📂Quantum Mechanics

The Variational Principle in Quantum Mechanics

Theorem¹

Let the ground state energy of a given physical system be $E_{\text{gs}}$. For the system’s Hamiltonian operator $H$ and any normalized function $\psi$, the following holds.

$$ E_{\text{gs}} \le \braket{H} = \braket{\psi | H | \psi} $$

Here $\braket{H}$ is the expectation value in the state $\psi$. This is called the variational principle.

Explanation

The variational principle is one of the methods for approximating the ground state energy when it cannot be determined exactly.

Suppose we know the Hamiltonian $H$ of the system, but the Schrödinger equation is too complicated to solve and we cannot obtain the ground state energy. In this case, using the variational principle we can find an upper bound of $E_{\text{gs}}$, so we can make a rough estimate of the ground state energy. This is because if we pick as a trial function a wave function $\psi$ presumed to be reasonably similar to the ground state and compute the expectation value, that value is guaranteed to be no smaller than $E_{\text{gs}}$, no matter how poor the choice is.

$$ E_{\text{gs}} \le \braket{H} = \braket{\psi | H | \psi} $$

In practice, one takes a trial wave function $\psi_{b}$ containing an adjustable parameter $b$ and minimizes $\braket{H}$ with respect to $b$ to obtain a bound as low as possible. Since equality holds when $\psi$ is the actual ground state, the more the trial wave function resembles the ground state, the better the approximation obtained. Of course, in reality we cannot know how close it actually is to the ground state, and moreover it can only be used to find the ground state.

Proof

Since $H$ is a Hermitian operator, the set $\left\{ \psi_{n} \right\}$ of its eigenfunctions is a 🔒(26/07/18)complete set. That is, any normalized eigenfunction $\psi$ can be expressed as a linear combination of orthonormalized eigenfunctions.

$$ H \psi_{n} = E_{n} \psi_{n}, \qquad \braket{\psi_{m} | \psi_{n}} = \delta_{mn} $$

$$ \psi = \sum_{n} c_{n} \psi_{n} $$

Since $\psi$ is normalized, the coefficients satisfy the following.

$$ 1 = \braket{\psi | \psi} = \sum_{m} \sum_{n} c_{m}^{\ast} c_{n} \braket{\psi_{m} | \psi_{n}} = \sum_{n} \left| c_{n} \right|^{2} $$

Meanwhile, the expectation value of $H$ is as follows.

$$ \braket{H} = \braket{\psi | H | \psi} = \sum_{m} \sum_{n} c_{m}^{\ast} c_{n} E_{n} \braket{\psi_{m} | \psi_{n}} = \sum_{n} E_{n} \left| c_{n} \right|^{2} $$

Now, since the ground state energy is by definition the lowest energy eigenvalue, we have $E_{\text{gs}} \le E_{n}$ for all $n$. Since $\left| c_{n} \right|^{2} \ge 0$, we obtain the following.

$$ \braket{H} = \sum_{n} E_{n} \left| c_{n} \right|^{2} \ge \sum_{n} E_{\text{gs}} \left| c_{n} \right|^{2} = E_{\text{gs}} \sum_{n} \left| c_{n} \right|^{2} = E_{\text{gs}} $$

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