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In QM and QFT, I have seen some equations where they have just the derivative and/or the gradient without specifying what it is acting on.

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Taken from wiki.

This does not make sense to me since I learned that gradients and derivatives can only act on functions and not exist by themselves (unless they are vector components). This seems to be going against the grain of what I learned.

Can someone please help me understand how can we just put a derivative without specifying what it is acting on?

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    $\begingroup$ Yes, these are differential operators, they act on functions $\endgroup$
    – Kosm
    Sep 4 at 21:44
  • $\begingroup$ @Kosm I perfectly know that they act on functions. Please refer to my comment under doublefelix's answer. $\endgroup$
    – Tachyon
    Sep 4 at 21:57
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People call it an operator. It is a thing which would map a function to another function. It's a lot like how a matrix maps a vector to another vector. You can invent lots of operators, like

$$2$$

or

$$2 + 2x$$

or

$$2 + 2x +\frac{d}{dx}$$

which, when acting on a function $f(x)$, would give

$$2 f(x) + 2x f(x) + \frac{df}{dx}$$

by definition.

There are some things to get used to, like how does multiplying operators work? For example $(\frac{d}{dx})(\frac{d}{dx}) = \frac{d^2}{dx^2}$. How did I figure that out? Just put a function at the end and then simplify, then remove the function from the end; remember that the operator closest to the function acts first. Also, some valid operations like "squaring the function" can't be written in this notation, but it rarely comes up anyway as most operators of interest only depend linearly on the function.

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    $\begingroup$ @Tachyon Think of rotation matrix, it will act on a vector, but you can write it as a matrix by itself. $\endgroup$ Sep 4 at 22:03
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    $\begingroup$ These quantum mechanical operators are meant to act on a function with the same variable as the operator (fancy people might say "in the same basis"). So to see how it works, you have to act it on a function of $x$. In the context of quantum mechanics, that function of $x$ is usually a wave function $\psi(x)$. $\endgroup$ Sep 4 at 22:14
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    $\begingroup$ But yes, if it were to act on a function of another variable $f(w)$, then only the $V(x)$ term would survive and the other term would give 0. That doesn't mean that the operator is zero, it just means that Operator($f(w)$) = 0. Similarly there can be a matrix $M$ which acts on a vector $v_0$ to give zero, even though $M \neq 0$. $\endgroup$ Sep 4 at 22:16
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    $\begingroup$ But you have a point - TECHNICALLY, if you want to get very precise, to specify an operator it is necessary to specify the space of functions which it acts on. This is something which mathematicians do and physicists almost never do. In the context of Quantum Mechanics, operators are understood to act on wave functions, which are functions of position and time. $\endgroup$ Sep 4 at 22:18
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    $\begingroup$ There are occasional circumstances where issues arise if you do not specify the domain of an operator, e.g. proving that the momentum operator is self-adjoint. However, specifying domains of operators and working at that level of precision takes a lot more time and energy. Most physicists simply don't know much functional analysis and avoid or ad-hoc the occasional issue which comes up where it might be useful. $\endgroup$ Sep 4 at 22:31
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I am ignoring $\frac{\partial}{\partial t}$, since it does not involve the issue that confuses you.

You are right to be puzzled by the abuse of notation, $\hat p \sim -i\hbar \frac{d}{dx}$, but your teacher should have made this very clear right from the start. What it means is that this is a representation in the coordinate picture, which is to say $$ \hat p = -i\hbar\int dx ~~ |x\rangle \frac{d}{dx}\langle x|. $$ Acting on any state in Hilbert space, it yields, e.g., $$ \hat p |\psi\rangle= -i\hbar\int dx~~ |x\rangle \frac{d}{dx} \psi(x). $$

(Sometimes, informally, people summarize the above in code, abusively, as $ \hat p_x \psi(x)= -i\hbar \frac{d}{dx} \psi(x)$, which apparently confused you.)

From the definition above, you may also easily derive $$ \hat p = \int d p ~~ |p\rangle p \langle p|. $$ This is a reminder of the logical power of the Dirac notation, and underscores the conceptual service he's offered the theory.

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    $\begingroup$ Thank you, this makes total sense to me. In essence, it is just a shorthand way to say that the operator acts on the wavefunction without specifying the wavefunction. At the end of the way, every operator acts on something, even if it is not written. $\endgroup$
    – Tachyon
    Sep 4 at 22:23
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If you are familiar with linear algebra then you can think of matrices as operators acting on vectors to produce some other vector (or in the following example the same vector but with a scalar) $$A\vec v=a\vec v$$

Here $A$ is a matrix (or an operator) and $a$ is a constant and $\vec v$ is a vector. How is this related to your question? In quantum physics it is very common to have operators acting on wave functions. These operators usually contain derivative(s) which by themselves does nothing but when you apply them to a wave function (in the previous case a vector) they can produce something new. For example the momentum operator $\vec p=-i\hbar\nabla$ which in the above example would be "$A$" acts on some wave function $\vec v$ then the momentum would be the eigenvalue $a$. $\textbf{One can talk about $\vec p$ without it needing to act on a wave function.}$

After rephrased question:

Can someone please help me understand how can we just put a derivative without specifying what it is acting on?

Let me define $y$ as following: $$y=\frac{d}{dx}$$ How does one interpret this? One can say that this is an operator that does nothing at the moment. But we could talk about what happens if you put $y$ infront of $x^2$ then you would say it would be equal to $2x$. There is nothing special about having an equation with an operator not acting on something but it could be very interesting to talk about what happens when we apply it.

$\textbf{If you see operators as machines then obviously one can talk about different drilling}$ $\textbf{machines without needing to drill holes as we are talking about them.}$

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  • $\begingroup$ So it is a matter of notation and not computation when writing operators like that, like "One can talk about p without it needing to act on a wave function.". At the end of the day, you would need to put in a wavefunction in order to solve? $\endgroup$
    – Tachyon
    Sep 4 at 22:05
  • $\begingroup$ I would not say it is notation or computation but rather it is simply talking about operators. If you see operators as machines then obviously one can talk about different drilling machines without needing to drill holes as we are talking about them. I'm not sure what you mean by your last sentence but at the end of the day then one need to put a operator to act on a wave function for it to do something. $\endgroup$
    – ludz
    Sep 4 at 22:08
  • $\begingroup$ I just mean that you need the wavefunction in order to solve anything about the system. You need to put one in order to do that. That's all. $\endgroup$
    – Tachyon
    Sep 4 at 22:26
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They (and $\hat{H}$ in the equation shown) are differential operators: something that when applied to a function produces another function.

Differential operators are in some sense another "data type" than functions and scalars, although one can of course see them as higher-order functions that map functions to other functions.

The notation is often a bit of abuse of notation (just consider how the del symbol is used in several ways to mean different but related operators: $\nabla f$, $\nabla\cdot \mathbf{v}$, $\nabla \times \mathbf{v}$), but the meaning should usually be clear enough from context.

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