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I'm reading Weinberg's Lectures on Quantum Mechanics and in chapter 3 he discusses invariance under Galilean transformations in the general context of non-relativistic quantum mechanics. Being a symmetry of nature (if we forget about relativity), Galilean boosts (particular case of Galilean transformations) should be represented by a linear unitary operator $U(\mathbf{v})$ which is taken to satisfy:

$$ U^{-1}(\mathbf{v}) \mathbf{X}_H (t) U(\mathbf{v}) = \mathbf{X}_H (t) + \mathbf{v} t $$

where $\mathbf{X}_H (t)$ is the Heisenberg picture operator for the position of any given particle. This is basically what we understand by a Galilean boost. Now, I have a few questions:

  1. Taking the time derivative of the previous expression and using $\dot{\mathbf{X}}_H(t) = i \left[H,\mathbf{X}_H(t)\right]/ \hbar$, we can conclude that, at $t=0$ (in this case Schrödinger and Heisenberg picture operators coincide): $$ i\left[U^{-1}(\mathbf{v}) H U(\mathbf{v}), \mathbf{X}\right] = i \left[H, \mathbf{X}\right] + \hbar \mathbf{v} $$ From here Weinberg says that necessarily $U^{-1}(\mathbf{v}) H U(\mathbf{v}) = H + \mathbf{P}\cdot\mathbf{v}$. I can prove that if this is the case (where $\mathbf{P}$ is the total momentum operator defined as the generator of spatial translations), the previous commutation relation holds, but couldn't we have a different form of the transformed Hamiltonian still satisfying the previous commutation relation?

  2. In any case, it is clear that although being a symmetry of nature the transformation $U(\mathbf{v})$ doesn't commute with the Hamiltonian. This is also the case for its generator $\mathbf{K}$, where $U(\mathbf{v}) = \exp(-i \mathbf{v} \cdot \mathbf{K})$. This is an exception to the general rule that the generators of symmetries commute with $H$, and Weinberg argues that the reason is that $\mathbf{K}$ is associated with a symmetry which depends explicitly on time, as the effect on $\mathbf{X}_H (t)$ shows. The question is then, shouldn't $U(\mathbf{v})$ be somehow time dependent, so that we can't simply take the time derivative in the first expression I wrote by considering that it only acts on $\mathbf{X}_H (t)$? This would mean that the whole derivation in 1 is fallacious...

  3. And now the question that is bothering me the most. Following the argument presented in the previous two points, I want to understand what $U(\mathbf{v})$ really does (this should solve question 2, showing that it has no time dependence). For simplicity, I will consider a one-particle system. My physical intuition tells me that, if $\Phi_{\mathbf{x},t}$ is an eigenstate of $\mathbf{X}_H(t)$ with eigenvalue $\mathbf{x}$, then we should have: $$ U(\mathbf{v}) \Phi_{\mathbf{x},t} = \Phi_{\mathbf{x}+\mathbf{v}t,t} $$ But this leads me to some contradictions. First of all, at $t=0$ this equation means that $ U(\mathbf{v}) \Phi_{\mathbf{x}} = \Phi_{\mathbf{x}} $, and since the $\{\Phi_{\mathbf{x}}\}$ form a complete set of orthonormal states, the operator $U(\mathbf{v})$ should be the identity (which is a disaster, since obviously this operator must act non-trivially on, e.g., $\mathbf{X}_H (t)$). We could argue that there might be phases (depending on $\mathbf{x}$) in the previous equation, so I will show another problem I have found. Let $\{\Psi_{\mathbf{p},t}\}$ be a complete orthonormal set of momentum eigenstates at time $t$, so that we have the usual inner product: $$ (\Psi_{\mathbf{p},t},\Phi_{\mathbf{x},t}) = (2\pi \hbar)^{-3/2} \exp(-i \mathbf{p}\cdot\mathbf{x}/\hbar) $$ This equation is certainly true at $t=0$ and I assume it is also valid at time $t$ because it follows from properties of translations at a fixed time. Again using what a Galilean boost should be, I assume $U(\mathbf{v}) \Psi_{\mathbf{p},t} = \Psi_{\mathbf{p} + m \mathbf{v},t}$. Then, going to momentum space, we conclude: $$ U(\mathbf{v}) \Phi_{\mathbf{x},t} = (2\pi \hbar)^{-3/2} \int d^3 \mathbf{p} \exp(-i \mathbf{p}\cdot\mathbf{x}/\hbar) \Psi_{\mathbf{p} + m \mathbf{v},t} = \exp(i m \mathbf{v}\cdot\mathbf{x}/\hbar) \Phi_{\mathbf{x},t} $$ which contradicts our original idea $U(\mathbf{v}) \Phi_{\mathbf{x},t} = \Phi_{\mathbf{x}+\mathbf{v}t,t}$.

