3
$\begingroup$

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.

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

$\endgroup$
1
  • $\begingroup$ Sorry - I misread your question. My answer was incorrect so I've deleted it! Apologies for the confusion. $\endgroup$ Dec 31, 2013 at 16:55

1 Answer 1

3
$\begingroup$

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 :

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

$\endgroup$
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
    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
    Dec 31, 2013 at 21:20

Your Answer

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

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