I've read that, when expressing a quantum mechanical operator into a particular representation, the more-formally correct notation is $\rightarrow$ rather than $=$, e.g. the momentum operator expressed in position representation should be written as $$\mathbf{\hat{p}}\rightarrow-i\hbar\nabla\quad\text{rather than as}\quad\mathbf{\hat{p}}=-i\hbar\nabla$$

Even though most sources in literature (including my own lecturers) use the notation on the right, I've read that the one on the left is actually the "formally correct" one. This makes sense to me: the object on the LHS ($\mathbf{\hat{p}}$) is independent of representation, while the expression on the right necessarily depends on the representation chosen, so it does not really make sense to equate the two, since then we could combine $\mathbf{\hat{p}}=-i\hbar\nabla$ and $\mathbf{\hat{p}}=\mathbf{p}$ (the latter in momentum representation) to give $-i\hbar\nabla=\mathbf{p}$ which clearly does not make any sense.

This got me thinking -- isn't this "abuse" of notation also present in vectors? For example, if I write, say, $\mathbf{a}=\begin{pmatrix}a_1&a_2&a_3\end{pmatrix}^T$, isn't the object on the LHS independent of basis, while the object on the RHS the decomposition/projection of $\mathbf{a}$ onto the chosen basis? What I'm trying to say is, shouldn't I have written $\mathbf{a}\rightarrow \begin{pmatrix}a_1&a_2&a_3\end{pmatrix}^T$ instead? I guess a way to avoid this problem would be to write $\mathbf{a}=a_1\mathbf{e}_1+a_2\mathbf{e}_2+a_3\mathbf{e}_3$; I believe it makes sense to place an equality here as the RHS explicitly contains information regarding the basis, i.e. it would make sense to equate it to another representation of $\mathbf{a}$ in a different basis.

To avoid misunderstandings, I just want to know if my doubt makes sense/is justified, since it has really never been brought up in my education yet. And yes I know it is a really pedantic point and nothing really profound, but sometimes these things pop up into my head and they just bother me a little :)

  • $\begingroup$ I've never seen that before. Could you provide some references? $\endgroup$ – cmk May 20 at 20:48

Let's start by distinguishing the classical observable $\mathbf{p}$ from the operator $\mathbf{\hat{p}}$ on a Hilbert space in quantum mechanics. The possible empirical values of $p_i$ are the eigenvalues of $\hat{p}_i$. What happens when we pass from classical results to their quantum-mechanical ones? We go from $\frac{d}{dt}\mathbf{p}=-\nabla V$ to $\frac{d}{dt}\langle\psi|\mathbf{p}|\psi\rangle=-\langle\psi|\nabla\hat{V}|\psi\rangle$, and from $\frac{p^2}{2m}+V=H$ to $\frac{\hat{p}^2}{2m}\psi+\hat{V}\psi=\hat{H}\psi$ with $\hat{H}=i\hbar\frac{\partial}{\partial t}$. We can only replace $\mathbf{\hat{p}}$ afterwards with $-i\hbar\nabla_\mathbf{x}$ in either equation if $\psi$ is in $\mathbf{\hat{x}}$-space. In fact, in $\mathbf{\hat{p}}$-space $\mathbf{\hat{x}}=i\hbar\nabla_\mathbf{p}$.

$\mathbf{\hat{p}}$ is independent of representation, while the expression on the right necessarily depends on the representation chosen

They both transform as vectors.

shouldn't I have written $\mathbf{a}\rightarrow\begin{pmatrix}a_1&a_2&a_3\end{pmatrix}^T$ instead?

The frequent statement $\mathbf{\hat{p}}=-i\hbar\nabla_\mathbf{x}$ really means $\mathbf{\hat{p}}\psi=-i\hbar\nabla_\mathbf{x}\psi$ constrains $\psi$, rather than that the operators are literally equal. This is an important point to understand, which doesn't have an analogy in vectors with non-operator-valued entries such as $\mathbf{a}$. The motive for using $\rightarrow$ isn't the mere fact vector entries vary with basis.


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