Momentum and infinitesimal translation

Question

My problem is all about this previous question. I'm trying to understand the reasoning behind the definition of the momentum operator in quantum mechanics. Sakurai tells me that for the infinitesimal translation of the previously cited question: $$X=x+dx$$ $$P=p$$ I have the following generating function for this transformation: $$F(x,P)=xP+pdx$$ Ok, lets verify that, I know that this is a type-2 generating function, so the following must hold: $$p=\frac{\partial F(x,P)}{\partial x}$$ $$X=\frac{\partial F(x,P)}{\partial P}$$ I also know from the cited previous question that i can write $dx$ as $\varepsilon f(x)$. Ok, so I get: $$p=\frac{\partial }{\partial x}[xP+p\varepsilon f(x)]=P+p\varepsilon f'(x)$$ $$X=\frac{\partial }{\partial P}[xP+p\varepsilon f(x)]=x$$ then: $$p=P+p\varepsilon f'(x) \ \ \Rightarrow \ \ P=p(1-\varepsilon f'(x))$$ and so i get the following transformation: $$X=x$$ $$P=p(1-\varepsilon f'(x))$$ This is not what I was expecting according to Sakurai (to be precise Sakurai's book titled: Modern Quantum Mechanics, at page 44). So Question one: Why I get this result?

But lets suppose that I don't have this problem and the calculation turns out fine, then I sill have another couple of problems: we all know that in QM the operator for infinitesimal translation is: $$T(dx)=1-iKdx$$ where 1 represents the identity matrix. Sakurai states that this strongly resembles the upper mentioned generating function $F$, so he states that we can speculate that the operator $K$ and the momentum $p$ are correlated in some way. But in one case, the QM case, the operator $K$ appears in the formula for the infinitesimal translation, however in the classical case $p$ appears in the generating function for the translation and not in the translation formula itself. Furthermore the resemblance is strong because Sakurai states that $xP$ is the generating function for the identity. This makes the resemblance even more convoluted to my eyes. So Question two: Why this reasoning about the correlation between $K$ and $p$ holds?

One last thing: of course knowing the formula for an infinitesimal translation we can find the formula for a finite translation (in QM): $$T(\Delta x)=\exp\left(-\frac{ip\Delta x}{\hbar}\right)$$ this is completely fine for me, however sometimes the argument is made that the fact that we can write the finite translation operator in this way is proof/definition that $p$ is the generator of the infinitesimal translation. Question three: Is this good reasoning?

I truly hope that I made myself clear. This problems are bugging me a lot. I know that part of my question has been partially covered in the other question I cited, but I hope that this still qualifies as a non duplicate question.

Answer 1

There are two issues here. The first is that if $dx$ is expressed as a function of $x$, then that means that the coordinate change corresponds to adding a position dependent shift to the position coordinate. Adding a constant shift is what is prescribed here, so $dx$ has no $x$ dependence. If it would make you more comfortable, you could write it as $X = x + a$ instead.

The second issue that you are running into is that during the first phase of the procedure, you write $$p = \frac{\partial F}{\partial x}$$ $$X = \frac{\partial F}{\partial P}$$ but these are, in general, implicit equations for the functions $p=p(X,P)$ and $x=x(X,P)$. In this case, $$p(X,P) = \frac{\partial}{\partial x}\big[x(X,P)\cdot P + p(X,P) dx\big] = P$$ $$X = \frac{\partial}{\partial P}\big[x(X,P)\cdot P + p(X,P) dx\big] = x(X,P) + \frac{\partial p}{\partial P}dx = x(X,P)+dx$$ where in the second line, we have used the fact that $p(X,P)=P$ from the first line. These relations can be (trivially) inverted to give $$P(x,p) = p$$ $$X(x,p) = x + dx$$

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Why does this reasoning about the correlation between $K$ and $p$ hold?

Sakurai claims that $$F =xP + p dx $$ $$T = 1 + (-iK) dx$$ look similar in the sense that $$F= (\text{identity generating function}) + p dx$$ $$T= (\text{identity operator}) + (-iK) dx$$

which suggests that $-iK$ might be a good choice for the momentum operator. This is meant to be merely suggestive. If you learn about the differential geometry of Hamiltonian mechanics, you find that the momentum is indeed the generator of spatial translations in a very precise sense. For now, consider it a plausibility argument.

