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I really should know this off by heart (this is my field...) but I never really grasped the difference between the total wavefunction of a system and the wavefunctions of particles within it, so it only just dawned on me that perhaps the total energy of a system was simply the sum of the energies of the individual particles.

It's true, isn't it?

By 'energy' here, I really mean $\langle E \rangle$. So would the expectation value of the total energy equal the sum of the expectation values of all the component particles? Or is there some conditionality to it? I.e. in a quantum computer, if the states of two qubits are opposites, then the expectation value of the total energy would be the sum of the component energies for each possible scenario $|0\rangle|1\rangle$ and $|1\rangle|0\rangle$, multiplied by the probability of each.

Took me a long time to learn to ask the stupid questions.

[Edit: brief mathsing for two qubits:

$ \hat{H} \Psi = E \Psi $

$ \Psi = \psi_m \psi_n $

$ \psi_m = a_m |0\rangle + b_m |1\rangle \quad\quad\mathrm{(ditto}~~\psi_n \mathrm{)} $

$ \Psi = a_m a_n |0\rangle |0\rangle + a_m b_n |0\rangle |1\rangle \dots $

$ \langle E \rangle = a_m a_n E_{00} + a_m b_n E_{01} \dots $

or time-independently

$ \langle E(t) \rangle = a_m(t) a_n(t) E_{00}(t) + a_m(t) b_n(t) E_{01}(t) $ ]

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2 Answers 2

The expectation value of the total energy of a quantum system is not always the sum of the expectation values of the energy of each particle. It's certainly not true if you can't write the total wavefunction as the product of two wavefunctions corresponding to each particle. You also need to be able to write the energy as a sum of the energies of each particle. That is, if $|\Psi\rangle {}= |\psi_1\rangle\ldots|\psi_N\rangle$ and $\hat{E} = \hat{E}_{p1} + \ldots + \hat{E}_{pN}$ then you get $\langle\hat{E}\rangle {}={} \langle\psi_1|\hat{E}_{p1}|\psi_1\rangle+\ldots+\langle\psi_N|\hat{E}_{pN}|\psi_N\rangle$. I don't think you can get that result otherwise.

In your example, however, you do something different. You are expressing your state as a linear combination on a base of eigenstates, and then get $\langle\hat{E}\rangle$ as the weighted mean of the energies of each vector in your base. And of course, that's how quantum mechanics calculates expectation values. In the same notation as the previous paragraph, if you now express your state in the base of eigenstates ($\hat{E}|\psi_i\rangle{}=E_i|\psi_i\rangle$) you have $|\Psi\rangle {}= a_1|\phi_1\rangle+\ldots+a_N|\phi_N\rangle$ and now the energy can be written as $\langle\hat{E}\rangle {}={} \langle\phi_1|E_1|\phi_1\rangle+\ldots+\langle\phi_N|E_N|\phi_N\rangle$. This result is true always, but it's very different from the previous one.

Notice that $\hat{E}_{pk}$ is the hamiltonian operator for particle $k$, while $E_i$ is the $i$-eth eigenvalue of the hamiltonian operator of the complete system.

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Let's look at the simplest case, very similar to what you tried to work out. Each particle has a two dimension space of states e.g. $H_1=\{\alpha|0\rangle_1+\beta|1\rangle_1:\alpha,\beta\in\mathbb C\}$ and $H_2=\{\alpha|0\rangle_2+\beta|1\rangle_2:\alpha,\beta\in\mathbb C\}.$

Then the space of states for the system of two particles is four dimensional $H=H_1\otimes H_2$ and in fact, $$H=\{\alpha|0\rangle_1\otimes|0\rangle_2+\beta|0\rangle_1\otimes|1\rangle_2+\gamma|1\rangle_1\otimes|0\rangle_2+\delta|1\rangle_1\otimes|1\rangle_2:\alpha,\beta,\gamma,\delta\in\mathbb C\}.$$

The space of states can then also be represented by vectors $[\alpha,\beta,\gamma,\delta]^T.$ The Hamiltonian acts on the states of the system, the four dimensional system state space. And since it is Hermitian, it is diagonalizable with real eigenvalues that are the diagonal elements. But we have no idea how those states that diagonalize the system Hamiltonian are related to single particle states. It is entirely possible that the state $[0,1,-1,0]^T$ ($|0\rangle_1\otimes|1\rangle_2-|1\rangle_1\otimes|0\rangle_2$) has an eigenvalue that is distinct from all the other eigenvalues.

An example Hamiltonian could send $|0\rangle_1\otimes|1\rangle_2-|1\rangle_1\otimes|0\rangle_2$ to itself, send $|0\rangle_1\otimes|1\rangle_2+|1\rangle_1\otimes|0\rangle_2$ to its own negative and send $|0\rangle_1\otimes|0\rangle_2$ and $|1\rangle_1\otimes|1\rangle_2$ to zero. In that case the highest energy state of the system is a state where the individual particles are entangled, so don't have their own state. And there isn't even a thing called the energy of just one particle (maybe the energy of particles by themselves are zero, and the Hamiltonian of the system is entirely an interaction energy).

So in general, you can't write the wavefucntion of a system as the product of wavefucntions for single particles, you can't ever get a state like $|0\rangle_1\otimes|1\rangle_2-|1\rangle_1\otimes|0\rangle_2$ that way. And in general, the full Hamiltonian of the system is not the sum of a Hamiltonian that acts on just one particle plus a Hamiltonian that acts on just the other particle, it also can represent an interaction between the two. Interactions between particles are sometimes ignored when first introducing the very basics, and sometimes ignored when talking about fundamental properties such as distinguishability and spin-statistics. But interactions can in general exist, and it just means the Hamiltonian is a function of the whole system.

So you err if you assume that a general state of the system is a product of states of the individual particles (it can be a non unique sum of products of single particles states where the non uniqueness comes from a choice of basis for the single particle states). You err to assume that the Energy (which can include interaction energy) can be defined on a single particle state.

So in general, when you have a state of the system, there isn't a state for each particle and even if their were, there might not be an energy for each single particle state. So the energy isn't the sum of the energies of the two things (the single particle energies) that don't exist for the two things (the single particles states) that also don't exist. In a particular situation it might, but in general it doesn't.

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