# Calculation of Matrix Element in “Old-Fashioned Perturbation Theory”

I would like to better under the manipulations/formalism applied in order to evaluate the following matrix element from Schwartz "Quantum Field Theory and the Standard Model" (Eq. 4.16)

$$\quad V _ { n i } ^ { ( R ) } = \left\langle \psi _ { e } ^ { 3 } \phi ^ { \gamma } \psi _ { e } ^ { 2 } | V | \psi _ { e } ^ { 1 } \psi _ { e } ^ { 2 } \right\rangle \, . \tag{1}$$

In this problem we are trying to calculate the transition matrix to first non-trivial order:

$$T _ { f i } = V _ { f i } + \sum _ { n } V _ { f n } \frac { 1 } { E _ { i } - E _ { n } } V _ { n i } + \cdots$$

where the interaction is between two scalar electrons defined by the interaction potential

$$V = \frac{1}{2} e \int d^3 x \psi_e (x) \phi(x) \psi_e (x)$$

with a scalar field operator for any scalar particle being defined by

$$\phi(\vec x) := \int \frac {d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}} \left( a_p e^{i \vec p \vec {x}} + a_p^\dagger e^{- i \vec{p} \vec{x}} \right) \, .$$

There are two possible intermediate states based on the time-ordering of the relevant Feynman diagram (which I do not have a picture of). In (1) we are attempting to calculate the retarded interaction amplitude between the initial state $$| \psi_e^1 \psi_e^2 \rangle$$ and the intermediate state $$| \psi_e^3 \phi^{\gamma} \psi_e^2 \rangle$$ where electron 1 emits a photon and electron 2 feels the effect of the photon at an earlier time. The following commutation relations are also assumed

$$\left[ a _ { k } , a _ { p } ^ { \dagger } \right] = ( 2 \pi ) ^ { 3 } \delta ^ { 3 } ( \vec { p } - \vec { k } ) \quad \left[ a _ { k } ^ { \dagger } , a _ { p } ^ { \dagger } \right] = 0 \quad \left[ a _ { k } , a _ { p } \right] = 0 \, .$$

Note that my questions are labeled with a (Qi).

Schwartz then proceeds to do the following from (1)

$$V _ { n i } ^ { ( R ) } = \left\langle \psi _ { e } ^ { 3 } \phi ^ { \gamma } \psi _ { e } ^ { 2 } | V | \psi _ { e } ^ { 1 } \psi _ { e } ^ { 2 } \right\rangle = \left\langle \psi _ { e } ^ { 3 } \phi ^ { \gamma } | V | \psi _ { e } ^ { 1 } \right\rangle \left\langle \psi _ { e } ^ { 2 } | \psi _ { e } ^ { 2 } \right\rangle = \left\langle \psi _ { e } ^ { 3 } \phi ^ { \gamma } | V | \psi _ { e } ^ { 1 } \right\rangle \\ = \frac{e}{2} \int d^3x \langle \phi^{\gamma} | \phi(\vec{x}) |0 \rangle \langle \psi_e^3 | \psi_e(\vec{x})^2 | \psi_e^1 \rangle \, .$$

My issue is with the first step

$$\left\langle \psi _ { e } ^ { 3 } \phi ^ { \gamma } \psi _ { e } ^ { 2 } | V | \psi _ { e } ^ { 1 } \psi _ { e } ^ { 2 } \right\rangle = \left\langle \psi _ { e } ^ { 3 } \phi ^ { \gamma } | V | \psi _ { e } ^ { 1 } \right\rangle \left\langle \psi _ { e } ^ { 2 } | \psi _ { e } ^ { 2 } \right\rangle$$

as I do not understand how he is allowed to seperate the matrix element like this. (Q1) In particular, how is he allowed to move the state $$\langle \psi_e^2 |$$ across $$V$$ with seemingly no effect? Why is he allowed to do this? Can someone show me explicit justification for this step?

