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Does it make any sense to talk about energy of any one particle in an interacting system? For example if we talk about a system of two coupled quantum harmonic oscillators of same mass and frequency, the Hamiltonian of the system is: \begin{align*} \hat{H}= & \frac{\hat{P}_{1}^{2}}{2m}+\frac{\hat{P}_{2}^{2}}{2m}+\frac{m\omega^{2}\hat{X}_{1}^{2}}{2}+\frac{m\omega^{2}\hat{X}_{2}^{2}}{2}+k\hat{X_{1}}\hat{X_{2}} \end{align*} But in the $X_{+},X_{-}$ basis the Hamiltonian can be written as, \begin{align*} \hat{H}= & \frac{\hat{P}_{+}^{2}}{2m}+\frac{\hat{P}_{-}^{2}}{2m}+\frac{m\omega_{-}^{2}\hat{X}_{+}^{2}}{2}+\frac{m\omega_{-}^{2}\hat{X}_{-}^{2}}{2} \end{align*}

Now when we talk about the first excited state, that would be: \begin{align*} \left|\psi_{1}\right\rangle & =\left|0_{+}\right\rangle \left|1_{-}\right\rangle \end{align*} with energy eigen value, \begin{alignat*}{1} E_{1} & =\frac{\hbar\omega_{+}}{2}+\frac{3\hbar\omega_{-}}{2} \end{alignat*}. In this state, can we comment about the energy of any one of the particle only? If we go on to measure the energy of anyone of the particles, what would be the measurement outcome and can we attribute this measurement outcome to just the particle under consideration?

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  • $\begingroup$ Sure, thank you! $\endgroup$ Commented Mar 14 at 15:51
  • $\begingroup$ You have assumed $k >0$. $\endgroup$
    – my2cts
    Commented Apr 1 at 7:13
  • $\begingroup$ The two forms of $H_0$ differ by an interaction. $\endgroup$
    – my2cts
    Commented Apr 1 at 7:14

2 Answers 2

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If we call the uncoupled Hamiltonian \begin{align*} \hat{H}_0= & \frac{\hat{P}_{1}^{2}}{2m}+\frac{\hat{P}_{2}^{2}}{2m}+\frac{m\omega^{2}\hat{X}_{1}^{2}}{2}+\frac{m\omega^{2}\hat{X}_{2}^{2}}{2}\,, \end{align*} then it's not too difficult to show that $[k\hat{X_{1}}\hat{X_{2}}, \hat{H}_0]\neq0$ and therefore $[\hat{H}_0,\hat{H}]\neq0$. This means that the two Hamiltonians do not share an eigenbasis, and hence there is at least one eigenvector of $\hat{H}$ (call it $\lvert E\rangle$) that is not an eigenvector of $\hat{H}_0$. This means that this particular state of definite energy $E$ is not a state of definite energy of the uncoupled system, which means that the two particles do not individually have definite energies.


To see more details, let's consider the following. The ground state for the full Hamiltonian \begin{align*} \hat{H}_0= & \frac{\hat{P}_{1}^{2}}{2}+\frac{\hat{P}_{2}^{2}}{2}+\frac{\hat{X}_{1}^{2}}{2}+\frac{\hat{X}_{2}^{2}}{2} + k\hat{X}_1\hat{X}_2\,, \end{align*} is given by $$ \psi^{(k)}_0(x_1,x_2)=\frac{\sqrt[8]{1-k^2}}{\sqrt{\pi }} \exp \left( -\frac{\sqrt{1-k}}{4} \left(x_1-x_2\right)^2-\frac{\sqrt{k+1}}{4} \left(x_1+x_2\right)^2 \right)\,. $$ One can immediately see that this state is necessarily entangled between the $x$ and $y$ degrees of freedom unless $k=0$, where this expression reduces to $$ \frac{1}{\sqrt{\pi}}e^{-x^2/2}e^{-y^2/2}\,. $$ That is, it cannot be written as a product.

Then, to measure the "energy of one of the particle", that requires one to have an operator in mind that corresponds to the energy of that particle. The Hamiltonian cannot be written as the sum of two terms, one involving $X_1$ and the other involving $X_2$, and so this doesn't really work. However, we could salvage this by saying that the energy of particle $i$ is just $$ \hat{H}_j = \frac{1}{2}\hat{P}_j^2 + \frac{1}{2}\hat{X}_j^2\,, $$ and ask the question of what would be the result of the measurement of $\hat{H}_j$ if the two-particle system was in the ground state of $\hat{HG}$. It's not too hard to show that $\psi^{(k)}_0(x_1,x_2)$ above is necessarily a linear combination of products $\psi_{n_1}(x_1)\psi_{n_2}(x_2)$ of the eigenstates of the decoupled system ($\hat{H}_0$). Thus, there is necessarily a spread in the measurements of $\hat{H}_1$ and $\hat{H}_2$ when the state is the ground state $\psi^{(k)}_0(x_1,x_2)$ of the coupled system.

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    $\begingroup$ @HypnoticZebra I've updated some things in the post. $\endgroup$
    – march
    Commented Mar 15 at 17:55
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It depends on what you are looking for. If you want to move away from mathematical rigor and more to situations in daily life where you need practical communication, then I think you will find many physicists will talk as if they are separable.

For example, suppose we have helium in the first excited state and then fire a photon at it to put it in a higher excited state. We know the way to think about it is to think of the system as a whole. But as limited beings, we think in levels of simplicity. (Notice you used a toy model for simplicity and did not include a fully realistic Hamiltonian of a particular system.)

We imagine that the original scenario has one electron in the 1s state and the other in the 2s state (ignoring spin). With our simple thinking, we then imagine that the photon excites one of the electrons and leaves the other untouched. But we know it's not really true.

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