0

I am trying to prove that the position operator in momentum space is $i\hbar \partial/\partial p$ but my derivation is missing one sign. Can someone spot the error?

Start with

$$<\hat x> = \int\limits_{-\infty}^{\infty} \psi^*(x) x \psi(x) \mathrm{d}x$$

Taking the fourier transform of both $\psi(x)$

$$ \frac{1}{2\pi} \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} \psi^*(\bar k_x) e^{i\bar k_xx} \, \mathrm{d} \bar k_x x \int\limits_{-\infty}^{\infty} \psi(k_x) e^{-i k_xx} \mathrm{d} k_x \mathrm{d} x $$

Taking integration by parts on the right most integral yields

$$ \frac{1}{2\pi} \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} \psi^*(\bar k_x) e^{i\bar k_xx} \, \mathrm{d} \bar k_x x * \frac{1}{ix} \int\limits_{-\infty}^{\infty} e^{-i k_xx} \frac {\partial \psi(k_x)}{\partial k_x} \mathrm{d} k_x \mathrm{d} x $$

We see the middle term gets canceled out.

Now using the following identity

$$ \delta(k_x-\bar k_x) = \frac{1}{2\pi} \int\limits_{-\infty}^{\infty} e^{-ix(k_x-\bar k_x)} \, \mathrm{d} x $$

we arrive at

$$ \frac{1}{i} \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} \psi^*(\bar k_x) \delta(k_x-\bar k_x) \mathrm{d} \bar k_x \frac {\partial \psi(k_x)}{\partial k_x} \mathrm{d} k_x $$

$$ \frac{1}{i}\int\limits_{-\infty}^{\infty} \psi^*(k_x) \frac {\partial \psi(k_x)}{\partial k_x} \mathrm{d} k_x $$

Very lastly, we use the following identity

$$\frac{\partial}{\partial k_x} = \frac{\hbar \partial}{\partial p_x}$$

We prove

$$ \frac{\hbar}{i} \int\limits_{-\infty}^{\infty} \psi^*(k_x) \frac {\partial \psi(k_x)}{\partial p_x} \mathrm{d} k_x $$

So

$$ x = \frac{\hbar \partial}{i\partial p_x}$$

So close, not quite right. Because the actual answer is $$ x = \frac{-\hbar \partial}{i\partial p_x}$$ or $$ x = \frac{i\hbar \partial}{\partial p_x}$$

Can someone spot the mistake. Thanks!

Qmechanic
  • 201,751
Fraïssé
  • 1,734

1 Answers1

0

How to prove that the position operator in momentum is iℏ∂/∂p

Apply the following useful result.

If

$$F(k) = \int_{-\infty}^{\infty} f(x)e^{-ikx}dx$$

then

$$\int_{-\infty}^{\infty} xf(x)e^{-ikx}dx = \int_{-\infty}^{\infty} f(x)\left(i\frac{\partial}{\partial k} e^{-ikx}\right)dx = i\frac{\partial}{\partial k}\int_{-\infty}^{\infty} f(x)e^{-ikx}dx = i\frac{\partial}{\partial k}F(k)$$

Alfred Centauri
  • 59,560
  • isn't that true that we don't have to solve home problems? – Sofia Feb 09 '15 at 16:22
  • @Sofia, isn't it true that this isn't (remotely) a solution to the problem but just a well known result from Fourier analysis? – Alfred Centauri Feb 09 '15 at 21:59
  • @Sofia, please get a grip. http://idioms.thefreedictionary.com/get+a+grip – Alfred Centauri Feb 10 '15 at 00:29
  • @Sofia, I don't have the slightest idea what you're talking about. – Alfred Centauri Feb 10 '15 at 20:59
  • the law says that home exercises should not be solved in entirety, just be given guide lines. This law should be respected by everybody. – Sofia Feb 10 '15 at 21:09
  • @Sofia, in my judgement, my answer is simply a helpful hint. If it is your judgement that my answer constitutes an entire solution, be aware that I substitute no one's judgement for my own. – Alfred Centauri Feb 10 '15 at 23:00
  • Alfred, in one of my answers I said bye. Let's close this issue. I only called your attention on the law. And despite the fact that it seems to me that you solved his exercise, I don't ask you to submit to my judgement. Let's close this issue, you are busy, I am busy. Let's put an end. – Sofia Feb 11 '15 at 11:40
  • @Sofia, you started this nonsense and I've called you out on it. This seems to be a pattern of yours; making some absurd claim and then running away from the push back. That's not going to fly with me. – Alfred Centauri Feb 11 '15 at 13:21
  • how aggressive you are. *Please take a grip, calm down. Aggresivity won't go with me. I don't run away. What you expected, that I begin to insult you as you insult me? I still consider that you fully answered the question, but the issue was discussed enough, and I hope that you'll be more careful in the future. My comments aren't nonsense, you please don't play the psychologist with me - you don't have the possibility to pattern me, whatever you dislike is not absurd. And with this, the issue is closed*. – Sofia Feb 11 '15 at 14:01
  • @Sofia, any reasonable person reading these comments is likely to come to the conclusion that the one losing one's calm is you. Why do you diminish yourself like this? – Alfred Centauri Feb 11 '15 at 15:07
  • Everybody just calm down...1. This isn't even remotely a homework question, I was reading a book on my trip to Alberta and thought it would be fun to prove this, 2. The question is trivial, I'm missing a sign somewhere, 3. I needed to see steps because the result is well known, so this answer is helpful – Fraïssé Feb 13 '15 at 00:24