3
$\begingroup$

I am reading the book Topics in Advanced Quantum Mechanics by Holstein. In Chapter 3, section 3 he discusses the Aharonov-Bohm effect, but before doing so he discusses the ordinary double slit experiment. This book is based on the path integral approach and what he does is as follows: First we assume that the two slits are the same distance $d_0$ from the particle source (located at $x_i$), and that the first slit, at a point $x_1$, is a distance $d_1$ from the detection point and the second slit at $x_2$ is a distance $d_2$ from the detection point, call it $x_f$. Now the action for the classical particle is simply the free particle action $S = \int_0^{t_f}dt\frac 12 m \dot x^2$. Holstein then states, that assuming that the classical path is dominant, the contribution to the path integral from the path through slit 1 is $$\exp(i\int_0^{t_f}dt\frac 12 m \dot x^2)\approx \exp(i\frac{2\pi d_0}{\lambda}+i\frac {2\pi d_1}{\lambda}),\tag{3.3}$$ and similarly for the path through slit 2, just replacing $d_1$ with $d_2$, and where $\lambda$ is the de Broglie wavelength.

My question is, how is this derived? Holstein simply states it, offering virtually no justification. I understand how, using this formula, one can derive the diffraction pattern for the double slit experiment, but I don't understand how this formula is derived.

Edit: $\hbar =1$ throughout.

$\endgroup$
1

2 Answers 2

1
$\begingroup$

Let's focus on the path from the source to the slits. Assuming that the particle's speed $v$ is constant, the phase accrued is $\phi = \frac{m v^2 t}{2\hbar} = \frac{p d_0}{2\hbar}$, where we used $d = v t$ and $p = mv$ (Note that I have added an $\hbar=h/2\pi$ to the denominator. This factor should also be in the denominator of your exponential).

Since the de Broglie wavelength is given by $\lambda = h/p$, the phase $\phi$ can be rewritten as $\phi = \pi d_0/ \lambda$. Note that this is off by a factor of 2 from what we expect from your result. I'm not sure where I could have dropped a factor of 2 in the above derivation. It may be that for Holstein's purposes, this factor is inconsequential, so he ignored it.

$\endgroup$
2
  • $\begingroup$ Holstein approximates uses natural units, by the way, I clarified that in my question. It probably is true that he just tossed the factor of 1/2 out for convenience. Somehow, I neglected to use the basic formula of $d= vt$, which was the main problem. Thanks for the answer. $\endgroup$ Aug 12, 2022 at 18:38
  • $\begingroup$ Sorry, the word approximates in that first sentence is a typo. $\endgroup$ Aug 12, 2022 at 19:11
0
$\begingroup$
  1. It seems Holstein is conflating their set-ups/sources. Before eq. (3.3) Holstein talks about a single particle time-dependent picture using classical action $$S~=~\frac{m}{\color{red}{2}}\frac{d^2}{t}~=~\frac{pd}{\color{red}{2}},\qquad p~=~\hbar k;$$ and afterwards a time-independent plane wave picture, where only the position-part of the phase $$\phi~=~k d, \qquad k~=~\frac{2\pi}{\lambda},$$ matters. The LHS of eq. (3.3) is written in the former action formulation, while the RHS of eq. (3.3) refers to the latter spatial phase formulation, i.e. the equality does not hold.

  2. Compare e.g. Holstein's result $$ I(\theta)~=~4I_0\cos^2\frac{\Delta\phi}{2} ~=~4I_0\cos^2\left( \pi\delta \frac{\sin\theta}{\lambda}\right) \tag{3.10} $$ for the double slit experiment. Eq. (3.10) is clearly in the latter picture.

References:

  1. B.R. Holstein, Topics in Advanced Quantum Mechanics, 1991; p. 112 eq. (3.3).
$\endgroup$
3
  • $\begingroup$ If the equality does not hold (even approximately), then is the resulting discussion wrong? $\endgroup$ Aug 14, 2022 at 13:19
  • $\begingroup$ It seems Holstein changed horses in midstream: The set-ups before and after are both possible and physically consistent in their own right, but mutually incompatible. $\endgroup$
    – Qmechanic
    Aug 14, 2022 at 13:31
  • $\begingroup$ Ah okay. The book does have numerous errors, so I guess that's to be expected. Thanks for the answer. $\endgroup$ Aug 14, 2022 at 14:07

Your Answer

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

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