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In the momentum representation, the position operator acts on the wavefunction as

$$\langle k|X_i|\psi\rangle = i\frac{\partial}{\partial k_i}\psi(k)\tag{1}$$

Now we want under rotations $U(R)$ the position operator to transform as

$$U(R)^{-1}\mathbf{X}U(R) = R\mathbf{X}$$

Therefore we should get,

$$\langle \mathbf{k}|U(R)^{-1}X_iU(R)|\psi\rangle = i\sum_{j}R_{ij}\frac{\partial}{\partial k_j}\psi(\mathbf{k})\tag{2}$$

However when trying to evaluate the LHS of 2) I get a different answer. Here is my working,

$$ \begin{align} \langle\mathbf{k}|U(R)^{-1} X_i U(R) |\psi\rangle &=\langle R\mathbf{k}|X_i U(R) |\psi\rangle \\ &=\int {\rm d}^3 k' \langle R\mathbf{k}|X_i|\mathbf{k'}\rangle\langle\mathbf{k'}|U(R)|\psi\rangle \\ \end{align} $$ Using the definition of $X_i$ acting on a wave function we get,

$$\langle\mathbf{k}|U(R)^{-1} X_i U(R) |\psi\rangle =\int {\rm d}^3 k' i \frac{\partial}{\partial(Rk)_i} \delta^3(R\mathbf{k} - \mathbf{k'}) \psi(R^{-1}\mathbf{k'})$$

Integrating by parts and evaluating the integral using the delta function we get,

$$\langle\mathbf{k}|U(R)^{-1} X_i U(R) |\psi\rangle = -i\frac{\partial}{\partial(R\mathbf{k})_i} \psi(\mathbf{k}) $$

Which is different from (2). Where am I going wrong?

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marked as duplicate by Qmechanic quantum-mechanics Nov 28 '17 at 15:42

This question was marked as an exact duplicate of an existing question.

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    $\begingroup$ What is the momentum-space representation of $U(R)$? $\endgroup$ – probably_someone Nov 27 '17 at 16:23
  • $\begingroup$ Thanks for your comment. Please see the edited question. I have added my working but I get the wrong answer. Where am I going wrong? $\endgroup$ – Prag1 Nov 28 '17 at 6:46
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    $\begingroup$ Possible duplicate of Evaluating the action of the position operator on momentum space representation (Incorrect Result) $\endgroup$ – Prahar Nov 28 '17 at 12:59
  • $\begingroup$ Hi Prag1. Please don't repost a closed question in a new entry. Instead, you are supposed to edit the original question within the original entry. $\endgroup$ – Qmechanic Nov 28 '17 at 15:43
  • $\begingroup$ Hi Qmechanic, This is the original question which I edited. I have deleted the duplicate question. Please allow this question to be answered. $\endgroup$ – Prag1 Nov 28 '17 at 15:52