I believe this might stem from my view of the multi-particle states. Formally, I would view $$| \psi_e^2 \psi_e^3 \phi^{\gamma} \rangle$$ as something like

$$\quad | \psi_e^2 \psi_e^3 \phi^{\gamma} \rangle \equiv | \psi_e^2 \rangle \otimes | \psi_e^3 \rangle \otimes | \phi^{\gamma} \rangle \propto a(\vec{p}_2) |0 \rangle \otimes a(\vec{p}_3) |0 \rangle \otimes a(\vec{p}_{\gamma}) |0 \rangle \equiv a(\vec{p}_2) a(\vec{p}_3) a(\vec{p}_{\gamma}) | 0 \rangle \, \tag{2}$$

where the photon and electron states "live" in different Fock spaces. (Q2) Is this a correct viewpoint?

Additionally, I believe this could just be the easiest way to evaluate the matrix element, in the sense that one has the freedom of breaking up the inner products, which is why he does it. However this would mean one should also be able to obtain the same answer by evaluating the matrix element as

$$\quad \left\langle \psi _ { e } ^ { 3 } \phi ^ { \gamma } \psi _ { e } ^ { 2 } | V | \psi _ { e } ^ { 1 } \psi _ { e } ^ { 2 } \right\rangle = \frac{e}{2} \int d^3x \langle \phi^{\gamma} | \phi(\vec{x}) |0 \rangle \langle \psi_e^3 |\psi_e( \vec{x})^2 | \psi_e^2 \rangle \langle \psi_e^2 | \psi_e^1 \rangle \, \tag{3}$$

or even

$$\quad \left\langle \psi _ { e } ^ { 3 } \phi ^ { \gamma } \psi _ { e } ^ { 2 } | V | \psi _ { e } ^ { 1 } \psi _ { e } ^ { 2 } \right\rangle = \frac{e}{2} \int d^3x \langle \phi^{\gamma} | \phi(\vec{x}) |0 \rangle \langle \psi_e^2 |\psi_e( \vec{x})^2 | \psi_e^2 \rangle \langle \psi_e^3 | \psi_e^1 \rangle \, \tag{4}$$ .

(Q3) Is this reasoning correct? How one partitions operators and states in the matrix element is arbitrary? Or is there some rule in which the field operators must act on specific states based on the description of the interaction, i.e. time ordering, which I am unaware of?

However, in evaluating (3) I obtain, modulo some normalization factors,

$$\frac{e}{2} \int d^3x \langle \phi^{\gamma} | \phi(\vec{x}) |0 \rangle \langle \psi_e^3 |\psi_e( \vec{x})^2 | \psi_e^2 \rangle \langle \psi_e^2 | \psi_e^1 \rangle = e (2 \pi)^3 \delta( \vec{p}_2 - \vec{p}_3 - \vec{p}_{\gamma})\delta( \vec{p}_1 - \vec{p}_2) + e (2 \pi)^3 \delta( \vec{p}_1 - \vec{p}_2 ) \delta( \vec{p}_{\gamma})$$

where I used the fact $$\langle 0 | \psi_e(\vec{x})^2 | 0 \rangle = 1$$. For (4) I obtain

$$\frac{e}{2} \int d^3x \langle \phi^{\gamma} | \phi(\vec{x}) |0 \rangle \langle \psi_e^2 |\psi_e( \vec{x})^2 | \psi_e^2 \rangle \langle \psi_e^3 | \psi_e^1 \rangle = e (2 \pi)^3 \delta( \vec{p}_3 - \vec{p}_1) \delta(\vec{p}_{\gamma})$$

where I've used the fact $$\langle \psi_e^2 | \psi_e(\vec{x})^2 | \psi_e^2 \rangle = 1$$. I am not sure, in either case, where one would go from here to retrieve the correct answer of $$e (2\pi)^3 \delta( \vec{p}_1 - \vec{p}_3 - \vec{p}_{\gamma})$$ with all these delta functions laying around under no integrals. (Q4) Assuming these are correct, how can the correct answer be obtained from here? Specifically, how should I view these delta functions under no integrals?

