A Hermitian Matrix $H$ acting on $|0\rangle, \cdots, |N \rangle$,entries $\langle j+1 |H |j \rangle= \langle j| H | j+1 \rangle = \sqrt{(N-j)(j+1)}/2$

Question

Let $N$ be a positive integer. Consider a Hamiltonian (a Hermitian matrix) acting on orthogonal basis states $|0 \rangle, \cdots, |N \rangle$ (Dirac notation), for which the non-zero matrix entries are $\langle j+1 |H |j \rangle= \langle j| H | j+1 \rangle = \sqrt{(N-j)(j+1)}/2$. A paper I am reading states without proof that

i) $e^{-i \pi H}|0 \rangle = |N \rangle$

ii) $||H || = N/2$ (not sure which matrix norm is being used, although it could possibly be the trace norm)

I am focused mostly on trying to understand i) for now. In the case where $N=1$, then $H$ is a matrix of dimension $2 \times 2$ of the form : $H = \frac {1} {\sqrt{2}} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end {bmatrix}$, where $H^k = \begin{cases} \frac {1} {\sqrt{2}}^k\begin{bmatrix} 0 & 1 \\ 1 & 0 \end {bmatrix} \text{if $k \mod 2 = 1$} \\ \frac {1} {\sqrt{2}}^k I_{2\times 2} \text{ if $k \mod 2 = 0$}\end{cases}$.

Expanding $e^{-i \pi H} = I -i \pi H - \pi^2 \frac {1}{4}I + \cdots$

I'm not sure how to continue more generally for any $N$ and how to conclude $i)$. Insights appreciated.

Answer 1

The first claim isn't quite right. The correct statement should be $e^{-i\pi H}|0\rangle=(-i)^N|N\rangle$.

In view of user Yukinooo's question, I believe there is a simple physics-oriented proof of the statement. Anyway, here is a matrix-theoretic proof. For $j=1,2,\ldots,N$, let $h_j=\sqrt{(N+1-j)j\,}$. Then $$ H=\frac12\pmatrix{0&h_1\\ h_1&0&h_2\\ &h_2&\ddots&\ddots\\ &&\ddots&0&h_N\\ &&&h_N&0}. $$ It is known that $2H$ is similar to the Kac matrix $$ K=\pmatrix{0&1\\ N&0&2\\ &N-1&\ddots&\ddots\\ &&\ddots&0&N\\ &&&1&0} $$ via the similarity transform $2H=D^{-1}KD$ where $$ D=\operatorname{diag}(d_0,\ldots,d_N),\ d_k = \frac1{k!}\prod_{i=1}^k h_i. $$ $K$ can be diagonalised as $K=V\Lambda V^{-1}$ where $$ \Lambda = \operatorname{diag}\left(N,\,N-2,\,N-4,\ldots,\,-(N-4),\,-(N-2),\,-N\right) $$ and $V$ is the zero-indexed Krawtchouk matrix $$ V_{ij} = \sum_{k=0}^N(-1)^k\binom{j}{k}\binom{N-j}{i-k} $$ with the property that $V^2=2^NI$. One may verify that $SV|0\rangle=V|N\rangle$ where $$ S=\operatorname{diag}\left(1,-1,1,-1,\ldots,(-1)^N\right). $$ Thus \begin{aligned} e^{-i\pi H}|0\rangle &=D^{-1}\exp(-i\pi K/2)D|0\rangle\\ &=D^{-1}\exp(-i\pi K/2)|0\rangle\ \text{(because $d_0=1$)}\\ &=D^{-1}V\exp(-i\pi \Lambda/2)V^{-1}|0\rangle\\ &=D^{-1}V\left((-i)^NS\right)V^{-1}|0\rangle\\ &=(-i)^ND^{-1}VSV^{-1}|0\rangle\\ &=(-i)^ND^{-1}V^{-1}SV|0\rangle\ \text{(because $V^2=2^NI$)}\\ &=(-i)^ND^{-1}V^{-1}V|N\rangle\ \text{(because $SV|0\rangle=V|N\rangle$)}\\ &=(-i)^ND^{-1}|N\rangle\\ &=(-i)^N|N\rangle\ \text{(because $d_N=1$)}. \end{aligned} Moreover, since $H$ is similar to $K/2$, its largest-sized eigenvalues are $\pm N/2$. Therefore (as $H$ is real symmetric) its spectral norm is $\|H\|_2=\rho(H)=N/2$.