0
$\begingroup$

In Adiabatic Quantum Computing, a Hamiltonian $H$ is evolved for time $T$ according to $$H(t) = (1-t/T)H_{0} + (t/T)H_{P} $$ where $H_{0}$ is an initial Hamiltonian, and the ground state of $H_{P}$ encodes the solution to a problem of interest.

If this Hamiltonian is evolved slowly enough, the adiabatic theorem claims the system will always be in the lowest energy state (i.e. for each $t$ with $0 \leq t \leq T$, the qubits will be in the state which minimises $H(t)$). When the ground state is not a superposition state (is a basis state), calculating the energy should be easy. But when in superposition, the system will have no objective energy; its energy state will be something of the form: $$ \sum_{s \in S} \alpha_{s}E_{s}$$ where $S$ denotes the set of all basis states (that we can find the qubits in when measured), $|\alpha_{s}|^{2}$ denotes the probability we measure the system in basis state $s$, while $E_{s}$ denotes the energy of the system when in basis state $s$.

But how do you determine which one of these (superposition) energy states is the ground state; I mean, how do you say that the energy of superposition state $A$ is less than the energy of superposition state $B$ if these energies are not scalars?

I am a computer scientist learning about Quantum Computing btw, so I apologize if there are errors in my argument.

Improve this question
$\endgroup$
3
  • $\begingroup$ How is your question related to adiabatic computing? I'd suggest removing the unnecessary part and focusing on the essence of your question. $\endgroup$
    – Norbert Schuch
    Commented Jan 2, 2017 at 13:27
  • $\begingroup$ I didn't make it entirely clear how this relates to adiabatic quantum computing. I was wondering how the ground state could be a superposition state in an adiabatic quantum computer. $\endgroup$
    – Alex Michael
    Commented Jan 2, 2017 at 22:29
  • $\begingroup$ My point is the opposite: You should remove everything unnecessary from the question. Try to focus on one point you want to clarify. This will be good for your understanding (you'll have to think the topic through) and likely also give you more attention (I guess I'm not the only one not reading lengthy & digressing posts in their entirety). $\endgroup$
    – Norbert Schuch
    Commented Jan 2, 2017 at 22:49

1 Answer 1

Reset to default
1
$\begingroup$

The energy is always a scalar, I am not sure why you think that's not the case. We can compare the energies of any two stationary states since they are both positive numbers.

The other issue I think you're having here is a misconception about the structure of vector spaces. A basis in which the Hamiltonian is diagonal is always available and the ground state is always the basis vector in that set with the lowest eigenvalue for $H$. It is true that sometimes there are ground state degeneracies, but that is not really relevant to this discussion.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Thanks for your reply Bobak. So you're saying the energy is a scalar even when the system is in superposition? This would seem to contradict a quantum system having no objective energy value when in superposition, but rather just a probability that it will be in each possible energy when measured $\endgroup$
    – Alex Michael
    Commented Jan 2, 2017 at 10:53
  • 1
    $\begingroup$ I'm saying a superposition of energy states is not the ground state unless all of those states in the superposition have the exact same energy, which happens to be the lowest eigenvalue in the spectrum. In such a case, you have again one and only one energy associated with the state. If the ground state is non-degenerate, then exactly one basis vector in the Hamiltonian's eigenbasis will correspond to the ground state, not a superposition. $\endgroup$
    – Something Different
    Commented Jan 2, 2017 at 19:47
  • $\begingroup$ Thanks a lot Bobak. Would this not imply that a system which evolves adiabatically would rarely be in a superposition state: for it would be rare for all states in a superposition to have the exact same energy, meaning the ground state will rarely be a superposition state - so will be a basis state, implying that if the system is in its ground state at all times it will rarely be in superpositon? This would also imply that an adiabatic quantum computer barely uses superpositon $\endgroup$
    – Alex Michael
    Commented Jan 2, 2017 at 22:26
  • 1
    $\begingroup$ Hi Alex, I think you've got the gist, but you're forgetting that a basis vector in one basis will generally be a superposition in another basis. For instance, consider the harmonic oscillator. The eigenbasis is spanned by number states, eigenvectors of the operator $a^\dagger a$. The ground state is $|0\rangle$, the state annihilated by $a$. But this state is not an eigenvector of $x$ or $p$. So the ground state is a superposition of momentum eigenstates or position eigenstates, but it is a basis vector of energy eigenstates. $\endgroup$
    – Something Different
    Commented Jan 2, 2017 at 23:02
  • $\begingroup$ Hmmmm, so would that mean the energy gap (between ground and first exited states) would be arbitrarily small in that case - because if the ground state is a superposition of momentum and position eigenstates, one could always modify this state only an arbitrarily small amount to create a state whose energy was only marginally higher (than the ground state)? $\endgroup$
    – Alex Michael
    Commented Jan 3, 2017 at 0:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.