At the end of Section 9 in Dirac's Principles of Quantum Mechanics (p. 34), there is a sentence that is very confusing to me. I am hoping that someone can explain whatever it is that I am missing. The argument leading up to this is as follows.

Suppose $\xi$ is a real (self-adjoint) linear operator that satisfies the "algebraic equation" $$\phi(\xi) \equiv \xi^n + a_1 \xi^{n-1} + a_2 \xi^{n-2} + \ldots + a_n = 0 \, ,$$ where the $a_i$ are complex numbers, and it is meant by $\phi(\xi)=0$ that $\phi(\xi)$ acting on any ket or bra produces zero. Further suppose that this algebraic equation is the "simplest" such equation satisfied by $\xi$. This may be factored as $$\phi(\xi) = (\xi - c_1)(\xi - c_2)\ldots(\xi - c_n) \, ,$$ for some numbers $c_r$. Define $\chi_r(\xi)$ by $$ \phi(\xi) = (\xi - c_r) \chi_r(\xi) \, ,$$ that is, $\chi_r(\xi)$ is the "quotient" of $\phi(\xi)$ and the factor $(\xi - c_r)$. It may be shown that the $c_r$ are the unique eigenvalues of the real operator $\xi$, from which it follows that $\chi_r(c_r) \neq 0$. Consider then the expression $$ f(\xi) \equiv \sum_r \frac{\chi_r(\xi)}{\chi_r(c_r)} - 1 \, . $$ If one inserts any eigenvalue into this expression, say $c_s$, then all the terms in the sum will be zero except for $r=s$, in which case the expression becomes $$ \frac{\chi_s(c_s)}{\chi_s(c_s)} - 1 = 1 - 1 = 0 \, .$$ Now comes the sentence in question:

Since, however, the expression [that is, $f(\xi)$?] is only of degree $n-1$ in $\xi$, it must vanish identically.

There must be something important that I am missing. Because, if the statement "$\phi(\xi) = 0$ is the simplest algebraic equation satisfied by the $\xi$" means that there is no polynomial in $\xi$ of degree less than $n$ that will produce zero when applied to any ket or bra, how can $f(\xi)$ vanish when it is only of degree $n-1$?

Earlier in the text, that statement was made that, for an arbitrary ket $| P \rangle$

... $\chi_r(\xi) | P \rangle$ cannot vanish for every $| P \rangle$, as otherwise $\chi_r(\xi)$ itself would vanish, and we should have $\xi$ satisfying an algebraic equation of degree $n-1$, which would contradict the assumption that [$\phi(\xi) = 0$] is the simplest equation that $\xi$ satisfies.

What am I missing?

  • 4
    $\begingroup$ Where's the contradiction? What you present as a contradiction is precisely the argument here - "Vanish identically" means that $f = 0$, i.e. it's not actually "a polynomial of degree n-1", but the zero polynomial. $\endgroup$
    – ACuriousMind
    Commented Sep 24, 2021 at 16:03
  • $\begingroup$ Sorry, but I don't understand. How is $f$ not a polynomial of degree n-1? $\endgroup$
    – tneulinger
    Commented Sep 24, 2021 at 16:07
  • 1
    $\begingroup$ It has no dependence on ξ. It is identically 0, regardless of kets and bras, as the author stated. Have you tried a simple example? $\endgroup$ Commented Sep 24, 2021 at 16:41
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/127804/2451 , physics.stackexchange.com/q/137094/2451 $\endgroup$
    – Qmechanic
    Commented Sep 24, 2021 at 17:02
  • 1
    $\begingroup$ Thank you. I get it now. $\endgroup$
    – tneulinger
    Commented Sep 24, 2021 at 18:41

2 Answers 2


Perhaps this was obvious to everyone replying to this post, but it was the comment of Cosmas Zachos that cleared it up for me. I will present that argument here in greater detail just in case it helps someone later.

