# Use of component tensor product in quantum mechanics?

1. Mar 28, 2014

### 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 ?

2. Mar 28, 2014

### 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.

3. Mar 28, 2014

### jk22

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