Generator of translations involving TIME

1. Jan 31, 2014

AxiomOfChoice

I'm in one spatial dimension. I want to know what the operator is that will turn $\psi(x)$ into $\psi(x + \frac{vt}{2})$, where $t$ is, yes, time. I know that, usually, we say that (assuming $\hbar = 1$) $e^{ipa}$ is the generator of translations by $a$, but I was wondering if things were different once we put in a $t$.

2. Jan 31, 2014

The_Duck

This transformation is called a boost (though I don't know why you have a factor of 1/2 in from of the v). The position operator $x$ is the generator of boosts. For example, a particle with some definite momentum $p_0$ has wave function $\psi(x) = e^{i p_0 x}$. If we act on this wave function with the operator $e^{i \Delta p x}$, where $\Delta p$ is some number, then we get the wave function $e^{i (p_0 + \Delta p) x}$. The particle's momentum has been shifted by $\Delta p$: it has been boosted.

To nitpick the terminology, the "generator" is the operator that appears in the exponential, not the exponential itself. So we say that the generator of spatial translations is the momentum operator $p$ because we can translate a wave function by a displacement $a$ by acting on it with the operator $e^{i p a}$.

3. Jan 31, 2014

ChrisVer

Is boost time-dependent?

I would try myself to use a Taylor expansion, and see what's going on... Prolly would find a problem with
$\frac{dψ}{dx}$ which I would have to write as $\frac{∂ψ}{∂x}+ \frac{∂t}{∂x}\frac{∂ψ}{∂t}= \frac{∂ψ}{∂x}+ \frac{2}{u} \frac{∂ψ}{∂t}$ (since you gave the velocity as $u/2$

Last edited: Jan 31, 2014
4. Jan 31, 2014

AxiomOfChoice

You're right, and I'm usually aware of that. I just misspoke. Thanks!

5. Feb 1, 2014

vanhees71

This is a bit more subtle. In fact the generator of the Galileo boosts is
$$\hat{\vec{K}}=m \hat{\vec{x}}-t \hat{\vec{p}}.$$
Since this operator is explicitly (!) time dependent, it does not commute with the Hamiltonian. Of course, it's conserved, i.e., the operator representing its time derivative,
$$\frac{1}{\mathrm{i}\hbar} [\hat{\vec{K}},\hat{H}]+\partial_t \vec{K} = \frac{1}{\mathrm{i}\hbar} [\hat{\vec{K}},\hat{H}] + \hat{\vec{p}}.$$
A careful analysis of the transformation properties of the wave function in the position representations yields its behavior under boosts,
$$\Psi_{\sigma}(t,\vec{x}) \rightarrow \Psi_{\sigma}'(t,\vec{x})=\exp \left (-\frac{\mathrm{i}}{\hbar} m \vec{v} \cdot \vec{x} - \frac{\mathrm{i} m}{2\hbar} \vec{v}^2 t \right ) \Psi_{\sigma}(t,\vec{x}+\vec{v} t).$$
Here, I've also included the possibility that the particle has non-zero spin $s$. Thus the wave function is spinor valued with the spinor index taking the values $\sigma \in \{-s,-s+1,\ldots,s-1,s \}$.

Note that the Galilei boost involves a phase factor in addition to the naive transformation rule for the wave function you have assumed in the beginning (despite this strange factor of 1/2, which I don't understand at all).

A very detailed treatment of Galilei transformations in quantum mechanics (involving a lot of subtle group-reprentation theory, including the fact that the mass is to be interpreted as a central charge of the quantum version of the Galilei group) can be found in

Ballentine, Quantum Mechanics, a Modern Development.

6. Feb 1, 2014

AxiomOfChoice

The $1/2$ is present in the particular application I'm considering. For our purposes, I could have just as easily left it off. Sorry for the confusion it seems to have created!

7. Feb 1, 2014

AxiomOfChoice

So, my understanding from this is that, if I am in a frame of reference $\mathcal O'$ that is moving with respect to my initial frame $\mathcal O$ with a velocity of $-vt$, and my particle has mass $m$ and wave function in $\mathcal O$ given by $\psi(x)$ then the wave function in $\mathcal O'$ isn't just $\psi(x+vt,t)$; it's

$e^{-imv^2t/(2\hbar)}e^{-imvx/\hbar}\psi(x+vt,t).$

Is that correct? Are there any additional assumptions I need to place on $\psi$ for this to work?

8. Feb 1, 2014

vanhees71

No. You can even prove that this is the correct rule for the Galilei boost by showing that, if $\Psi$ fulfills the Schrödinger equation so does $\Psi'$, and becomes clear, why you necessarily need the phase factor.

There's a famous paper by Inönü and Wigner that the unitary representations of the orginal Galilei group do not lead to physically reasonable dynamics (or a physically interpretable wave equation):

I. Inönü, E. P. Wigner, Representations of the Galilei group, Nuovo. Cim. 9, 705 (1952)
http://rd.springer.com/article/10.1007/BF02782239