7
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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? $\endgroup$ – AccidentalFourierTransform Jan 15 '17 at 10:20
  • 1
    $\begingroup$ Also, there seems to be something missing in $e^{(i/\hbar)(E_M-E_n-ikc\hbar)} $ (dimensions are inconsistent). $\endgroup$ – AccidentalFourierTransform Jan 15 '17 at 10:20
  • $\begingroup$ 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. $\endgroup$ – M. M. R. Jan 15 '17 at 10:35
  • $\begingroup$ ah you are right sorry, there is an error in that expression. It should have been $e^{(i/\hbar)(E_M-E_n-kc\hbar)}$ $\endgroup$ – M. M. R. Jan 15 '17 at 10:38
  • 1
    $\begingroup$ the dimensions are still inconsistent. You need something with units of time ($\mathrm e^{iEt/\hbar}$). $\endgroup$ – AccidentalFourierTransform Jan 15 '17 at 10:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.