Galilean transformation in non-relativistic quantum mechanics

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?

• 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/… – udrv Nov 11 '16 at 19:02
• @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! – Alex V. Nov 14 '16 at 11:10
• Welcome and good luck. – udrv Nov 14 '16 at 23:25

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 $\hbar=1$), $$i\partial_t\Big(e^{iK_j(t)v_j}\Psi_t\Big) = H\Big(e^{iK_j(t)v_j}\Psi_t\Big)$$ From that we can derive, $$i\partial_t K_j(t)=[H,K_j(t)],$$ 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. $$e^{iK_j(t)v_j}x_k e^{-iK_j(t)v_j}=x_k+v_j t\delta_{jk},\quad e^{iK_j(t)v_j}p_k e^{-iK_j(t)v_j}=p_k+m v_j \delta_{jk}$$ whereas simple shift of coordinates changes coordinates but not momentum because it's generated by momentum operator that commutes with itself, $$e^{ip_jv_j t}x_ke^{-ip_jv_j t}=x_k+v_j t\delta_{jk},\quad e^{ip_jv_j t}p_ke^{-ip_jv_j t}=p_k$$
For non-relativistic particle Galilean boost generator can be written in the form, $$K_j(t)=tp_j+mx_j$$ 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. $$e^{iK_j(t)v_j}\psi_t(x)=e^{imx_jv_j+tv_j\partial_j}\psi_t(x)$$ Use Baker-Campbell-Hausdorff formula to rewrite it in the form, $$e^{i\frac{mv_j^2}{2}t}e^{imx_jv_j}e^{tv_j\partial_j}\psi_t(x)=e^{i\frac{mv_j^2}{2}t}e^{imx_jv_j}\psi_t(x+vt)$$ For $t=0$ we reproduce your result $$\psi_0(x)\mapsto e^{imx_jv_j}\psi_0(x)$$