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First, let us look at some theory regarding diagonalization. A linear transformation on an n-dimensional vector space (or equivalently, an n-by-n matrix), is diagonalizable if and only if the sum of the dimensions of its eigenspaces is equal to n. Since matrices and linear transformation are equivalent, let us concentrate on an n-by-n matrix, say A. We can assume that it is the matrix representation of some linear transformation (on an n-dimensional vector space, say V) with respect to some basis.
Now, suppose λ1, λ;2, ..., λk are distinct eigenvalues of A. Note that k need not be equal to n, as some eigenvalues may be repeated in the set of all eigenvalues, and therefore, when considering only distinct eigenvalues, the number may be less than n. There is an eigenspace corresponding to each eigenvalue, and each of these eigenspaces can be represented by a set of (one or more) linearly independent vectors. These are the eigenvectors. If x is an eigenvector corresponding to the eigenvalue λ, then:
Ax = ...
This solution clearly assesses theory regarding diagonalization. References are also listed to promote research.
State Energy: Consider an electron with spin magnetic moment u_s
Consider an electron with spin magnetic moment u_s in a strong magnetic field B_z in the z direction. The potential for an electron with spin magnetic moment u_s in a magnetic field B is V=-u_s . B
where u_s = -((g_s)(u_B))/(hbar) . S
Thus the Hamiltonian is H_0 = ((g_S)(u_B))/(hbar) . B . S = ((g_s)/2)(u_B) . B . sigma
and g_s = 2 for the electron. The eigenstates of H_0 are just the spin up and spin down states, | up > and | down >, with energies E_+ = (u_B)(B_z) and E_- = -(u_B)(B_z).
Suppose we are in the spin up state, | up >, and we add the small magnetic field (B_x)(x hat) with B_x << B_z. Take B = (B_x)(x hat) + (B_z)(z hat), and calculate the exact energies.View Full Posting Details