2
$\begingroup$

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?

$\endgroup$
10
  • 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

2
$\begingroup$

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 $f(\xi) = 0$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.