So I am really lost here... Any help, especially with question 3?

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    $\begingroup$ Before anything else, you may want to see if your questions are answered by the very nice approach to Galilei transformations in Fonda & Ghirardi's "Symmetry Principles in Quantum Physics", Sec. 2.5, pgs.83-89: scribd.com/doc/30539019/… $\endgroup$ – udrv Nov 11 '16 at 19:02
  • $\begingroup$ @udrv Very helpful reference! I found there the full answer to my question, and I will definitely keep an eye on it for further clarification on symmetries in quantum mechanics since it seems to be a pretty complete book! $\endgroup$ – Alex V. Nov 14 '16 at 11:10
  • $\begingroup$ Welcome and good luck. $\endgroup$ – udrv Nov 14 '16 at 23:25
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First, about about symmetries of the theory. Let's work in the Schroedinger picture where the only dependence on time of operators is explicit. That $K_j(t)$ generates symmetry of Hamiltonian $H$ means that transformed wavefunction satisfies the same Schroedinger equation (I'll take $\hbar=1$), \begin{equation} i\partial_t\Big(e^{iK_j(t)v_j}\Psi_t\Big) = H\Big(e^{iK_j(t)v_j}\Psi_t\Big) \end{equation} From that we can derive, \begin{equation} i\partial_t K_j(t)=[H,K_j(t)], \end{equation} that generalizes the usual condition of vanishing commutator for time-independent operators. It is equivalent to vanishing of $\frac{d}{dt}K_j(t)$ in the Heisenberg picture.

Now the answer for your third question is that your physical intuition is not correct. The reason is that the Galilean boost changes both coordinate AND momentum. \begin{equation} e^{iK_j(t)v_j}x_k e^{-iK_j(t)v_j}=x_k+v_j t\delta_{jk},\quad e^{iK_j(t)v_j}p_k e^{-iK_j(t)v_j}=p_k+m v_j \delta_{jk} \end{equation} whereas simple shift of coordinates changes coordinates but not momentum because it's generated by momentum operator that commutes with itself, \begin{equation} e^{ip_jv_j t}x_ke^{-ip_jv_j t}=x_k+v_j t\delta_{jk},\quad e^{ip_jv_j t}p_ke^{-ip_jv_j t}=p_k \end{equation}

For non-relativistic particle Galilean boost generator can be written in the form, \begin{equation} K_j(t)=tp_j+mx_j \end{equation} which can be obtained as limit $c\rightarrow\infty$ of the Lorentzian boost. It can be easily checked that it satisfies the symmetry generator condition for Hamiltonian $H=\frac{\vec{p}^2}{2m}$. You can also check that it transforms $x$ and $p$ correctly.

Now how it transforms the wavefunction in the coordinate representation. \begin{equation} e^{iK_j(t)v_j}\psi_t(x)=e^{imx_jv_j+tv_j\partial_j}\psi_t(x) \end{equation} Use Baker-Campbell-Hausdorff formula to rewrite it in the form, \begin{equation} e^{i\frac{mv_j^2}{2}t}e^{imx_jv_j}e^{tv_j\partial_j}\psi_t(x)=e^{i\frac{mv_j^2}{2}t}e^{imx_jv_j}\psi_t(x+vt) \end{equation} For $t=0$ we reproduce your result \begin{equation} \psi_0(x)\mapsto e^{imx_jv_j}\psi_0(x) \end{equation}

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