I'm reading the Dirac's book about QM and I am finding troubles understanding a proof of a theorem (in my edition at page 32), which says that "there are so many eigenkets of $\xi$ that any ket whatever can be expressed as a sum of such eigenkets". I can't understand the expression

$$ \sum_r {{\chi_r (\xi)} \over {\chi_r (c_r)}} - 1 $$

Then Dirac says that if $c_s$ is substituted for $\xi$, every term vanishes except for the one where $r = s$. My reasoning led me to the expression:

$$ \sum_r {{(\xi - c_1)(\xi - c_2) \cdots (\xi - c_{r-1})(\xi - c_{r+1}) \cdots (\xi - c_{n-1})(\xi - c_n)} \over {(c_r - c_1)(c_r - c_2) \cdots (c_r - c_{r-1})(c_r - c_{r+1}) \cdots (c_r - c_{n-1})(c_r - c_n)}} - 1$$

As I understand, there is no $\xi - c_r$ in $\chi_r (\xi)$ as for definition $\chi_r (\xi) = {\phi(\xi) \over (\xi - c_r)}$. For the same motivations $\chi_r (c_r)$ has no term $(c_r - c_r)$. The author says to substitute $c_s$ to $\xi$ in the expression, so I might have to do:

$$ \sum_r {{(c_s - c_1)(c_s - c_2) \cdots (c_s - c_{r-1})(c_s - c_{r+1}) \cdots (c_s - c_{n-1})(c_s - c_n)} \over {(c_r - c_1)(c_r - c_2) \cdots (c_r - c_{r-1})(c_r - c_{r+1}) \cdots (c_r - c_{n-1})(c_r - c_n)}} - 1 $$

So I tried, if $s \neq r$ then:

$$ \sum_r {{(c_s - c_1)(c_s - c_2) \cdots (c_s - c_s) \cdots (c_s - c_{r-1})(c_s - c_{r+1}) \cdots (c_s - c_{n-1})(c_s - c_n)} \over {(c_r - c_1)(c_r - c_2) \cdots (c_r - c_s) \cdots (c_r - c_{r-1})(c_r - c_{r+1}) \cdots (c_r - c_{n-1})(c_r - c_n)}} - 1 $$

hence $c_s - c_s$ in $\chi_r (c_r)$ is equal to zero, then the sum will be $-r \neq 0$. The case $s = r$ gives me some problem:

$$ \sum_r {{(c_r - c_1)(c_r - c_2) \cdots (c_r - c_{r-1})(c_r - c_{r+1}) \cdots (c_r - c_{n-1})(c_r - c_n)} \over {(c_r - c_1)(c_r - c_2) \cdots (c_r - c_r) \cdots (c_r - c_{r-1})(c_r - c_{r+1}) \cdots (c_r - c_{n-1})(c_r - c_n)}} - 1 $$

There is no $c_r - c_r$ in $\chi_r (\xi)$ term as it has been simplified, so $\chi_r (\xi)$ is never $0$. But the denominator contains the term $c_r - c_r = 0$. How can it be possible to have a null term in the denominator?

Could you please explain this sum, and how Dirac intended it?

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    $\begingroup$ There is no $c_r-c_r$ in the denominator. $\chi_r$ has had that factor taken out. $\endgroup$ – RogerJBarlow Feb 24 at 15:16
  • $\begingroup$ @CosmasZachos Thank you for your answer, but I actually don't know Sylvester's formula and Frobenius invariants in deep, I started study qm as hobbyist, so I might have missed some important math piece, thanks anyway for your time $\endgroup$ – Luca Mattioni Feb 24 at 15:20
  • $\begingroup$ @RogerJBarlow Thank you too, if I understood rightly there is no $c_r - c_r$ for the same reasoning on which there is no such factor in $\chi_r (\xi)$, right? $\endgroup$ – Luca Mattioni Feb 24 at 15:22
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    $\begingroup$ Yes. The $\chi_r$ is the the $\phi$ with that term factored out. Reading Dirac's book is the best way to understand QM: keep going! $\endgroup$ – RogerJBarlow Feb 24 at 17:11
  • $\begingroup$ @RogerJBarlow The best way, to be sure--preternaturally insightful; but not the kindest of introductions for the novice, by now... $\endgroup$ – Cosmas Zachos Feb 25 at 20:53

So, to simplify a bit, we are here given an array of numbers $c_1\dots c_n$ (they are not operators!) that are all distinct, and we are asked to consider the smooth, ordinary function $$ \phi(x) =(x-c_1)(x-c_2)\dots(x-c_n).$$ This has a root of order 1 at every $c_k$ and thus we can also define the function missing the $k$th root, $$\chi_k(x)=\phi(x)/(x-c_k)=(x-c_1)\dots(x-c_{k-1})(x-c_{k+1})\dots(x-c_n).$$ This function can be filled in at this point where it is technically undefined at $x=c_k$ with some non-zero value $\chi_k(\c_k),$ known to be non-zero because the original numbers were all distinct.

