Definition of Generating Function and Momentum

Question

At page 44 of Modern Quantum Mechanics by Sakurai it's stated that given the following formula for the translation operator ($T$): $$T(d\vec{x}')=1-i\vec{K} \cdot d\vec{x}'$$ (where for some unknown reason $1$ represents the identity matrix $\mathbb{I}$) we can identify $\vec{K}$ with the momentum operator $\vec{p}$. It's stated that we can operate this identification because in classical mechanics the momentum is the generator of translations. My question is: is there a straightforward way to define what a generating function (for a transformation) is? And then how can we easily show that the momentum is the generating function of translation in both the classical and quantum mechanical scenario?

EDIT: just to clarify: I know that we choose this particular formula for the translation operator $T$ because we can prove that this form has all the proprieties that we want and also because this formula is made in such a way that we can interprete $\vec{K}$ as the generator of translation. My problem regards the definition of generating function and the application of this definition to the context of momentum. Is there a way to define what a generating function is in an intuitive manner?

Answer 1

I'm going to interpret your question as asking more about an intuitive understanding of generators rather than the translation operator in particular, I hope it answers your question. I'll try to keep quantum mechanics out of it and rather speak of generating a transformation. Most of this can be explained (and has been, in the answer linked in probably_someone's comment) using the more formal mathematics of Lie Groups and Lie Algebras but I think there is some benefit to also doing it less rigorously.

Essentially the situation is the following: let's say you have a nice continuous transformation (of one parameter, say, $a$) that is well-behaved and -- most importantly -- reduces to the identity transformation when $a=0$. For example, translation would be such a transformation. You could "translate" a function $f(x)$ by some distance $a$, and you'd get $\mathcal{T}_a f(x) = f(x+a).$ Obviously, if we translated it by $a=0$, we're fundamentally "doing nothing", and the "do-nothing" transformation is the identity.

Generating Rotations in 2D

I'll first do a slightly more complicated example: that of rotations in 2D. You should be aware that you can rotate a "vector" $(x,y)^T$ using the rotation matrix:

$$\begin{pmatrix}x^\prime\\y^\prime\end{pmatrix} = \begin{pmatrix}\cos{\theta} & -\sin{\theta}\\ \sin{\theta}&\cos{\theta}\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix},$$

where $\theta$ is our parameter and $\theta = 0 \implies$ the matrix is just $\mathbb{I}$, the identity (or "do-nothing") matrix. Now what if we rotated by a very tiny angle $\epsilon\ll1$? We can just stick to the first non-zero terms in the expansion of sin and cos, and our "rotation matrix" would just become

$$\mathcal{R}(\epsilon) = \begin{pmatrix}\cos{\epsilon} & -\sin{\epsilon}\\ \sin{\epsilon}&\cos{\epsilon}\end{pmatrix} = \begin{pmatrix}1 & -\epsilon\\ \epsilon&1\end{pmatrix} + \mathcal{O}(\epsilon^2),$$

where we can ignore $\mathcal{O}(\epsilon^2)$ if $\epsilon$ is sufficiently small, and so

$$\mathcal{R}_\text{inf}(\epsilon) \approx \mathbb{I} + \epsilon \begin{pmatrix}0 & -1\\ 1&0\end{pmatrix} = \mathbb{I} + \epsilon M $$

Now the question is this: let's say you didn't know the rotation matrix, but all you knew was the matrix $M$. Could you use this "infinitesimal" transformation to "generate" a complete rotation of some angle $\theta$? Well, it turns out, yes! You could take the entire rotation angle $\theta$, and break it up into $n$ pieces of $\epsilon = \theta/n$. Now a rotation by $\theta$ would just be the same as $n$ rotations by $\epsilon$.

$$\mathcal{R}(\theta) = \left(\mathcal{R}\left(\frac{\theta}{n}\right) \right)^n$$

Then, you could take the limit $n \to \infty$, so that $\epsilon\to 0$ while $\theta = n\epsilon$ is finite. But then we could just replace $\mathcal{R}$ with $\mathcal{R}_\text{inf}$, and so

$$\mathcal{R}(\theta) = \lim_{n\to\infty} \left(\mathbb{I} + \frac{\theta}{n}\begin{pmatrix}0&-1\\1&0\end{pmatrix} \right)^n = \lim_{n\to\infty} \left(\mathbb{I} + \frac{\theta}{n}M\right)^n = e^{\theta M},$$

where in the last step we used definition of the exponential "function", except that it's a function of the matrix $M$.

It's a fun exercise to show that since $M^2 = -\mathbb{I}$, we have a formula very much like the famous Euler formula: $$e^{\theta M} = \mathbb{I}\times \cos{\theta} + M \times \sin{\theta} = \begin{pmatrix}\cos{\theta} & -\sin{\theta}\\ \sin{\theta}&\cos{\theta}\end{pmatrix}.$$

We have reconstructed (or "generated") the rotation matrix by just composing an infinite number of infinitesimal rotations!

Generating Translations

The translation operator is defined by $\mathcal{T}_a f(x) = f(x+a)$, and if $a \ll 1$,

$$\mathcal{T}_\text{inf} f(x) \approx f(x) + a \frac{\text{d}f}{\text{d}x} = \left( \mathbb{I} + a \frac{\text{d}}{\text{d}x} \right) f(x)$$

Just as before, we can see that a total translation of $a$ can be broken into $n$ successive translations of $a/n$, and so

$$\mathcal{T}_a = \lim_{n\to\infty}\left( \mathbb{I} + \frac{a}{n} \frac{\text{d}}{\text{d}x} \right)^n f(x) = e^{a\frac{\text{d}}{\text{d}x} }f(x).$$

Thus, the translation operator is $\mathcal{T}_a$, and it is generated by an operator proportional to $\frac{\text{d}}{\text{d}x}$, which from Quantum Mechanics is the momentum operator (up to some constants).

Answer 2

Generating function is not the correct word. The idea is simply Taylor theorem. For an analytic function $f$ we have

\begin{align} (T_a f) & := f(x+a) \\ & = \sum_{n=0}^\infty \frac{a^n}{n!} f^{(n)} (x) \\ &= \exp{ \left ( a \frac{d}{dx} \right )} f \end{align}

So we say that the derivative (or $i$ times $-i \frac{d}{dx} $) Is the generator of translation.