I've been trying to show that the generator of boosts can be written in operator form as can be seen here, as:

$$ B = \sum_i m_i x_i(t) - t \sum_i p_i $$

As a reminder the transformation rules are:

$$ x\to x+Vt\\ p\to p+mV $$

In order to accomplish this I used procedure shown here, and I seem to run into inconsistency issues, which I can't figure out.


In short, if we have a multi particle state described by:

$$ \left|\psi\right\rangle=\left|p_1,p_2,..,p_N\right\rangle $$ then a boost should act as: $$ T_V\left|p_1,p_2,..,p_N\right\rangle=\left|p_1+m_1V,p_2+m_2V,..,p_N+m_NV\right\rangle $$

In the position base we get:

$$ \begin{split} &T_V\left|x_1,x_2,..,x_N\right\rangle=\left({1\over 2\pi}\right)^{-N/2}\int dp_1dp_2.. dp_N\prod_{i=1}^N e^{-ip_ix_i}T_V\left|p_1,p_2,..,p_N\right\rangle\\ &=\left({1\over 2\pi}\right)^{-N/2}\int dp_1dp_2.. dp_N\prod_{i=1}^N e^{-ip_ix_i}\left|p_1+m_1V,p_2+m_2V,..,p_N+m_NV\right\rangle\\ &=\left({1\over 2\pi}\right)^{-N/2}\int dp_1'dp_2'.. dp_N'\prod_{i=1}^N e^{-i\left(p_i'-m_iV\right)x_i}\left|p_1',p_2',..,p_N'\right\rangle\\ &=e^{i\sum_{i=1}^Nm_iVx_i}\left({1\over 2\pi}\right)^{-N/2}\int dp_1'dp_2'.. dp_N'\prod_{i=1}^N e^{-ip_i'x_i}\left|p_1',p_2',..,p_N'\right\rangle\\ &=e^{iV\sum_{i=1}^Nm_ix_i}\left|x_1,x_2,..,x_N\right\rangle=e^{iMVX_{cm}}\left|x_1,x_2,..,x_N\right\rangle \end{split} $$

Similarly we can obtain:

$$ \begin{split} &T_V\left|p_1,p_2,..,p_N\right\rangle=\left({1\over 2\pi}\right)^{-N/2}\int dx_1dx_2.. dx_N\prod_{i=1}^N e^{ip_ix_i}T_V\left|x_1,x_2,..,x_N\right\rangle=\\ &=\left({1\over 2\pi}\right)^{-N/2}\int dx_1dx_2.. dx_N\prod_{i=1}^N e^{ip_ix_i}\left|x_1+Vt,x_2+Vt,..,x_N+Vt\right\rangle\\ &=\left({1\over 2\pi}\right)^{-N/2}\int dx_1'dx_2'.. dx_N'\prod_{i=1}^N e^{i\left(x_i'-Vt\right)p_i}\left|x_1',x_2',..,x_N'\right\rangle\\ &=e^{i\sum_{i=1}^N-Vtp_i}\left({1\over 2\pi}\right)^{-N/2}\int dx_1'dx_2'.. dx_N'\prod_{i=1}^N e^{ip_ix_i'}\left|x_1',x_2',..,x_N'\right\rangle\\ &=e^{-iV\sum_{i=1}^Np_i}\left|p_1,p_2,..,p_N\right\rangle=e^{-iVtP}\left|p_1,p_2,..,p_N\right\rangle \end{split} $$

This would Suggest that our proposed generator is indeed the generator of boosts.

my main issue

There are two ways to formulate my problem:

  1. My first step in the derivation assumed that the boost changes the momentum state, while the second step shows that the momentum basis is supposedly an eigenbasis of the boost. The same issue is true for the position.

  2. I assumed that the transformation changes the coordinates, and momenta. This is the basic concept of Galilean relativity, how come I'm getting that any of them is an eigenvector of the boost?


