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Let's first determine the stationary states of the Hamiltonian in order to determine the time evolution of a set of vectors. The problem's vectors are not the provided Hamiltonian's eigen states, so we must determine the Hamiltonian's potential energies (also known as its eigen values), and then we must be able to expand the problem's vectors based on the Hamiltonian's eigenvectors. The following equation for eigen values begins with.

$\widehat{H}|蠄鉄�=E|蠄鉄�鈫�\left[\begin{array}{ccc}a& 0& b\\ 0& c& 0\\ b& 0& a\end{array}\right]|蠄鉄�=\left[\begin{array}{ccc}E& 0& 0\\ 0& E& 0\\ 0& 0& E\end{array}\right]|蠄鉄�$

From which follows:

$\left[\begin{array}{ccc}a鈭�E& 0& b\\ 0& c鈭�E& 0\\ b& 0& a鈭�E\end{array}\right]|蠄鉄�=0$

Only if the determinant of the system matrix is equal to zero can this system of equations have a non-trivial solution. By requiring the latter, we obtain:

$\begin{array}{c}(a鈭�E)\left[\right(c鈭�E\left)\right(a鈭�E\left)\right]鈭�b^2(c鈭�E)=0\\ {(a鈭�E)}^2(c鈭�E)鈭�b^2(c鈭�E)=0\\ ({(a鈭�E)}^2鈭�b^2)(c鈭�E)=0\\ (a鈭�E鈭�b)(a鈭�E+b)(c鈭�E)=0\end{array}$

We factored the expression using difference of squares in the last line. We get the following energy since any of the three brackets might equal zero:

$\begin{array}{l}{E}_1=a鈭�b\\ E_2=c\\ E_2=a+b\end{array}$

The Hamiltonian's necessary eigenvectors must then be identified. In order to do this, we must first enter the obtained energies into the system of equations, and we must do this for each of the three cases of obtained energies.

$E_1=a鈭�b$

The system of equations for the given energy reads:

$\left[\begin{array}{ccc}b& 0& b\\ 0& c鈭�a+b& 0\\ b& 0& b\end{array}\right]|{蠄}_1鉄�鈫�\left[\begin{array}{ccc}b& 0& b\\ 0& c鈭�a+b& 0\\ b& 0& b\end{array}\right]\left[\begin{array}{l}{c}_1\\ c_2\\ c_3\end{array}\right]=0$

Let's focus on the second row first. It implies that:

$\begin{array}{c}(c鈭�a+b)c_2=0\\ c_2=0\end{array}$

The following is the result of looking at the first of the third row (they are identical):

$\begin{array}{c}b{c}_1+b{c}_3=0\\ c_1=鈭�c_3\end{array}$

Since the system of equations is degenerate, one of the coefficients is arbitrary. With that in mind,

$c_1=1$

Therefore, the first eigenvector reads:

$|{蠄}_1鉄�=\left[\begin{array}{c}1\\ 0\\ 鈭�1\end{array}\right]$

However, since this vector has not been normalized, we must. The first eigenvector after this vector has been trivially normalized is:

$|{蠄}_1鉄�=\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 0\\ 鈭�1\end{array}\right]$

$E_2=c$

The system of equations for the given energy reads:

$\left[\begin{array}{ccc}a鈭�c& 0& b\\ 0& 0& 0\\ b& 0& a鈭�c\end{array}\right]|{蠄}_2鉄�鈫�\left[\begin{array}{ccc}a鈭�c& 0& b\\ 0& 0& 0\\ b& 0& a鈭�c\end{array}\right]\left[\begin{array}{l}{d}_1\\ d_2\\ d_3\end{array}\right]=0$

From the first and the third row we get:

$\begin{array}{c}(a鈭�c)d_1+b{d}_3=0\\ b{d}_1+(a鈭�c)d_3=0\end{array}$

The only possible way to satisfy this equation is to set $d_1=d_3=0.$1

The last equation for the second row is trivial, meaning $0鈰�d_2=0.$

This means that we can set$d_2$to be any number.

We choose $d_2=1$for the vector to be normalized. Therefore, the second eigenvector is:

$|{蠄}_2鉄�=\left[\begin{array}{l}0\\ 1\\ 0\end{array}\right]$

$E_3=a+b$

The set of equations for the specified energy is as follows::

$\left[\begin{array}{ccc}鈭�b& 0& b\\ 0& c鈭�a鈭�b& 0\\ b& 0& 鈭�b\end{array}\right]|{蠄}_3鉄�鈫�\left[\begin{array}{ccc}鈭�b& 0& b\\ 0& c鈭�a鈭�b& 0\\ b& 0& 鈭�b\end{array}\right]\left[\begin{array}{l}{e}_1\\ e_2\\ e_3\end{array}\right]=0$

This is comparable to the first eigenvector example. The second row gives us:

$\begin{array}{c}(c鈭�a鈭�b)e_2=0\\ e_2=0\end{array}$

Next, by studying the first element of the third row (which is identical), we discover:

$\begin{array}{c}鈭�b{e}_1+b{e}_3=0\\ e_1=e_3\end{array}$

Therefore, the normalized third eigenvector looks like this without the first component's negative sign:

$|{蠄}_3鉄�=\frac{1}{\sqrt{2}}\left[\begin{array}{l}1\\ 0\\ 1\end{array}\right]$

Now prepared to determine the temporal evolution of the states specified in the problem after obtaining all the eigen values (i.e., permissible energies of the system) and eigenvectors (i.e., possible states of the Hamiltonian with well-defined energies).

