# Path Integral on Feynman Hibbs: Interaction of EM field and matter, how can we get to equation (9.68) from (9.67)?

On Feynman Hibbs "Quantum Mechanics and Path Integrals", the equation (9.67) describe the transition amplitude of the matter (for example an atom) to go from the state $M$ to the state $M$ when it iteracts with the electromagnetic field, considering vacuum to vacuum transitions only. This equation is

$$\lambda^{(1)}_{MM} = \frac{i}{\hbar}\sum_{\mathbf{k}}\int_{t_a}^{t_b}\mathcal{D}\mathbf{x}(t)\psi^{*}_M(\mathbf x_b)e^{(i/\hbar)S_{\text{mat}}} \frac{i\pi}{kc} \\ \times \int_{t_a}^{t_b}\int_{t_a}^{t_b}dtds[\bar j_{1, \mathbf k}(t)\bar j^*_{1, \mathbf k}(s)+\bar j_{2, \mathbf k}(t)\bar j^*_{2, \mathbf k}(s)]e^{-ikc|t-s|}\psi_M(\mathbf x_a) \tag{1}$$

Where $\psi_n(\mathbf x)$ are the states describing the matter.

Considering polarization 1 only, I call

$$I_1 = \sum_{\mathbf{k}}\frac{i\pi}{kc} \int_{t_a}^{t_b}\int_{t_a}^{t_b}dtds[\bar j_{1, \mathbf k}(t)\bar j^*_{1, \mathbf k}(s)]e^{-ikc|t-s|}$$

calling $t \equiv t_c$ and $s \equiv t_d$, and considering $t_a < t_d < t_c < t_b$ I can write (thanks also to appendix (A.12))

$$I_1 = \sum_{\mathbf{k}}\frac{2i\pi}{kc} \int_{t_a}^{t_b}dt_c\int_{t_a}^{t_c}dt_d[\bar j_{1, \mathbf k}(t_c)\bar j^*_{1, \mathbf k}(t_d)]e^{-ikc(t_c-t_d)}$$

Now I think I understood how to get the following expression for $\lambda^{(1)}_{MM}$ still considering polarization 1 only (this is the same expression of appendix:Notes at page 365)

\begin{align} \lambda^{(1)}_{MM} & = \sum_n\sum_{\mathbf{k}} A_{\mathbf{k}} \int_{t_a}^{t_b}dt_c\int_{t_a}^{t_c}dt_d f_{\mathbf{k}}(t_c) f^*_{\mathbf{k}}(t_d)e^{-(i/\hbar)E_M(t_b-t_c)}e^{-(i/\hbar)E_n(t_c-t_d)}e^{-(i/\hbar)E_M(t_d-t_a)}e^{-ikc(t_c-t_d)} \\ & = \sum_n\sum_{\mathbf{k}} A_{\mathbf{k}} e^{-(i/\hbar)E_M(t_b-t_a)}\int_{t_a}^{t_b}dt_c\int_{t_a}^{t_c}dt_d f_{\mathbf{k}}(t_c) f^*_{\mathbf{k}}(t_d)e^{(i/\hbar)(E_M-E_n-\hbar kc)(t_c-t_d)} \qquad \qquad \quad (2) \end{align}

where $A_{\mathbf{k}} =i(2\pi i)/(\hbar kc)$ and $f_{\mathbf{k}}(t_c)$ $f^*_{\mathbf{k}}(t_d)$ are the matrix elements

\begin{align} f_{\mathbf{k}}(t_c) & = \int dx_c\psi^{*}_M(\mathbf x_c) \bar j_{1,\mathbf k}(t_c) \psi_n(\mathbf x_c) \\ f^*_{\mathbf{k}}(t_d) & = \int dx_d\psi^{*}_n(\mathbf x_d) \bar j^*_{1,\mathbf k}(t_d) \psi_M(\mathbf x_d) \end{align}

because what matters is that they are functions of $t_c$ and $t_d$.

I can rewrite (2) as

$$\lambda^{(1)}_{MM} = -\frac{i}{\hbar}\Delta E_{\text{my}} e^{-(i/\hbar)E_MT}$$

Where $T=t_b-t_a$ and

$$\Delta E_{\text{my}} = -\sum_{\mathbf k}\sum_n \frac{2\pi i}{kc}\int_{t_a}^{t_b} dt_c \int_{t_a}^{t_c} dt_d f_{\mathbf{k}}(t_c) f^*_{\mathbf{k}}(t_d) e^{(i/\hbar)(E_M-E_n-kc\hbar)(t_c-t_d)} \tag 3$$

Now I can't understand how to get the expression (9.68). I mean on Feynman hibbs I read (still considering polarization 1 only)

With large values of T we get

$$\lambda^{(1)}_{MM} = -\frac{i}{\hbar}(\Delta E) T e^{-(i/\hbar)E_MT}$$

where

$$\Delta E = -\sum_{\mathbf k}\sum_n \frac{2\pi i}{kc}\left[f_{\mathbf{k}} f^*_{\mathbf{k}}\right]\int_{0}^{\infty} d\tau \ e^{(i/\hbar)(E_M-E_n-kc\hbar)\tau} \tag{9.68}$$

Now if $f_{\mathbf{k}}(t_c)$ and $f^*_{\mathbf{k}}(t_d)$ were not functions of time, then I could solve the double integral of (3). Changing variable $t_d = t_c -\tau$ in the second integral, I get

$$\int_{t_a}^{t_b} dt_c \int_{0}^{t_c-t_a} d\tau \ e^{(i/\hbar)(E_M-E_n-kc\hbar)\tau}$$

And now I interpret "large values of $T = t_b-t_a$" as $|t_a| \gg |t_c|$ and $t_b \gg |t_c|$ (like sending $t_a \to -\infty$ and $t_b \to \infty$) therefore \begin{align} \int_{t_a}^{t_b} dt_c \int_{0}^{t_c-t_a} d\tau \ e^{(i/\hbar)(E_M-E_n-kc\hbar)\tau} & \simeq \int_{t_a}^{t_b} dt_c \int_{0}^{|t_a|} d\tau \ e^{(i/\hbar)(E_M-E_n-kc\hbar)\tau} \\ & \simeq T\int_{0}^{+\infty} d\tau \ e^{(i/\hbar)(E_M-E_n-kc\hbar)\tau} \end{align} Therefore my $\Delta E_{\text{my}} = (\Delta E)T$, and I can write the same expression for $\lambda^{(1)}_{MM}$

My question now is, is it correct try to remove the dependence on time of the matrix element of the current? Perhaps with the condition "large values of $T$" do I remove this dependency?

• I applaud you for your effort in formatting this, but I fear that the cluttered notation is intimidating. Do you think you could reduce this into a "minimal working example"? Can you remove all the parts that are not strictly necessary? – AccidentalFourierTransform Jan 15 '17 at 10:20
• Also, there seems to be something missing in $e^{(i/\hbar)(E_M-E_n-ikc\hbar)}$ (dimensions are inconsistent). – AccidentalFourierTransform Jan 15 '17 at 10:20
• You are right, the notation is pretty awful, I just tried to follow that of the book though. Now I rewrite It, also because I think I can handle the first part. – M. M. R. Jan 15 '17 at 10:35
• ah you are right sorry, there is an error in that expression. It should have been $e^{(i/\hbar)(E_M-E_n-kc\hbar)}$ – M. M. R. Jan 15 '17 at 10:38
• the dimensions are still inconsistent. You need something with units of time ($\mathrm e^{iEt/\hbar}$). – AccidentalFourierTransform Jan 15 '17 at 10:41