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however sometimes the argument is made that the fact that we can write the finite translation operator in this way is proof/definition that 𝑝 is the generator of the infinitesimal translation.

That is more or less the definition of a generator. To be more precise, if you have some Lie group $G$ which represents a set of transformations, then it will possess a corresponding Lie algebra $\frak{g}$ whose elements roughly correspond to infinitesimal translations. $\frak{g}$ is in particular a vector space which can be equipped with a basis $\{A_i\}$, and the elements of that basis are called the generators of the group because an arbitrary$^\dagger$ group element $g$ can be expressed as $$g=\exp\left[-i\sum_i c_i A_i\right]$$ for some constants $c_i$.

In this case, the group $G$ is the group of translation operators. Since a finite translation can be expressed as $$g=\exp\left[-i \left(\frac{\Delta x}{\hbar}\right)p\right]$$

then that implies that $p$ is a generator of the group.

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$^\dagger$ At least, in a neighborhood of the identity element

Answer 2

First of all, I don't know why would yo do this with canonical transformations as it can be a bit confusing, and I think the notation may be making things even more difficult. A very basic derivation of what you're looking for is the following.

You are looking for translations, then you want an operator $T_{\vec{a}}$ which acts on a vector in the following way \begin{equation} \vec{r} \longrightarrow \vec{r}'= \vec{r}+\vec{a} \end{equation} for some vector $\vec{a}$. Note that the inverse operator of $T_{\vec{a}}$, say $T_{\vec{a}}^{-1}$ substracts the same quantity from $\vec{r}'$. For QM we prefer unitary operators, namely $\mathcal{U}_T$ so it maintains the magnitude of inner products.

Let's now think of the unitary operator of an infinitesimal translation, $\mathcal{U}_T(\delta \vec{a})$ (let's use the delta notation instead of $dx$). In this fashion, we're looking for something that acts the following way on a wavefunction $\psi(\vec{r})$ \begin{equation} \begin{cases} \mathcal{U}_T(\delta \vec{a})\psi(\vec{r})=\psi'(\vec{r}) \\ \psi'(\vec{r})=\psi(\vec{r}-\delta\vec{a}) \end{cases} \Rightarrow \mathcal{U}_T(\delta \vec{a})\psi(\vec{r})=\psi(\vec{r}-\delta\vec{a})=\psi(T_{\delta\vec{a}}^{-1}\vec{r}) \end{equation}

This may be a bit confusing. I have to point out we took what is called an $\mathbf{active\ transformation}$, where the system is moved and the reference frame stays the same. Another way to think of it is that the transformed wavefunction in the transformed point is equal to the original wavefunction in the original point: \begin{equation} \psi'(\vec{r}+\vec{a})=\psi(\vec{r}) \end{equation}

If we now expand the following term (as $\delta\vec{a}$ is an infinitesimal quantity) \begin{equation} \psi(\vec{r}-\delta\vec{a})=\psi(\vec{r})-\delta\vec{a}\cdot \nabla\psi \end{equation} we can find a relation between the general expression of an infinitesimal translation operator, which for some (vectorial) generator $\vec{q}$ can be expressed as \begin{equation} \mathcal{U}_T(\delta \vec{a})=\mathbb{I}-i\delta\vec{a}\cdot \vec{q} \end{equation}

To find this relation we have to remember that \begin{equation} \mathcal{U}_T(\delta \vec{a})\psi(\vec{r})=\psi(\vec{r}-\delta\vec{a}) \end{equation}

Combining these last three relations you will be able to find what is the relation between the generator $\vec{q}$ and the momentum $\vec{p}$. If you write everything in components it'll be easier to identify the momentum operator around.

To answer your first question, as I told you, $p$ can't appear in the expression of $F(x,P)$, mainly because as you did, you will derivate wrongly. So, as $p\equiv P$, you have $$F(x,P)=xP+p\delta x=xP+P\delta x=P(x+\delta x)$$

With this expression and the derivative constraints you stated above (which are correct), you get $X=x+\delta x$ and the problem arises with the calculation of $p$.