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I've been trying to learn quantum mechanics from a formal point of view, so I picked up Dirac's book. In the fourth edition, 33rd page, starting from this:$$\xi|\xi'\rangle=\xi'|\xi'\rangle$$ (Where $\xi$ is a linear operator and all the other $\xi'$'s are eigen-(value|ket)s.)

,and this:$$\phi(\xi)=a_1\xi^n+a_2\xi^{n-1}\cdots a_n=0$$ (where $\phi$ is an algebraic expression) He has deduced $$\phi(\xi)|\xi'\rangle=\phi(\xi')|\xi'\rangle$$

I understand that the LHS is a linear operator acting on a ket, while the RHS is a ket multiplied by a number.

What I don't get is how the step is justified. He seems to have applied $\phi$ to both sides. But shouldn't that give $$\phi(\xi|\xi'\rangle)=\phi(\xi'|\xi'\rangle)$$?

This expression makes no sense as it is, as I doubt that you can apply an algebraic expression on a ket (I'm not sure of this, but to me, $|A\rangle^2$ &c makes no sense, as I don't think that you can multiply a ket with a ket and get another ket)

So how did he deduce the expression?

The context: Dirac is proving that an eigenvalue $\xi'$ of $\xi$ must satisfy $\phi(\xi')=0$ if $\phi(\xi)=0$.

Oh (no need to answer this if you don't want), and is there any reason for Dirac introducing the confusing notation that eigen-whatevers of an operator should be denoted by the same symbol? Usually, different types of variables (eg matrices, vectors, numbers) use different classes of symbols (capital letters, letters with overbars, and lowercase letters respectively).

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  • $\begingroup$ Did you consider the possibility of a typo? $\endgroup$
    – Antillar Maximus
    Feb 19, 2012 at 15:50
  • $\begingroup$ @AntillarMaximus it's not a typo, just a confusing notation which looks like onr $\endgroup$
    – Manishearth
    Feb 20, 2012 at 1:35

4 Answers 4

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$$\xi |\xi'\rangle = \xi'|\xi'\rangle$$ so $$\xi^2 |\xi'\rangle = \xi'^2|\xi'\rangle$$ continuing like this you see that applying any power of $\xi$ to $|\xi'\rangle$ just multiplies $|\xi'\rangle$ by $\xi'$ to that power

So any sum of powers of $\xi$ applied to $|\xi'\rangle$ just ends up multiplying $|\xi'\rangle$ by that polynomial in $\xi'$

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Probably the notations are confusing you. What you have is an operator $A$ and an eigenvector $v$ with eigenvalue $\lambda$. If $\phi(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots +c_0$. Then $v$ is an eigenvecotr for the operator $\phi(A)$ with eigenvalue $\phi(\lambda)$.

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The core of OP's question(v1) seems to be the following.

What does $\hat{A}^2|\psi \rangle$ mean? Does it e.g. mean $\hat{A}|\psi \rangle\otimes\hat{A}|\psi \rangle $?

Answer: No, it is defined as $\hat{A}(\hat{A}|\psi \rangle)$.

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  • $\begingroup$ Actually i'd understood that part :p. $\endgroup$
    – Manishearth
    Feb 19, 2012 at 14:29
  • $\begingroup$ Well, I still think this is the core issue. The rest is just straightforward manipulation with eigenvectors, eigenvalues, etc. $\endgroup$
    – Qmechanic ♦
    Feb 19, 2012 at 15:00
  • $\begingroup$ The 'straightforward' bit isn't so straightforward when you're new to this stuff =P. The same-symbol-for-eigenstuff compounds with the fact that all these variables have no concrete meaning (I haven't reached that part) or visualization. I like to look at them as matrices, but manipulating them is still a bit confusing. $\endgroup$
    – Manishearth
    Feb 19, 2012 at 15:33
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The operator $\hat\xi$ acts on the eigenvector $\left|\xi'\right>$ to give a simple multiple, $\hat\xi\left|\xi'\right>=\xi'\left|\xi'\right>$. Hence, the operator $\hat\xi^2$ acts on the same eigenvector (of $\hat\xi$) to give $\hat\xi^2\left|\xi'\right>=\hat\xi\xi'\left|\xi'\right>=(\xi')^2\left|\xi'\right>$. In other words, $\left|\xi'\right>$ is an eigenvector of $\hat\xi^2$ with eigenvalue $(\xi')^2$. All polynomial functions of a single operator have the same set of eigenvectors, with eigenvalues determined by what that function is, which can, with restrictions, be extended to all functions.

Yes, the notation $\left|\xi'\right>$ can be problematic, but careful definitions make it manageable.

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