Your question concerns the value of $\chi_k(c_\ell)/\chi_k(c_k)$. You appear to expand it out to find 1 or so? It is not 1, it is the Kronecker delta, $$ \chi_k(c_\ell)/\chi_k(c_k)=\delta_{k\ell}=\{1\text{ if } k =\ell\text{ else } 0\}.$$ As Cosmas says, your problem appears to be getting too abstract too quickly and we may prefer to do this once by example. So consider the numbers [-1,2,3] for our starting list. Then we have three parabolas $\chi_1(x)=(x-2)(x-3)$, $\chi_2(x)=(x+1)(x-3)$, $\chi_3(x)=(x+1)(x-2)$. Working through this expression for $k=2$, say, we would find that $$ \chi_2(c_1)=(c_1+1)(c_1-3)=({-1}+1)({-1}-3)=(0)({-4})=0,\\ \chi_2(c_2)=(c_2+1)(c_2-3)=(2+1)(2-3)=(3)({-1})={-3},\\ \chi_2(c_3)=(c_3+1)(c_3-3)=(3+1)(3-3)=(4)({0})={0}. $$ So for these ratios we found $\chi_2(c_1)/\chi_2(c_2)=0/({-3})=0$, and $\chi_2(c_2)/\chi_2(c_2)=1$ more trivially (dividing any number by itself yields 1), and $\chi_2(c_3)/\chi_2(c_2)=0/({-3})=0$ again.

Learning is pain. Or more precisely, an abstraction provides a sort of pain relief, it organizes a chaos for us, and we can only learn abstractions when we have first been exposed to the pain that demands the organizing. Learning the general rule may require doing a lot of examples. So I have exposed you to a bit of pain, writing all of that stuff out, and now we can start to organize that pain.

Once we see the particular examples, we can now see that by construction $c_\ell$ is defined to be a root for any polynomial $\chi_k$ so long as $k\ne \ell,$ because it is a root of $\phi$ but it is not the root that $\chi_k$ eliminated, unless $k=\ell.$ and this idea that we're going to divide by $\chi_k(c_k)$ produces the Kronecker delta by removing the only piece of information that was left!

In turn you have the expression $$ f(x) ={-1} + \sum_{k=1}^n\frac{\chi_k(x)}{\chi_k(c_k)}, $$you can now appreciate the claim that $$ f(c_\ell) ={-1} + \sum_{k=1}^n\delta_{k\ell}={-1} +1=0.$$


You might well be crushed by out-of-control formal abstraction. This you learned from your linear algebra course, in all likelihood. Little will change if you take n =2: rerun all formulas and arguments for a daft 2×2 matrix ξ.

The characteristic (Cayley-Hamilton) polynomial equation is $$ \phi(\xi)= (\xi -c_1)(\xi_2-c_2)=0, ~~~\leadsto \\ \chi_r(\xi) ={ \phi(\xi)\over \xi-c_r} \implies \chi_1=\xi-c_2 ,~~~\chi_2=\xi-c_1. $$ The sum you are considering then only has two terms, $$ {\xi -c_2\over c_1-c_2} +{\xi -c_1\over c_2-c_1} . $$ As PAMD reminds you, plugging in either $c_1$ or $c_2$ for $\xi$ will always yield 1, which he takes out of the sum.

So the expression you start with always vanishes for all roots of the characteristic equation. You may now take n =3, or arbitrary n: the mechanics is identical!

Culture . In today's language, the matrices $\chi_r(\xi)/\chi_r(c_r)$ are the Frobenius covariants, which is to say the projection operators $P_r$ onto the eigenvector of (nondegenerate) eigenvalues $c_r$. As such, they are idempotent, $P_r P_r=P_r$ and complete, $\sum_r P_r=1\!\!1 $, Dirac's expression when acting on the vectors. Sylvester's formula, $$ f(\xi)=\sum_r f(c_r) P_r $$ utilizing them is used routinely in QM, e.g., $$ f(\hat x) = \int\!\! dx ~~|x\rangle f(x) \langle x| . $$

  • $\begingroup$ Thanks a lot, I really appreciate your effort! $\endgroup$ – Luca Mattioni Feb 24 at 15:59
  • $\begingroup$ you bet I do XD $\endgroup$ – Luca Mattioni Feb 24 at 16:12
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    $\begingroup$ Ah! You beat me to it with your greater concision $\endgroup$ – CR Drost Feb 24 at 16:21

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