Finally, concerning my naive use of the relations $$\langle \psi_e^2 | \psi_e(\vec{x})^2 | \psi_e^2 \rangle = 1$$ and $$\langle 0 | \psi_e(\vec{x})^2 | 0 \rangle = 1$$, I have reservations about this. Explicitly calculating each element I obtain

$$\langle \psi_e^2 | \psi_e(\vec{x})^2 | \psi_e^2 \rangle = 1 + 2 \omega_{p_2} \langle 0 | \psi_e(\vec{x})^2 | 0 \rangle$$

and

$$\langle 0 | \psi_e(\vec{x})^2 | 0 \rangle = \iint \frac{ d^3 p d^3 p'}{ (2 \pi)^6 } \frac{1}{\sqrt{4 \omega_{p} \omega_{p'} }} \left ( e^{i ( \vec{p} + \vec{p}') \cdot \vec{x} } \langle 0 | a(\vec{p}) a( \vec{p}') | 0 \rangle + e^{i ( \vec{p} - \vec{p}') \cdot \vec{x}} \langle 0 | a(\vec{p}) a^{\dagger}( \vec{p}') | 0 \rangle + e^{i ( - \vec{p} + \vec{p}') \cdot \vec{x} } \langle 0 | a^{\dagger} (\vec{p}) a( \vec{p}') | 0 \rangle + e^{- i ( \vec{p} + \vec{p}') \cdot \vec{x}} \langle 0 | a^{\dagger}(\vec{p}) a^{\dagger} ( \vec{p}') | 0 \rangle \right ) \\ = \iint \frac{ d^3 p d^3 p'}{ (2 \pi)^6 } \frac{1}{\sqrt{4 \omega_{p} \omega_{p'} }} \left ( e^{i( \vec{p} - \vec{p}') \cdot \vec{x} } \langle 0 | a(\vec{p}) a^{\dagger}(\vec{p}') |0 \rangle \right ) = \iint \frac{ d^3 p d^3 p'}{ (2 \pi)^3 } \frac{1}{\sqrt{4 \omega_{p} \omega_{p'} }} \delta( \vec{p} - \vec{p}') e^{i( \vec{p} - \vec{p}') \cdot \vec{x} } = \int \frac{ d^3 p}{ (2 \pi)^3 } \frac{1}{2 \omega_{p} } = ?$$

which means I really don't know what these two matrix elements are and therefore I am not sure the final form of (3) and (4) are correct. (Q5) What am I doing wrong in these other calculations?

If anyone would be willing, I'd greatly appreciate some feedback on these calculational issues. Thank you.

My main question (Q1) was concerned with explicitly showing the following equality

$$\left\langle \psi _ { e } ^ { 3 } \phi ^ { \gamma } \psi _ { e } ^ { 2 } | V | \psi _ { e } ^ { 1 } \psi _ { e } ^ { 2 } \right\rangle = \left\langle \psi _ { e } ^ { 3 } \phi ^ { \gamma } | V | \psi _ { e } ^ { 1 } \right\rangle \left\langle \psi _ { e } ^ { 2 } | \psi _ { e } ^ { 2 } \right\rangle \tag{1}$$

which I will show and afterward address the pathologies of my succeeding questions. We want to evaluate

$$\left\langle \psi _ { e } ^ { 3 } \phi ^ { \gamma } \psi _ { e } ^ { 2 } | V | \psi _ { e } ^ { 1 } \psi _ { e } ^ { 2 } \right\rangle = \frac{e}{2} \int d^3x\left\langle \psi _ { e } ^ { 3 } \phi ^ { \gamma } \psi _ { e } ^ { 2 } | \psi_e(x) \phi(x) \psi_e(x) | \psi _ { e } ^ { 1 } \psi _ { e } ^ { 2 } \right\rangle$$

where (Q2) $$| \psi_i \rangle = \sqrt{ 2 \omega_{ p_i}} a^{\dagger}( \vec{p}_i) |0\rangle$$ and we can interpret

