I don't understand a particular statement in the QFT book by Klauber. The particular page I'm having difficulty on is page 67 of chapter 3 (PDF link).

The big picture is that the author wishes to investigate what the (operator) solutions to the Klein-Gordon equation, $\phi(x)$ and $\phi^\dagger(x)$, do when acting on the vacuum state $|0\rangle$. As prep for this, he creates a "general single particle state" ("general" meaning non $\mathbf{k}$-eigenstate) by operating on the vacuum with the operator $$C\equiv\sum_\mathbf{k}A_\mathbf{k}a_\mathbf{k}^\dagger,\tag{3-108}$$ $$C|0\rangle=\sum_\mathbf{k}A_\mathbf{k}a_\mathbf{k}^\dagger|0\rangle=A_1|\phi_1\rangle+A_2|\phi_2\rangle+\cdots\equiv|\phi\rangle\tag{3-109}$$ Each $A_\mathbf{k}$ is just a number, the absolute value square of which represents the probability of finding the $\mathbf{k}$ eigenstate for the single particle.

The new state $C|0\rangle=|\phi\rangle$ is interpreted as a single particle state in a superposition of $\mathbf{k}$-eigenstates $|\phi_k\rangle$. The subscript $\mathbf{k}$ represents different momenta.

For probability/normalization arguments, the numbers $A_\mathbf{k}$ should obey $$\sum_\mathbf{k}\left|A_\mathbf{k}\right|^2=1.\tag{3-110}$$ I feel like I understand the above statements.

The author then introduces the "general single particle destruction operator" $$D\equiv\sum_\mathbf{k}a_k,\tag{3-111}$$ and shows that when applied to our general single particle state $|\phi\rangle$ above, the vacuum is (re)produced: $$\begin{eqnarray} D|\phi\rangle&=&\left(\sum_\mathbf{k}a_k\right)A_1|\phi_1\rangle+\left(\sum_\mathbf{k}a_k\right)A_2|\phi_2\rangle+\cdots\\ &=&A_1\underbrace{a_1|\phi_1\rangle}_{=|0\rangle}+A_1\underbrace{a_2|\phi_1\rangle}_{=0}+A_1\underbrace{a_3|\phi_1\rangle}_{=0}+\cdots+\\ &\ &+A_2\underbrace{a_1|\phi_2\rangle}_{=0}+A_2\underbrace{a_2|\phi_2\rangle}_{=|0\rangle}+ A_2\underbrace{a_3|\phi_2\rangle}_{=0}+\cdots+\\ &\ &+\cdots\\ &=&\underbrace{\left(A_1 + A_2 + \cdots\right)}_\text{can normalize = 1}|0\rangle. \end{eqnarray}\tag{3-112}$$ (Note the subtle but important differences in the underbraces; some are $0$ while others are $|0\rangle$.)

The part I am struggling with is understanding how the underbrace "can normalize = 1" at the end of $\text{(3-112)}$ can be true given $\text{(3-110)}$. It seems to me that the $A$ terms appearing at the end of $\text{(3-112)}$ are the same ones defined in the construction operator $C$ and normalized so that their absolute values squared sum to $1$. How can their just-plain sum also be of magnitude $1$? I know that one would *like * the underbraced term to sum to zero, but I don't see how that can be.

It was suggested I consider the quantity $\langle\phi|D^\dagger D|\phi\rangle$. Here is my attempt to calculate it.

$$ \begin{eqnarray} \langle\phi|D^\dagger D|\phi\rangle&=&\langle0|(A_1^\dagger+A_2^\dagger+\cdots)(A_1+A_2+\cdots)|0\rangle=\langle0|\sum_\mathbf{j}\sum_\mathbf{k}A_\mathbf{j}^\dagger A_\mathbf{k}|0\rangle\\ &=&\sum_\mathbf{j}\sum_\mathbf{k}A_\mathbf{j}^\dagger A_\mathbf{k}\underbrace{\langle0|0\rangle}_{=1}=\underbrace{\sum_\mathbf{j}\sum_\mathbf{k}A_\mathbf{j}^\dagger A_\mathbf{k}}_\text{Can't simplify}\ne1 \end{eqnarray} $$

  • $\begingroup$ Sorry - I misread your question. My answer was incorrect so I've deleted it! Apologies for the confusion. $\endgroup$ Commented Dec 31, 2013 at 16:55

1 Answer 1


Paragraphs "Creating a General Single Particle State (Discrete Solution Form) "($3.108 \to 3.110 $) and "Destroying a General Single Particle State (Discrete) " ($3.111 \to 3.112 $) are two independent paragraphs, are should not be mixed.

It is not possible to start with a normalized state $C\equiv\sum_\mathbf{k}A_\mathbf{k}a_\mathbf{k}^\dagger$, with $\sum_\mathbf{k}\left|A_\mathbf{k}\right|^2=1$, then applying the operator $D\equiv\sum_\mathbf{k}a_k$, and find that the resulting state $\sum\limits_i A_i|0\rangle$ is also normalized (except in the trivial case where there is only one term in the sum).

The main reason is that the operator $D$ is not unitary, so there is no reason why it should transform a normalized state into an other normalized state. Or said, differently :

$\sum_\mathbf{k}\left|A_\mathbf{k}\right|^2 \neq |\sum\limits_{k} A_\mathbf{k} |^2$

  • $\begingroup$ If this is the case, then $|\phi\rangle$ given in $\text{(3-109)}$ has to be different than $|\phi\rangle$ given in $\text{(3-112)}$. That is, the state in $\text{(3-112)}$ is not normalized. $\endgroup$
    – BMS
    Commented Dec 31, 2013 at 21:12
  • $\begingroup$ @BMS : $3.109$ is OK for paragraph "Destroying", but not $3.110$, that is : $\sum_\mathbf{k}\left|A_\mathbf{k}\right|^2 \neq |\sum\limits_{k} A_\mathbf{k} |^2$. So, if the initial state $|\psi\rangle$ is normalized, the final state $D|\psi\rangle$ is not normalized, and if the final state is normalized, this means that the initial state is not normalized (except trivial exceptions) $\endgroup$
    – Trimok
    Commented Dec 31, 2013 at 21:20

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