# Contradictory statements in Dirac's "Principles of Quantum Mechanics"?

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?

• 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. Commented Sep 24, 2021 at 16:03
• Sorry, but I don't understand. How is $f$ not a polynomial of degree n-1? Commented Sep 24, 2021 at 16:07
• 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? Commented Sep 24, 2021 at 16:41
• Commented Sep 24, 2021 at 17:02
• Thank you. I get it now. Commented Sep 24, 2021 at 18:41

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.

• Thanks for your comment. However, how can one see that $f$ vanishes in general? Commented Sep 24, 2021 at 17:28
• @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 $\beta(z_i)=0,i=1,\ldots,n \implies \beta(z)=0$. Commented Sep 24, 2021 at 17:35
• 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. Commented Sep 24, 2021 at 18:22