# Translation operator to higher order

IN QM, the space translation operator, or generator of Translations is set to be Ie-ie/h_bar*P up to order e.

Now my question is what is the physical justification of only going up to order e and do the higher order terms have any meaning?

Is it correct to say(if we are only including up to order e) that we are doing a constant linear translation(equal distances in equal times) and hence assuming there is no change in how fast we do it?

• – Qmechanic Sep 24 '13 at 18:05

The full translation operator corresponding to a translation by a vector $\mathbf a$ is given by the operator exponential \begin{align} T_{\mathbf a} = e^{-i\mathbf a\cdot\mathbf P/\hbar} \end{align} If the translation that you want to perform is small, namely if $\mathbf a = \epsilon\mathbf n$ for $\epsilon$ small and $\mathbf n$ a unit vector, then one can approximate the translation operator by Taylor expanding the operator exponential to first order in $\epsilon$: \begin{align} T_{\epsilon \mathbf n} = I - \epsilon\left(\frac{i\mathbf n\cdot \mathbf P}{\hbar}\right) +O(\epsilon^2) \end{align} The physical significance/intuition of the first order approximation is that it agrees with the full translation to good approximation provided $\epsilon$ is small.

Here is a motivation why the object you mention is called 'generator'.

Consider the exponential function for numbers (the algebra for operators is similar but when taking the limits one has to be more careful), for example:

$$e^{-i a p / \hbar} = \lim_{n \rightarrow \infty} \left(1 + \frac{-i a p / \hbar}{n}\right)^n = \lim_{n \rightarrow \infty} \left[\left(1 + \frac{-i a p / \hbar}{n}\right) \cdot \ldots \cdot \left(1 + \frac{-i a p / \hbar}{n}\right)\right]$$

One can see this as the term $1+\dfrac{-i a p/\hbar}{n}$ or $1 + \varepsilon \dfrac{-i p}{\hbar}$ (for $\varepsilon = a/n$) is 'generating' the exponential because 'applying' this term 'a large number of times' (in the limit where it becomes infinitesimally small) gives you the original function.

Physically, I would think of the 'operator' $1 + \varepsilon \dfrac{-i p}{\hbar}$ as being an infinitesimally small displacement of size $\varepsilon$ which when successively applied a large ($n \rightarrow \infty$) number of times, 'add up' to the overall translation $a$.

With $P_i = -i \hbar \frac{\partial}{\partial x^i}$, the translation operator is $e^{\large \frac{-i}{\hbar} P.a}$, where $a$ indicates the translation ($e^{\large \frac{-i}{\hbar} P.a}|x\rangle = |x+a\rangle$ and $e^{\large \frac{-i}{\hbar} P.a} \psi(x) = \psi(x-a))$ If you take an infinitesimal $a$, you have $e^{\large \frac{-i}{\hbar} P.a}= \mathbb{Id} - \frac{i}{\hbar}P.a$. But this represents only an infinitesimal translation. Here infinitesimal has the same meaning as considering an infinitesimal interval $dx$. You have $(\mathbb{Id} - \frac{i}{\hbar}P.a) \psi(x) = \psi(x) - a. \frac{\partial \psi}{\partial x}(x)$, which are the first terms of the development of $\psi(x-a)$ if $a$ is infinitesimal.