From the explicit construction of $f(\xi)$, that is $$ f(\xi) \equiv \sum_r \frac{\chi_r(\xi)}{\chi_r(c_r)} - 1 \, ,$$

it is clear that it is of order $n-1$ at most. However, it can also be seen that substitution of $\xi$ for any of the $n$ eigenvalue $c_s$ gives $f(c_s) = 0$. These are therefore the $n$ roots of $f(\xi)$. Factoring $f(\xi)$, we have that, for example,

$$f(\xi) = (\xi - c_1) \, g_1(\xi) \, , $$

where $g_1(\xi)$ is what remains after factoring out $(\xi - c_1)$. Continuing, we have

$$g_1(\xi) = (\xi - c_2) \, g_2(\xi) \, , $$ $$ \ldots $$ $$ g_{n-1}(\xi) = (\xi - c_n) \, g_n \, $$

where in the last line $g_n$ must be a constant because we have finished factoring. We thus have

$$ f(\xi) = (\xi - c_1)(\xi - c_2)\ldots(\xi - c_n) \, g_n \, . $$

However, it is clear from this expression that, unless $g_n = 0$, $f(\xi)$ will be a polynomial of degree $n$, which would contradict the earlier observation that it can only be of order $n-1$ at most. Therefore, it must be true that $g_n = 0$, so that $f(\xi) = 0$.


As a simple example, consider the operator $\xi = \pmatrix{1&0\\0&-1}$, which obeys the algebraic equation $\phi(\xi) := \xi^2 - 1 = 0$ (of course, the final constant is multiplied by the identity matrix). $\phi(\xi)$ can of course be factored into $\phi(\xi)=\big(\xi+1\big)\big(\xi-1\big)\equiv\big(\xi-c_1\big)\big(\xi-c_2\big)$ with $c_1=-1$ and $c_2= 1$. From there, $$\chi_1(\xi)=\xi-c_2 = \xi-1 \qquad \chi_2(\xi)=\xi-c_1 = \xi+1$$

$$\implies f(\xi) := \left[\frac{\chi_1(\xi)}{\chi_1(c_1)}\right] + \left[\frac{\chi_2(\xi)}{\chi_2(c_2)}\right] -1$$ $$= \left[\frac{\xi-1}{-2}\right] + \left[\frac{(\xi+1)}{2}\right] - 1 = 0 $$ which vanishes identically, as claimed. In particular, $f(\xi)$ is of degree at most $n-1$ in $\xi$ by explicit construction, but is in fact of degree zero.

  • $\begingroup$ Thanks for your comment. However, how can one see that $f$ vanishes in general? $\endgroup$
    – tneulinger
    Commented Sep 24, 2021 at 17:28
  • $\begingroup$ @aFo23 As ACM mentioned in their comment, the entire argument is that if $\phi$ - which is of degree $n$ - is the simplest (non-zero, because the first term is $\xi^n$) polynomial which annihilates every ket and $f$ - which is of degree at most $n-1$ also annihilates every ket, then the only possibility is that $f(\xi)=0$. You might consider it analogous to the argument that if a complex polynomial $\alpha(z)$ has zeroes $z_1,\ldots,z_n$ and $\beta(z)$ is has degree at most $n-1$, then $\beta(z_i)=0,i=1,\ldots,n \implies \beta(z)=0$. $\endgroup$
    – J. Murray
    Commented Sep 24, 2021 at 17:35
  • $\begingroup$ I am sorry, but why should $f$ also "annihilate" every ket? For one, in the book, it has not been proven that any arbitrary ket may be represented in terms of a "complete" basis formed by the $\chi_r(\xi) |P \rangle$. Why should $f(\xi) | P \rangle = 0$? Also, the train of logic in the book seems to be that it is first established that $f=0$, from which it follows that $f+1$ is the identity. $\endgroup$
    – tneulinger
    Commented Sep 24, 2021 at 18:22

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