$$| \psi_e^2 \phi_{\gamma} \psi_e^3 \rangle = \sqrt{ 2 \omega_{p_2} }a_e^{\dagger}( \vec{p}_2) |0_e \rangle \otimes \sqrt{ 2 \omega_{p} } a_{\gamma}^{\dagger}( \vec{p}) |0_{\gamma} \rangle \otimes \sqrt{ 2 \omega_{p_3} } a_e^{\dagger}( \vec{p}_3) |0_e \rangle \equiv 2 \sqrt{ \omega_{p_3} \omega_{p_2} } a_e^{\dagger}( \vec{p}_2) a_e^{\dagger}( \vec{p}_3) | 0_e \rangle \otimes \sqrt{ 2 \omega_{p} } a_{\gamma}^{\dagger}( \vec{p}) | 0 _{\gamma} \rangle \equiv \sqrt{8 \omega_{p_2} \omega_{p} \omega_{p_3} } a_{\gamma}^{\dagger}( \vec{p})a_e^{\dagger}( \vec{p}_2) a_e^{\dagger}( \vec{p}_3)| 0 \rangle,$$

with the understanding that operators in the photon Fock space commute with all operators in the electron Fock space, and the field operators corresponding to any particle are of the form

$$\phi ( \vec { x } ) : = \int \frac { d ^ { 3 } p } { ( 2 \pi ) ^ { 3 } } \frac { 1 } { \sqrt { 2 \omega _ { p } } } \left( a _ { p } e ^ { i \vec { p } \vec { x } } + a _ { p } ^ { \dagger } e ^ { - i \vec { p } \vec { x } } \right).$$

Through simply using the commutator relations

$$\left[ a _ { k } , a _ { p } ^ { \dagger } \right] = ( 2 \pi ) ^ { 3 } \delta ^ { 3 } ( \vec { p } - \vec { k } ) \quad \left[ a _ { k } ^ { \dagger } , a _ { p } ^ { \dagger } \right] = 0 \quad \left[ a _ { k } , a _ { p } \right] = 0$$

we can show the relation in (1) holds. Redefining the annihilation and creation operators in the field expressions to be $$a( \vec{p}) e^{i \vec{p} \cdot \vec{x} } \to a( \vec{p})$$ and $$a^{\dagger} ( \vec{p} ) e^{-i \vec{p} \cdot \vec{x} } \to a^{\dagger}( \vec{p})$$ the full expression we need to evaluate is of the form

$$\left\langle \psi _ { e } ^ { 3 } \phi ^ { \gamma } \psi _ { e } ^ { 2 } | V | \psi _ { e } ^ { 1 } \psi _ { e } ^ { 2 } \right\rangle = \frac{e}{2} \langle \phi_{\gamma} | \phi( \vec{x} ) | 0_{\gamma} \rangle \int \frac { d ^ { 3 } p d^3 p' } { ( 2 \pi ) ^ { 6 } } \frac { 4 \omega_{p_2} \sqrt{ \omega_{p_3} \omega_{p_1} } } { \sqrt { 4 \omega _ { p } \omega_{p'} } } \langle 0_e | a( \vec{p}_3) a( \vec{p}_2) \left( a(\vec{p}) + a^ { \dagger } (\vec{p}) \right) \left(a(\vec{p} ') + a^ { \dagger } (\vec{p }') \right) a^{\dagger} ( \vec{p}_1) a^{\dagger} ( \vec{p}_2) | 0_e \rangle \tag{2}.$$

Focusing on the inner product we use the commutation relations to move $$a( \vec{p}_2)$$ to the right to obtain

