In (3x3 ) hermitian model matrix , the P-parity opetaor has eigenvalues -1;+1;+1 . Suppose
I have an operator p having eigenvalues -1;-1;+1 . Can I take p as parity operator ? Give your valuable remark . IF p is not parity operator ; then what is this ? .
B.Rath
@ Biswanath Rath,
Your question one month ago was :
It seems there is a conflict with your current issue,
here we have a 3x3 matrix parity operator! Could you please, explain!
@Issam Kaddoura
My question one month back and now
are the same but reflected in different
tunes. My suggestion is that take any
complex hermitian matrix and verify the different properties of P ( parity)
If you are successful then write me.
Regards
B.Rath
Parity operator in the matrix form and in the Hilbert space of any dimension $n$ has the determinant $-1$. In contrary, the rotation as a matrix in any dimension $n$ has the determinant $+1$. Hence, for $n=3$, the 3x3 diagonal matrix with diagonal elements $-1, -1, +1$ can not be a parity operator. It is the rotation of 180 degrees around the rotational axis given by the unit vector $u = (0,0,1)^T$.
@Gabriel Oksa
Thank you for replying. Let me accept
your answer . Suppose I have q1,q2,q3,q4
having eigenvalues :1,-1,1 . Two positive and one negative .Further if q1=/= q2=/=q3 =/=q4 . Then which one I will accept and which three to reject ?
Reply
B.Rath
I am sorry I do not understand what you mean. Are q1, q2, q3 and q4 3-dimensional vectors? If so, what are their components? What does it mean "to accept"? I am not an expert in quantum mechanics.
@ Gabrael Oksa
Let me explain you. Suppose I find
q1;q2;q3;q4 are four different(3x3) matrices having eigenvalues +1,-1,+1.
with a view to find parity matrix to a
given matrix B(3×3). Please inform which q , I have to accept and why ?
Regards
B.Rath
If you have 4 different Hermitian (symmetric in real case) 3x3 matrices q1, q2, q3 and q4, and all have eigenvalues +1, +1 and -1, then they can differ only in eigenvectors, Because the eigenvectors of any Hermitian matrix are unitary (orthonormal in real case), these 4 matrices can differ only by unitary (orthogonal) similarities. It means there exist unitary matrices Q1, Q2, and Q3 such that:
q1 = Q1^H *q2* Q1 = Q2^H*q3*Q2 = Q3^H*q4*Q3
Using the properties of unitary matrices, you can easily find similarity transformations (by unitary matrices!) between any pair of given 4 matrices.
Now, if your problem is invariant w.r.t. similarity transformation by unitary matrices, than you can choose ANY of q1, q2, q3 and q4. If it is not invariant in the above sense, you must have additional criteria how to choose the appropriate matrix.
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