The first vector that we are interested in is:

$\left|蠁\right(t=0)鉄�=\left[\begin{array}{l}0\\ 1\\ 0\end{array}\right]$

To expand it in the basis of the Hamiltonian eigenvectors as:

$\left|蠁\right(t=0)鉄�=\underset{i}{鈭�}{c}_i|{蠄}_i鉄�$

In general case we would take a dot product of the initial vector with $j鈭�th$ eigen state to obtain the desired $c_j$coefficient, but in this case the case is trivial since:

$\left|蠁\right(t=0)鉄�=|{蠄}_2鉄�$

Therefore, the time evolution of the vector is:

$\begin{array}{c}\left|蠁\right(t)鉄�=e^{鈭�i\widehat{H}t/\mathrm{鈩�}}\\ \left|蠁\right(t=0)鉄�=e^{鈭�i\widehat{H}t/\mathrm{鈩�}}|{蠄}_2鉄�\\ =e^{鈭�i{E}_{n}t/\mathrm{鈩�}}|{蠄}_2鉄�\\ =e^{鈭�ict/\mathrm{鈩�}}|{蠄}_2鉄�\end{array}$

Therefore:

$\left|蠁\right(t)鉄�=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]$

The next vector is:

$\left|蠁\right(t=0)鉄�=\left[\begin{array}{l}0\\ 0\\ 1\end{array}\right]$

The Hamiltonian eigen basis must be expanded, as before. It is probably not difficult to immediately recognize the proper linear combination, but for the purpose of generality, we will establish the following formal approach.

$\left|蠁\right(t=0)鉄�=\underset{i}{鈭�}{c}_i|{蠄}_i鉄�$

Next, to find $c_1,c_2$and$c_3$we take multiply (i.e. taking a inner product) both sides by$c_j$What we have then is:

$鉄�{\mathrm{蠄}}_1鈭�\mathrm{蠁}(\mathrm{t}=0)鉄�=\underset{\mathrm{i}}{鈭�}{\mathrm{c}}_{\mathrm{i}}鉄�{\mathrm{蠄}}_{\mathrm{j}}鈭�{\mathrm{蠄}}_{\mathrm{i}}鉄�={\mathrm{c}}_1$

The last equality sign follows from the fact that the Hamiltonian eigen states are mutually orthogonal. Therefore, to calculate $c_1$:

$\begin{array}{c}{c}_1=鉄�{蠄}_1鈭�蠁(t=0)鉄�\\ =\frac{1}{\sqrt{2}}\left[\begin{array}{lll}1& 0& 鈭�1\end{array}\right]\left[\begin{array}{l}0\\ 0\\ 1\end{array}\right]\\ =鈭�\frac{1}{\sqrt{2}}\end{array}$

In similar fashion we obtain$c_2=0$and$c_3=1.$Therefore:

$\left|蠁\right(t=0)鉄�=\frac{1}{\sqrt{2}}(鈭�|{蠄}_1鉄�+|{蠄}_3鉄�)$

Therefore, the time evolution of the vector is:

$\begin{array}{c}\left|蠁\right(t)鉄�=e^{鈭�i\widehat{H}t/\mathrm{鈩�}}|蠁(t=0)鉄�=\frac{1}{\sqrt{2}}{e}^{鈭�i\widehat{H}t/\mathrm{鈩�}}(鈭�|{蠄}_1鉄�+|{蠄}_3鉄�)\\ =\frac{1}{\sqrt{2}}(鈭�e^{鈭�i{E}_{1}t/\mathrm{鈩�}}|{蠄}_1鉄�+e^{鈭�i{E}_{3}t/\mathrm{鈩�}}|{蠄}_3鉄�)\\ =\frac{1}{\sqrt{2}}(鈭�e^{鈭�i(a鈭�b)t/\mathrm{鈩�}}|{蠄}_1鉄�+e^{鈭�i(a+b)/\mathrm{鈩�}}|{蠄}_3鉄�)\end{array}$

The last equation can be phrased more clearly by explicitly stating the Hamiltonian eigen states as column vectors, even if technically we are finished:

$\begin{array}{c}\left|蠁\right(t)鉄�=\frac{1}{\sqrt{2}}\left(鈭�e^{鈭�i(a鈭�b)t/\mathrm{鈩�}}\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 0\\ 鈭�1\end{array}\right]+e^{鈭�i(a+b)/\mathrm{鈩�}}\frac{1}{\sqrt{2}}\left[\begin{array}{l}1\\ 0\\ 1\end{array}\right]\right)\\ =\frac{1}{2}{e}^{鈭�iat/\mathrm{鈩�}}\left(鈭�e^{ibt/\mathrm{鈩�}}\left[\begin{array}{c}1\\ 0\\ 鈭�1\end{array}\right]+e^{鈭�ibt/\mathrm{鈩�}}\left[\begin{array}{l}1\\ 0\\ 1\end{array}\right]\right)\\ =\frac{1}{2}{e}^{鈭�iat/\mathrm{鈩�}}\left[\begin{array}{c}鈭�e^{ibt/\mathrm{鈩�}}+e^{鈭�ibt/\mathrm{鈩�}}\\ 0\\ e^{ibt/\mathrm{鈩�}}+e^{鈭�ib)t/\mathrm{鈩�}}\end{array}\right]\end{array}$

We reach the answer for the time evolution by applying the definition of complex trigonometric functions.

$\left|蠁\right(t)鉄�=e^{鈭�iat/\mathrm{鈩�}}\left[\begin{array}{c}i\sin (bt/\mathrm{鈩�}0)\\ 0\\ \cos (bt/\mathrm{鈩�})\end{array}\right]$