In quantum mechanics if two quantities $A$ and $B$ are said to be coupled what does this actually mean?
I would guess that it means we have a term like $A\cdot B$ in the Hamiltonian but this is only a guess.
[Physics] The meaning of ‘coupling’
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Related Solutions
This problem appears interesting for the following reason. Let us write it down in Cartesian coordnates:
$$-\frac{1}{2}\left(\frac{\partial^2\psi}{\partial x^2}+\frac{\partial^2\psi}{\partial y^2}+\frac{\partial^2\psi}{\partial z^2}\right)+\frac{1}{2}(x\cdot S)^2\psi+L\cdot S\psi=E\psi$$
where I have introduced a 1/2 factor for later convenience. Now, I concentrate on x and I consider the operator
$$-\frac{1}{2}\frac{\partial^2}{\partial x^2}+\frac{1}{2}(x\cdot S)^2$$
One can introduce creation and annihilation operators in a similar way as for the harmonic oscillator
$$A_S=\frac{1}{\sqrt{2}}\left(\frac{\partial}{\partial x}+xS\right)$$
and the corresponding eigenvectors will be labeled as $|n,S\rangle$. The next step is to write down $L\cdot S=\frac{1}{2}(J^2-L^2-S^2)$ and we can restate this problem in the form
$$\left(A_S^\dagger A_S+\frac{1}{2}\right)\psi-\frac{1}{2}\left(\frac{\partial^2\psi}{\partial y^2}+\frac{\partial^2\psi}{\partial z^2}\right)+\frac{1}{2}(J^2-L^2-S^2)\psi=E\psi$$
Disclaimer at the bottom. I'm assuming we're working with the infinite (continuous position) case. A few things that may help you:
- In quantum mechanics, $\langle \hat{A} \rangle$ is usually defined to be $\langle \psi | \hat{A} | \psi \rangle=\langle \psi | \hat{A} \psi \rangle$. If you're not given $\psi$ you can't find $\langle \hat{A} \rangle$, which would explain your difficulty in finding a definition. It isn't referred to as the average, usually, it is referred to as the expectation value of the operator over psi.
- With regards to the position operator, the wikipedia page states that it is motivated by the fact that the expected x-position should be: $\langle \hat{x} \rangle=\int x |\psi(x)|^2 dx=\int \psi^* x \psi dx=\int \psi^* \hat{x} \psi dx=\langle \psi |\hat{x} |\psi\rangle$, where $\hat{x}$ is the operator defined in the position representation to be $\hat{x} \psi(x)=x \psi(x)$.
- I think there are other motivations for the position operator. One feature it has is that if $| X \rangle$ is an eigenvector of $\hat{x}$, then the inner product $\langle X | \psi \rangle$ gives you the value of $|\psi\rangle$ at position $|X\rangle$. $|X\rangle$ in the position representation in this case is a delta function, $X(x)=\delta(x-x_0)$, so that $\langle X|\psi \rangle=\psi(x_0)$, which is perfect because it picks out a sharp value of $\psi$ as desired.
The conceptual weirdness and the fact that I'm being so inspecific (in terms of $| \psi \rangle$ instead of $\psi(x)$) is because you're working in a vector space (of all $\psi$) and you can change basis at will, and so the position representation truly isn't fundamental (mathematically, at least - I'm not so sure about physically).
So, short version:
$\langle \hat{A} \rangle=\langle \psi | \hat{A} | \psi \rangle = \int \psi^*(x) \hat{A}\psi(x)dx $ (the last only holds in the position representation)
$\hat{x}$ is an operator (matrix in the finite dimensional case), its eigenvectors $|X\rangle$ are vectors, and you can use the inner product $\langle X | \psi \rangle$ to pick out the value of the wavefunction at position $|X\rangle$.
Disclaimer: I'm still on very, very, very shaky ground on QM, so please correct me if I'm wrong.
Best Answer
Let me preface by saying that "coupling" is a favorite physicist word that is perhaps best described linguistically than rigorously; it's deployed in a few different situations.
In general, we say that a coupling exists in quantum mechanics if the evolution of one part of the system depends on another quantity, which could be either classical or quantum. I'll give one example for each.
Suppose the Hamiltonian of a two-level system is an internal Hamiltonian $H_\mathrm{int}$ and an additional part that depends on some external parameter, maybe $\theta$: \begin{equation} H = H_\mathrm{int} + \theta \sigma_z \end{equation} Here the $\sigma_z$ was arbitrary. The point is that this system's evolution depends directly on the parameter $\theta$--maybe it's an external magnetic field, or some other feature of the environment. In this case, we would generally say that the system is "coupled to $\theta$." (You'd often see this in a metrological context, where we might be interested in using a quantum system coupled to an external parameter to measure the parameter.) In this case, there is only one quantum object, evolving under $H$.
Another common system--maybe a little more general--would be the evolution of two different variables both treated quantum mechanically. The idea here is that there would be some operator $A$ characterizing one observable of interest, and another $B$ characterizing a second. Then the Hamiltonian might be: \begin{equation} H = H_A + H_B + H_{AB} \end{equation} Where $H_A$ doesn't contain any term depending on $B$, $H_B$ doesn't contain any term depending on $A$, and $H_{AB}$ might have terms like $A \cdot B$, $A^2 B$, etc. The reason why this couples the system is that if we now evaluate Heisenberg equations of motion $\dot{A} = \frac{i}{\hbar} \left[ H, A \right]$ we'll find that the $H_{AB}$ term will put terms depending on $B$ into $\dot{A}$ and vice versa. Therefore, solving the equations of motion will require describing both $A$ and $B$. On the other hand, if the equations "decouple" or we do something to decouple them ourselves, we can usually find a solution for $A(t)$ that doesn't depend on $B$ and vice versa.
This is all paralleled in classical mechanics, by the way, where we would call two variables coupled if they appeared in each others' equations of motion.
EDIT: Peter Shor points out that objects can be "indirectly coupled," which is correct but would usually require me to introduce another variable $C$. I think the most general statement of being coupled/uncoupled is asking whether the equations of motion can be solved independently of each other.