Simplifying a tough experession of matrix product

Question

In the Dirac's $bra$ and $ket$ notation, $|x ~\rangle$ (pronounced as $ket~ x$ represents a column vector and $\langle y ~|$ (pronounced as $bra~y$) a row vector, such that $\langle x | y \rangle $ is the inner product. Consider the following:

$F(i,k) = \langle i | \Big(\sum\limits_{x=-L}^L |x-1\rangle \langle x|\Big)^{L-k} \Big(\sum\limits_{y=-L}^L |y+1\rangle \langle y|\Big)^{k}|0 \rangle$

Here, $L \in \{ 0,1,2,3, \dots \}$, $k \in \{ 0,1,2,3, \dots \}$, and $i \in \{ 0,1,2,3, \dots \}$, and $k \le L$. Also,

$| -L \rangle = \begin{pmatrix} 1\\0\\ \vdots \\ \end{pmatrix}_{2L+1,1} $, $\cdots$, $| L \rangle = \begin{pmatrix} \vdots \\0\\ 1 \\ \end{pmatrix}_{2L+1,1} $

What is the general form of $F(i,k)$?

Are the kets $|k\rangle$ for $k = 0,1,2,3,\dots$ orthonormal? — Ben Grossmann, Jun 20 '20 at 15:14
Yes. In fact in my question, $\langle i | j \rangle = \delta_{ij}$, i.e., are orthogonal unless $i=j$. — Rob, Jun 20 '20 at 15:19

Ben Grossmann · Accepted Answer · 2020-06-24T13:14:12.510

1

A helpful observation in simplifying this is that $$ \Big(\sum\limits_{x=-L}^L |x-1\rangle \langle x|\Big)^{k} = \Big(\sum\limits_{x=-L-1}^{L-1} |x\rangle \langle x+1|\Big)^{k} = \sum\limits_{x=-L-1}^{L-k} |x\rangle \langle x+k|. $$ We can similarly observe (or use the adjoint of the above) to find that $$ \Big(\sum\limits_{y=-L}^L |y+1\rangle \langle y|\Big)^{k} = \sum\limits_{y=-L}^{L-k+1} |y+k\rangle \langle y|. $$ With that established, it is straightforward to verify that $$ F(i,k) = \langle i+L-k|k\rangle = \delta_{i+L,2k}. $$

Proof of first equation: proceed by induction. We note that $$ \begin{align} \Big(\sum\limits_{x=-L-1}^{L-1} |x\rangle \langle x+1|\Big)^{k} &= \Big(\sum\limits_{x=-L-1}^{L-1} |x\rangle \langle x+1|\Big)\Big(\sum\limits_{x=-L-1}^{L-1} |x\rangle \langle x+1|\Big)^{k-1} \\ & = \Big(\sum\limits_{x=-L-1}^{L-1} |x\rangle \langle x+1|\Big)\Big(\sum\limits_{y=-L-1}^{L-k+1} |y\rangle \langle y+k-1|\Big)^{k-1} \\ & = \sum\limits_{x=-L-1}^{L-1} \sum\limits_{y=-L-1}^{L-k+1} |x\rangle \langle x+1|y\rangle \langle y+k-1| \\ & = \sum\limits_{x=-L-1}^{L-1} \sum\limits_{y=-L-1}^{L-k+1} \delta_{x+1,y} |x\rangle \langle y+k-1| \\ & = \sum\limits_{x=-L-1}^{L-k} |x\rangle \langle (x+1)+k-1| = \sum\limits_{x=-L-1}^{L-k} |x\rangle \langle x + k|. \end{align} $$

edited Jun 24 '20 at 13:14

answered Jun 20 '20 at 15:28

Ben Grossmann

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Could you be little detailed, please? – Rob Jun 20 '20 at 16:10
How did you reach at $\Big( \sum\limits_{-L-1}^{L-1} | x \rangle \langle x+1 | \Big)^k = \sum\limits_{-L}^{L-k} |x \rangle \langle x +k|$? – Rob Jun 20 '20 at 16:30
@Rob It's equivalent to the proof that I have written here (with $b = 1$). – Ben Grossmann Jun 20 '20 at 16:37
could you be kind to prove the first equation in your answer? – Rob Jun 24 '20 at 12:56
@Rob See my latest edit – Ben Grossmann Jun 24 '20 at 13:09
Thanks a lot!. One minor doubt. How did the upper limit iof $x$ change from $L-1$ to $L-k$? I was stuck at the same point! – Rob Jun 24 '20 at 13:22
@Rob Note that if $x > L-k$, then the corresponding $y = x+1$ would satisfy $y > L-k+1$. However, we never have $y > L - k + 1$ in the original sum. – Ben Grossmann Jun 24 '20 at 13:25
Wow! Thanks a ton!! – Rob Jun 24 '20 at 13:28

Simplifying a tough experession of matrix product

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