Finding the exponential of a matrix

Question

I am trying to find $\exp\left(\frac{i{\mu}t\mathbf{H}}{\sqrt{2}}\right)$ where

$$\mathbf{H}=\begin{pmatrix}0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{pmatrix}$$

I've got as far as figuring out that that, for $n=1,2,3,\dots$

$$ \mathbf{H}^{2n}=2^{n-1}\mathbf{H}^2=2^{n-1} \begin{pmatrix}1 & 0 & 1 \\ 0 & 2 & 0 \\ 1 & 0 & 1\end{pmatrix} $$

$$\mathbf{H}^{2n-1}=2^{n-1}\mathbf{H}=2^{n-1}\begin{pmatrix}0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{pmatrix}$$

but I'm not getting anywhere substituting this into the power series and I've no idea how to proceed. I've seen mention of Jordan form in other questions about this but I've never heard of it. If I need to learn it so-be-it, but I think there must be some other way as we have never been shown it in class.

Answer 1

Let$$T=\begin{pmatrix}1&1&-1\\-\sqrt2&\sqrt2&0\\1&1&1\end{pmatrix}.$$Note that the columns of $T$ form a basis of eigenvectors of $H$. Besides, the eigenvalues of $H$ corresponding to the first, the second, and the third column of $T$ are $-\sqrt2$, $\sqrt2$ and $0$; respectively. Therefore$$T^{-1}.H.T=\begin{pmatrix}-\sqrt2&0&0\\0&\sqrt2&0\\0&0&0\end{pmatrix}$$and so$$T^{-1}.\frac{i\mu t}{\sqrt2}H.T=\begin{pmatrix}-i\mu t&0&0\\0&i\mu t&0\\0&0&0\end{pmatrix},$$from which it follows that$$T^{-1}.\exp\left(\frac{i\mu t}{\sqrt2}H\right).T=\begin{pmatrix}e^{-i\mu t}&0&0\\0&e^{i\mu t}&0\\0&0&1\end{pmatrix}.$$Therefore,\begin{align}\exp\left(\frac{i\mu t}{\sqrt2}H\right)&=T.\begin{pmatrix}e^{-i\mu t}&0&0\\0&e^{i\mu t}&0\\0&0&1\end{pmatrix}.T^{-1}.\\&=\begin{pmatrix}\frac{1}{2} \left(1+e^{-i t \mu }\right) & 0 & \frac{1}{2} \left(-1+e^{-i t \mu }\right) \\ 0 & e^{-i t \mu } & 0 \\ \frac{1}{2} \left(-1+e^{-i t \mu }\right) & 0 & \frac{1}{2} \left(1+e^{-i t \mu }\right)\end{pmatrix}.\end{align}

Answer 2

The matrix exponential of a diagonalizable matrix $$H = SDS^{-1}$$ is particularly easy, since $$e^H = Se^DS^{-1}$$ and the matrix exponential of a diagonal matrix is just a matrix of the exponential of the diagonal entries.

Now your matrix $H$ is symmetric and so is guaranteed to be diagonalizable. Since the matrix is $3\times 3$ you're even guaranteed closed-form formulas, though you can easily get computational packages like Matlab, Mathematica, or even Wolfram Alpha to compute $S$ and $D$ for you.

Answer 3

\begin{align} \exp\left(\frac{i\mu t}{\sqrt{2}}H\right) &=\sum_{n=0}^\infty\frac1{n!}\left(\frac{i\mu t}{\sqrt{2}}H\right)^n\\ &=I+\sum_{n=1}^\infty\frac1{(2n)!}\left(\frac{i\mu t}{\sqrt{2}}H\right)^{2n} +\sum_{n=1}^\infty\frac1{(2n-1)!}\left(\frac{i\mu t}{\sqrt{2}}H\right)^{2n-1}\\ &=I+\sum_{n=1}^\infty\frac1{(2n)!}\left(\frac{i\mu t}{\sqrt{2}}\right)^{2n}2^{n-1}H^2 +\sum_{n=1}^\infty\frac1{(2n-1)!}\left(\frac{i\mu t}{\sqrt{2}}\right)^{2n-1}2^{n-1}H\\ &=I+\frac12\sum_{n=1}^\infty\frac1{(2n)!}\left(i\mu t\right)^{2n}H^2 +\frac1{\sqrt{2}}\sum_{n=1}^\infty\frac1{(2n-1)!}\left(i\mu t\right)^{2n-1}H\\ &=I+\frac12(\cos(e^{i\mu t})-1)H^2 +\frac1{\sqrt{2}}\sin(e^{i\mu t})H\\ &=\pmatrix{ \frac12(\cos(e^{i\mu t})+1)&\frac1{\sqrt{2}}\sin(e^{i\mu t})&\frac12(\cos(e^{i\mu t})-1)\\ \frac1{\sqrt{2}}\sin(e^{i\mu t})&\cos(e^{i\mu t})&\frac1{\sqrt{2}}\sin(e^{i\mu t})\\ \frac12(\cos(e^{i\mu t})-1)&\frac1{\sqrt{2}}\sin(e^{i\mu t})&\frac12(\cos(e^{i\mu t})+1)}. \end{align}