On page 32-33 of Dirac's book Principles of Quantum Mechanics, he considers the case when the linear, Hermitian$^1$ operator $\xi$ satisfies an algebraic equation

$$\phi(\xi)\equiv(\xi - c_1)(\xi - c_2)(\xi - c_3)...(\xi - c_n)=0\tag{18}$$

with the $c$'s being numbers, not assumed to be all different. Let the quotient when $\phi(\xi)$ is divided by $(\xi - c_r)$ be $\chi_r(\xi)$, so that

$$\phi(\xi)\equiv(\xi - c_r)\chi_r(\xi)\quad(r= 1,2,3,...,n)$$

After proving that the $c$'s are all different, he then says

Let $\chi_r(c_r)$ be the number obtained when $c_r$ is substituted for $\xi$ in the algebraic expression $\chi_r(\xi)$. Since the $c$'s are all different, $\chi_r(c_r)$ cannot vanish.

How is he able to conclude this?


$^1$ Note that a Hermitian operator is called a real operator in Dirac's non-standard terminology.

  • $\begingroup$ Because if $X_r(c_r)$ did vanish then it must be divisible by $(\xi-c_r)$ which would be a contradiction. $\endgroup$
    – lemon
    Commented Sep 24, 2014 at 18:26

2 Answers 2


If all the c's are different, then since $X_r(\xi)$ is the quotient when $\phi(\xi)$ is divided by $(\xi-c_r)$, $(\xi-c_r)$ cannot be a factor of $X_r(\xi)$ else two of the c's would equal $c_r$.

$X_r(c_r)$ is, by the remainder theorem (I imagine this has some other name), the remainder when $X_r(\xi)$ is divided by $(\xi-c_r)$. Since $(\xi-c_r)$ is not a factor of $X_r(\xi)$, $X_r(c_r)$ cannot be zero.


A very explicit argument:

$$\phi(\xi)=(\xi-c_1)\dots (\xi-c_{r-1})(\xi-c_r)(\xi-c_{r+1})\dots(\xi-c_n)$$

where $r\leq n$ and $c_k> c_l$ whenever $k>l$. We can order the $c_i$ like this because the $c_i$ are real.

Now, $$X_r(\xi)=(\xi-c_1)\dots (\xi-c_{r-1})(\xi-c_{r+1})\dots(\xi-c_n) $$

Clearly, the set of zeros of $X_r(\xi)$ is $\{c_i\ \bigl|\ i=1,\dots,r-1,r+1,\dots,n \}$. Note that this does not include $c_r$, so $X_r(c_r)$ does not vanish.


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