b_{ij}

hence in matrix form : \left(\begin{array}{cc} a-d & 0\\0 & -a+d\end{array}\right)=\lambda\left(\begin{array}{cc}a & b\\c & d\end{array}\right)

we have 2 cases :

\lambda\neq 0\Rightarrow b=c=0,a=-d\Rightarrow\lambda=2

the other case has eigenvalue 0.

Hence if we use the tensor product in component (instead of Kronecker product), we don't get an isomorphism for the eigenvalue, in other words the eigenvalues of the tensor product of operators is not the product of the eigenvalues.

Does anyone know if the tensor product has any meaning in quantum mechanics ?

 

[Bill_K]

Don't understand. The definition of the tensor product is
A \otimes B = \left( \begin{array}{cccc}a_{11}b_{11}&a_{11}b_{12}&a_{12}b_{11}&a_{12}b_{12} \\a_{11}b_{21}&a_{11}b_{22}&a_{12}b_{21}&a_{12}b_{22} \\a_{21}b_{11}&a_{21}b_{12}&a_{22}b_{11}&a_{22}b_{12} \\a_{21}b_{21}&a_{21}b_{22}&a_{22}b_{21}&a_{22}b_{22} \end{array} \right)
Substitute your form for A, and what you get is sure enough diagonal, with eigenvalues ±1.

 

[jk22]

Indeed. I made a mistake, the application of the tensor on the multivector is C_{ijkl}d_{jl}

Thanks for your help.