Finding the Ground State of a Hamiltonian Operator

[Matthollyw00d]

When given a Hamiltonian operator (in this case a 3x3 matrix), how do you go about find the ground state, when this operator is all that is given? By the SE when have $$H\Psi=E\Psi$$. I can easily solve for Eigenvalues/vectors, but which correspond to the ground state, or am I missing something?

 

[vela]

The ground state is the state with the lowest energy.

 

[Matthollyw00d]

So I achieve eigenvalues of $$\{0,1,2\}$$ (this is actually the eigenvalues I have for the problem), would the 0 or the 1 be the lowest state? I assumed 0, but can a ground state have 0 energy?

 

[vela]

It depends on the potential. You can always add a constant to the potential without changing anything physically, which would shift the total energy by the same amount. What you can't really have is the kinetic energy being 0 because there should always be some motion even in the ground state.

 

[Matthollyw00d]

Let $$H=\hbar\omega \[ \left( \begin{array}{ccc}<br /> 1 & i & 0 \\<br /> -i & 1 & 0 \\<br /> 0 & 0 & 1 \end{array} \right) \]$$
and let $$A =\hbar \left[ \begin{array}{ccc}<br /> 1 & 0 & i \\<br /> 0 & 1 & 0 \\<br /> -i & 0 & 1 \end{array} \right]$$.

Calculate the uncertainty relation $$\sigma_E \sigma_a$$ for a system in the energy ground state.

My problem is calculating $$\langle [H,A]\rangle$$.
The commutator $$[H,A]=\hbar^2\omega \left[ \begin{array}{ccc}<br /> 0 & 0 & 0 \\<br /> 0 & 0 & 1 \\<br /> 0 & -1 & 0 \end{array} \right]$$
And if relevant the eigenvalues of H are $$\hbar\omega \{0,1,2\}$$.
It says to calculate in the ground state, but I don't know the ground state, so how do I gather it from this data?

 

[fzero]

You have the eigenvalue for the ground state, now find the corresponding eigenvector. It will be a (non-zero) vector

$$v_0 = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$$

satisfying

$$H v_0 = 0.$$

 

[Matthollyw00d]

Yes, but $$\langle v_0|[H,A]|v_0 \rangle =0$$ which gives me a trivial inequality, which leads me to believe something is incorrect and hence why I posted here.