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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.

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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}

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    $\begingroup$ This is definitely a valid method, but I'd consider it of limited usefulness without some explanation of where $T$ comes from. (I mean, I know what you did there, but I'm not sure if someone learning about matrix algebra would necessarily know.) $\endgroup$ – David Z Oct 13 '17 at 2:38
  • $\begingroup$ @DavidZ I agree. What do you think now? $\endgroup$ – José Carlos Santos Oct 13 '17 at 6:14
  • $\begingroup$ Yeah, that is an improvement. $\endgroup$ – David Z Oct 13 '17 at 7:06
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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.

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\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}

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    $\begingroup$ It will look nicer if you rewrite in terms of $sin$ and $cos$ $\endgroup$ – mtheorylord Oct 13 '17 at 1:26

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