Consider, a quantum system has a hamiltonian with eigenstates $\{|\phi_1\rangle,|\phi_2\rangle,|\phi_3\rangle\}$ and associated eigenvalues $\{\lambda_a,\lambda_a,\lambda_b\}$. My notes state that any vector in the subspace $\{|\phi_1\rangle,|\phi_2\rangle\}$ has the corresponding eigenvalue $\lambda_a$.

This seems like an obvious statement, but would like to know how I could prove it (if there is a way to do so).


1 Answer 1


Suppose we have an arbitrary state $|\phi\rangle$ in the subspace $\{|\phi_1\rangle,|\phi_2\rangle\}$. What this means is that we can write $|\phi\rangle$ as a superposition (linear combination) of these two states. For example, $|\phi\rangle = a|\phi_1\rangle + b|\phi_2\rangle$ for which $|a|^2 + |b|^2 = 1$.

Now, we can find the eigenvalue of $|\phi\rangle$ by applying the Hamiltonian, as follows:

$H|\phi\rangle = H(a|\phi_1\rangle + b|\phi_2\rangle) = aH|\phi_1\rangle + bH|\phi_2\rangle = a\lambda_a|\phi_1\rangle + b\lambda_a|\phi_2\rangle = \lambda_a(a|\phi_1\rangle + b|\phi_2\rangle) = \lambda_a|\phi\rangle$

So, we have shown that any state in the subspace $\{|\phi_1\rangle,|\phi_2\rangle\}$ has eigenvalue $\lambda_a$.


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