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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!

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marked as duplicate by Danu, ACuriousMind, Qmechanic Mar 21 '15 at 14:36

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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

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

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  • $\begingroup$ isn't that true that we don't have to solve home problems? $\endgroup$ – Sofia Feb 9 '15 at 16:22
  • $\begingroup$ @Sofia, isn't it true that this isn't (remotely) a solution to the problem but just a well known result from Fourier analysis? $\endgroup$ – Alfred Centauri Feb 9 '15 at 21:59
  • $\begingroup$ @Sofia, please get a grip. idioms.thefreedictionary.com/get+a+grip $\endgroup$ – Alfred Centauri Feb 10 '15 at 0:29
  • $\begingroup$ @Sofia, I don't have the slightest idea what you're talking about. $\endgroup$ – Alfred Centauri Feb 10 '15 at 20:59
  • $\begingroup$ the law says that home exercises should not be solved in entirety, just be given guide lines. This law should be respected by everybody. $\endgroup$ – Sofia Feb 10 '15 at 21:09

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