1 Answer 1

  1. The action of the boost on momentum states specifies the matrix elements of $T_V$ in momentum representation, it does not "change the momentum representation of the state". In other words, from $$ T_V\left|p_1,p_2,..,p_N\right\rangle=\left|p_1+m_1V,p_2+m_2V,..,p_N+m_NV\right\rangle $$ it just follows that $$ \langle p'_1,p'_2,..,p'_N | T_V |p_1,p_2,..,p_N\rangle \equiv \langle p'_1,p'_2,..,p'_N |p_1+m_1V,p_2+m_2V,..,p_N+m_NV\rangle \\ = \prod_j \delta(p'_j - p_j - m_jV) $$ or $$ T_V = \int{dp_1dp_2.. dp_N\;|p_1+m_1V,p_2+m_2V,..,p_N+m_NV\rangle\langle p_1,p_2,..,p_N|} $$

  2. According to eqs.(3) and (5) of your 2nd source, the action of $T_V$ in position representation is $$ T_V|x_1,x_2,..,x_N\rangle = e^{i\sum_j{m_jVx_j}}|x_1,x_2,..,x_N\rangle $$ not $$ T_V|x_1,x_2,..,x_N\rangle = |x_1+Vt,x_2+Vt,..,x_N+Vt\rangle $$ as appears in your 2nd calculation. With the correct action the latter becomes $$ \begin{split} &T_V\left|p_1,p_2,..,p_N\right\rangle=\left({1\over 2\pi}\right)^{-N/2}\int dx_1dx_2.. dx_N\prod_{i=1}^N e^{ip_ix_i}T_V\left|x_1,x_2,..,x_N\right\rangle=\\ &=\left({1\over 2\pi}\right)^{-N/2}\int dx_1dx_2.. dx_N\prod_{i=1}^N e^{ip_ix_i}e^{i\sum_j{m_jVx_j}}|x_1,x_2,..,x_N\rangle\\ &=\left({1\over 2\pi}\right)^{-N/2}\int dx_1dx_2.. dx_N\prod_{j=1}^N e^{i\left(p_j+m_jV\right)x_j}\left|x_1,x_2,..,x_N\right\rangle\\ &=|p_1+m_1V,p_2+m_2V,..,p_N+m_NV\rangle \end{split} $$ as expected.

Note added in answer to comment: The form you have for the Galilei transform is actually only half of it, the momentum shift. The complete form is a product of the momentum shift $e^{i\sum_j{m_j\hat x_jV}}$ and the coordinate shift: $$ T_v = e^{i\sum_j{m_jV^2t}}e^{-i\sum{\hat p_jVt}}e^{i\sum_j{m_jV\hat x_j}}\\ \equiv e^{i\sum_j{m_jV^2t}}e^{-i\hat P Vt}e^{iMV\hat X} $$ where $\hat X$ and $\hat P$ are the center of mass position operator and the total momentum operator, respectively.

Notice that the position shift $e^{-i\hat P Vt}$ does not act on momentum eigenstates, while the momentum shift $e^{i\sum_j{m_j\hat x_jV}}$ does not act on position eigenstates. This explains both your existing attempt and its difference from the desired result.

You can find a very nice derivation and discussion of the transform in Fonda & Ghirardi's "Symmetry Principles in Quantum Physics", Sec.2.5, pgs.83-89.

  • $\begingroup$ I see what you mean. However the matrix elements of $T_V$ are determined by the transformation rules. Since the transformation rule for the coordinates is: $x\to x+Vt$, this should work. $\endgroup$
    – Yair M
    Jan 30, 2016 at 16:25
  • $\begingroup$ You are right to expect that both should work. Added a note about this in the answer. Details in the suggested ref. Hope it helps. $\endgroup$
    – udrv
    Jan 30, 2016 at 23:45
  • $\begingroup$ TThanks, and the reference you brought is great too. $\endgroup$
    – Yair M
    Jan 31, 2016 at 6:10

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