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# 419: The Variation Theorem in Dirac Notation

The recipe for calculating the expectation value for energy using a trial wave function is,

$\langle E \rangle = \langle \psi | \hat{H} | \psi \rangle \label{1}$

Now suppose the eigenvalues of $$\hat{H}$$ are denoted by $$|i \rangle$$. Then,

$\hat{H} |i \rangle = \varepsilon_{i} |i \rangle = |i \rangle \varepsilon _{i} \label{2}$

Next we write $$| \psi \rangle$$ as a superposition of the eigenfunctions $$|i \rangle$$,

$| \psi \rangle = \sum_{i} |i \rangle \langle I| \psi \rangle$

and substitute it into Equation \ref{1}.

$\langle E \rangle = \sum_{i} \langle \psi | \hat{H} | i \rangle \langle i | \psi \rangle$

Making use of Equation \ref{2} yields,

$\langle E \rangle = \sum_{i} \langle \psi |i \rangle \varepsilon_{i} \langle i | \psi \rangle$

After rearrangement we have,

$\langle E \rangle = \sum_{i} \varepsilon_{i} | \langle i| \psi \rangle |^{2}$

However, $$| \langle i | \psi \rangle |^{2}$$ is the probability that $$\varepsilon_{i}$$ will be observed, $$p_i$$.

$\langle E \rangle = \sum_{i} \varepsilon_{i} p_{i} \geq \varepsilon_{0}$

Thus, the expectation value obtained using the trial wave function is an upper bound to the true energy. In other words, in valid quantum mechanical calculations you can't get a lower energy than the true energy.

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