$$\langle 0_e | a( \vec{p}_3 ) \left ( a( \vec{p}_2) a( \vec{p}) + a( \vec{p}_2) a^{\dagger} ( \vec{p}) \right ) \left ( a( \vec{p}') + a^{\dagger} ( \vec{p}') \right ) a^{\dagger}( \vec{p}_1) a^{\dagger}( \vec{p}_2) | 0_e \rangle \\ = \langle 0_e | a( \vec{p}_3 ) \left ( a( \vec{p}) + (2 \pi)^3 \delta( \vec{p} - \vec{p}_2 ) + a^{\dagger} ( \vec{p}) \right ) \left ( a( \vec{p}_2) a( \vec{p}') + a( \vec{p}_2) a^{\dagger} ( \vec{p}') \right ) a^{\dagger}( \vec{p}_1) a^{\dagger}( \vec{p}_2) | 0_e \rangle \\ = \langle 0_e | a( \vec{p}_3 ) \left ( a( \vec{p}) + (2 \pi)^3 \delta( \vec{p} - \vec{p}_2 ) + a^{\dagger} ( \vec{p}) \right ) \left ( a( \vec{p}') + (2 \pi)^3 \delta( \vec{p}' - \vec{p}_2 ) + a^{\dagger} ( \vec{p}') \right ) a( \vec{p}_2) a^{\dagger}( \vec{p}_1) a^{\dagger}( \vec{p}_2) | 0_e \rangle \\ = \langle 0_e | a( \vec{p}_3 ) \left ( a( \vec{p}) + (2 \pi)^3 \delta( \vec{p} - \vec{p}_2 ) + a^{\dagger} ( \vec{p}) \right ) \left ( a( \vec{p}') + (2 \pi)^3 \delta( \vec{p}' - \vec{p}_2 ) + a^{\dagger} ( \vec{p}') \right ) a^{\dagger}( \vec{p}_1) a( \vec{p}_2) a^{\dagger}( \vec{p}_2) | 0_e \rangle \\ = \langle 0_e | a( \vec{p}_3 ) \left ( a( \vec{p}) + (2 \pi)^3 \delta( \vec{p} - \vec{p}_2 ) + a^{\dagger} ( \vec{p}) \right ) \left ( a( \vec{p}') + (2 \pi)^3 \delta( \vec{p}' - \vec{p}_2 ) + a^{\dagger} ( \vec{p}') \right ) a^{\dagger}( \vec{p}_1) \left ((2 \pi)^3 + a^{\dagger} ( \vec{p}_2) a( \vec{p}_2) \right ) | 0_e \rangle \\ = (2 \pi)^3 \langle 0_e | a( \vec{p}_3 ) \left ( a( \vec{p}) + (2 \pi)^3 \delta( \vec{p} - \vec{p}_2 ) + a^{\dagger} ( \vec{p}) \right ) \left ( a( \vec{p}') + (2 \pi)^3 \delta( \vec{p}' - \vec{p}_2 ) + a^{\dagger} ( \vec{p}') \right ) a^{\dagger}( \vec{p}_1) | 0_e \rangle \\$$

If we can show that all terms with a delta function vanish, then we recover $$(2 \pi)^3 \langle 0_e | a( \vec{p}_3) \psi_e(\vec{x}) ^2 a^{\dagger}( \vec{p}_1) | 0_e \rangle$$. There are 5 such terms:

$$\delta( \vec{p} - \vec{p}_2 ) \langle 0_e | a( \vec{p}_3) a( \vec{p}') a^{\dagger}( \vec{p}_1) | 0_e \rangle = \delta( \vec{p} - \vec{p}_2 ) \langle 0_e | a( \vec{p}') \left ( ( 2 \pi)^3 + a^{\dagger}( \vec{p}_1) a( \vec{p}_3) \right ) | 0_e \rangle = 0$$

$$\delta( \vec{p} - \vec{p}_2 ) \delta( \vec{p}' - \vec{p}_2 ) \langle 0_e | a( \vec{p}_3) a^{\dagger}( \vec{p}_1) | 0_e \rangle = \delta( \vec{p} - \vec{p}_2 ) \delta( \vec{p}' - \vec{p}_2 ) \langle 0_e | a^{\dagger}( \vec{p}_1) a( \vec{p}_3)| 0_e \rangle = 0$$

$$\delta( \vec{p} - \vec{p}_2 ) \langle 0_e | a( \vec{p}_3) a^{\dagger}( \vec{p}') a^{\dagger}( \vec{p}_1) | 0_e \rangle = \delta( \vec{p} - \vec{p}_2 ) \langle 0_e | \left ( ( 2 \pi)^3 + a^{\dagger}( \vec{p}_1) a( \vec{p}_3) \right ) a^{\dagger}( \vec{p}') | 0_e \rangle = 0$$

