Functions of a matrix

The projectors satisfy $P_k^2=P_k$ and $P_kP_l=0$ for $k\neq l$. Thus $A^2=\bigl(\sum_ka_kP_k\bigr)^2=\sum_ka_k^2P_k$, and by induction $A^n=\sum_ka_k^nP_k$; by linearity, this extends to every polynomial. Here, $A^2=16P_4+P_1=\frac{16}{3}J+\mathbf{1}-\frac13J=\mathbf{1}+5J$. Direct calculation gives $(\mathbf{1}+J)^2=\mathbf{1}+2J+J^2=\mathbf{1}+5J$: the two methods agree.

Since the eigenvalues are non-zero, $A$ is invertible, and

$$A^{-1}=\frac14P_4+P_1=\frac{1}{12}J+\mathbf{1}-\frac13J=\mathbf{1}-\frac14J .$$

Verification: $(\mathbf{1}+J)(\mathbf{1}-\frac14J)=\mathbf{1}+J-\frac14J-\frac14J^2=\mathbf{1}+\frac34J-\frac34J=\mathbf{1}$. Similarly,

$$\sqrt A=2P_4+P_1=\frac23J+\mathbf{1}-\frac13J=\mathbf{1}+\frac13J,$$

and $(\mathbf{1}+\frac13J)^2=\mathbf{1}+\frac23J+\frac19J^2=\mathbf{1}+\frac23J+\frac13J=\mathbf{1}+J=A$. This is the unique square root of $A$ whose eigenvalues are positive; others exist, for example $-\sqrt A$.

We have $(A-4\,\mathbf{1})(A-\mathbf{1})=(J-3\,\mathbf{1})J=J^2-3J=0$. This can also be seen from the spectral decomposition: $(A-4)(A-1)=\sum_k(a_k-4)(a_k-1)P_k$, and each factor vanishes on the spectrum. Now consider the Lagrange polynomials $L_4(x)=\frac{x-1}{4-1}$ and $L_1(x)=\frac{x-4}{1-4}$, which equal $1$ at one eigenvalue and $0$ at the other. We have $L_4(A)=\sum_kL_4(a_k)P_k=P_4$, hence

$$P_4=\frac{A-\mathbf{1}}{3}=\frac13J,\qquad P_1=\frac{4\,\mathbf{1}-A}{3}=\mathbf{1}-\frac13J,$$

in agreement with the statement. Finally, expanding the relation gives $A^2-5A+4\,\mathbf{1}=0$, or $A(5\,\mathbf{1}-A)=4\,\mathbf{1}$, hence $A^{-1}=\frac{5\,\mathbf{1}-A}{4}=\frac{4\,\mathbf{1}-J}{4}=\mathbf{1}-\frac14J$.

With $f(x)=e^{-i\omega tx}$, we obtain

$$e^{-i\hat Ht/\hbar}=e^{-4i\omega t}P_4+e^{-i\omega t}P_1=e^{-4i\omega t}\,\frac13J+e^{-i\omega t}\Bigl(\mathbf{1}-\frac13J\Bigr).$$

Since $\bra{e_1}J\ket{e_1}=1$ and $\bra{e_1}\mathbf{1}\ket{e_1}=1$, the amplitude for remaining in the initial state is

$$\bra{e_1}e^{-i\hat Ht/\hbar}\ket{e_1}=\frac13e^{-4i\omega t}+\frac23e^{-i\omega t}=\frac{e^{-i\omega t}}{3}\Bigl(e^{-3i\omega t}+2\Bigr),$$

and the probability is

$$P_{11}(t)=\frac19\bigl|e^{-3i\omega t}+2\bigr|^2=\frac19\bigl(5+4\cos3\omega t\bigr).$$

It oscillates between $1$ and $\frac19$, at angular frequency $3\omega$, equal to the difference between the two eigenvalues of $\hat H$ divided by $\hbar$: this is a beat between the two energy levels. For example, this model describes a particle that can move from one site to either of the other two sites of a triangle.