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Here is the definition of the Noether momentum in my script.

$$I = \left.\frac{\partial L}{\partial \dot{x}} \frac{d x}{d \alpha} \right|_{\alpha=0} = \frac{\partial L}{\partial \dot{x}} = m \dot{x} = p_x.$$ But I don't understand exactly what this vertical line with $\alpha=0$ means, I would have interpreted it as $\frac{dx}{d0}$ now.

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  • $\begingroup$ Well it is not really different from the notation $f^\prime\vert_{x=0}$, for a function $f:\mathbb R\longrightarrow \mathbb R$ and means that you perform the derivative of $f$ and evaluate it at $0$. $\endgroup$
    – Tobias Fünke
    Commented Dec 5, 2022 at 14:39
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    $\begingroup$ @TobiasFünke Please don't put answers in comments. $\endgroup$
    – Massimo Ortolano
    Commented Dec 5, 2022 at 14:40
  • $\begingroup$ @MassimoOrtolano You're right. But I did not want to expand this further and hoped someone else will do so in an answer, which actually just happened. $\endgroup$
    – Tobias Fünke
    Commented Dec 5, 2022 at 14:42

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It means "compute the previous expression and, after that, set $\alpha = 0$". So you will compute the derivative with respect to $\alpha$, and only then plug $\alpha = 0$.

Let $f = f(x)$. These two mean the same thing: $$\left.\frac{df}{dx}\right|_{x=0} = f'(0).$$ Sometimes the bar notation just happens to be convenient.

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