$$\delta( \vec{p}' - \vec{p}_2 ) \langle 0_e | a( \vec{p}_3) a( \vec{p}) a^{\dagger}( \vec{p}_1) | 0_e \rangle = \delta( \vec{p}' - \vec{p}_2 ) \langle 0_e | a( \vec{p}) \left ( ( 2 \pi)^3 + a^{\dagger}( \vec{p}_1) a( \vec{p}_3) \right ) | 0_e \rangle = 0$$

$$\delta( \vec{p}' - \vec{p}_2 ) \langle 0_e | a( \vec{p}_3) a^{\dagger}( \vec{p}) a^{\dagger}( \vec{p}_1) | 0_e \rangle = \delta( \vec{p}' - \vec{p}_2 ) \langle 0_e | \left ( ( 2 \pi)^3 + a^{\dagger}( \vec{p}_1) a( \vec{p}_3) \right ) a^{\dagger}( \vec{p}) | 0_e \rangle = 0$$

We then obtain the resulting inner product

$$(2 \pi)^3 \langle 0_e | a( \vec{p}_3 ) \left ( a( \vec{p}) + a^{\dagger}( \vec{p} )g \right ) \left ( a( \vec{p}') + a^{\dagger}( \vec{p}') \right ) a^{\dagger}( \vec{p}_1) | 0_e \rangle$$

Inserting this result back into (2) we obtain

$$( 2 \pi)^3 \frac{e}{2} \langle \phi_{\gamma} | \phi( \vec{x} ) | 0_{\gamma} \rangle \int \frac { d ^ { 3 } p d^3 p' } { ( 2 \pi ) ^ { 6 } } \frac { 4 \omega_{p_2} \sqrt{ \omega_{p_3} \omega_{p_1} } } { \sqrt { 4 \omega _ { p } \omega_{p'} } } \langle 0_e | a( \vec{p}_3) \left( a(\vec{p}) + a^ { \dagger } (\vec{p}) \right) \left(a(\vec{p} ') + a^ { \dagger } (\vec{p }') \right) a^{\dagger} ( \vec{p}_1) | 0_e \rangle \\ = ( 2 \pi)^3 2 \omega_{p_2} \frac{e}{2} \langle \phi_{\gamma} | \phi( \vec{x} ) | 0_{\gamma} \rangle \langle \psi_e^1|\psi_e(\vec{x} )^2 | \psi_e^3 \rangle$$

Modulo the $$(2 \pi)^3 2 \omega_{p_2}$$, this is exactly what Schwartz obtains as given in Eq. 4.18 (2013 Ed.). Now, these constant may cancel and it is the case that I have overlooked the mechanism by which this occurs however, I believe it might also be the case that Schwartz left these constants out on purpose because when the calculation iss done using standard QFT methods, i.e. Feynman rules, these constants do not emerge when performing the same calculation. If one were to look at (1), by definition, the inner product $$\langle \psi_e^2 | \psi_e^2 \rangle \neq 1$$ but instead $$\langle \psi_e^2 | \psi_e^2 \rangle = (2 \pi)^3 2 \omega_{p_2}$$ which is consistent with the second quantization development in Schwartz (Ch. 2.3). He could also be using the standard normalization scheme in non-rel QM however this would change the form of the field operators so I do not think this is the case.

If anyone would like to clarify this point about the constants for completeness, I would much appreciate it.

To answer my question (Q3), I do not believe one can partition the initial/final states in any arbitrary manner - whatever manipulations one performs must be in accordance with the canonical commutation relations of raising and lowering operators. This also answers (Q4) with the addition that one can practically interpret $$\delta(0) := 1$$ when it is not under an integral, although one should always keep in mind this is not rigorous.

Concerning (Q5), I need to redo this calculation which does not seem all to relevant in hindsight to answering the main question I posed.

• Thanks for the question and the answer. People who are studying Schwartz's QFT may find this useful. – rainman Jan 10 '19 at 10:00