The Meaning of the Eigenvalue Equation in Quantum Mechanics
Definition
Matrix
Suppose an $n\times n$ matrix $A$ is given. The value $a$ satisfying the equation below is called an eigenvalue, and a nonzero $n\times 1$ vector $\mathbf{x}$ is called the eigenvector corresponding to $a$.
$$ A \mathbf{x} = a \mathbf{x} \tag{1} $$
Operator
Suppose an operator $A$ is given. The value $a$ satisfying the equation below is called an eigenvalue, and a nonzero $\ket{a}$ is called the eigenfunction corresponding to $a$.
$$ A \ket{a} = a \ket{a} \tag{2} $$
In particular, equation $(2)$ is called the eigenvalue equation, and the process of setting up the eigenvalue equation for a given operator $A$ to find the eigenvalues and eigenfunctions is called the eigenvalue problem.
Explanation
In physics, the term eigenfunction or eigenstate is used more often than the term eigenvector. As a notation for representing states, single angle brackets are used to write $\ket{a}$, which is read as [ket a]. This way of notation is called the Dirac notation.
Mathematical Properties
If $\ket{a}$ is the eigenfunction of $A$ corresponding to $a$, then for any constant $c$, $c\ket{a}$ is also an eigenfunction corresponding to $a$. In other words, a constant multiple of an eigenfunction is also an eigenfunction corresponding to the same eigenvalue. Of course, the converse does not hold in general; multiple eigenfunctions that are not constant multiples of one another may correspond to a single eigenvalue, and this is called degeneracy.
$$ \begin{align*} A(c\ket{a}) &= cA\ket{a} \\ &= ca\ket{a} \\ &= a(c\ket{a}) \end{align*} $$
Physical Interpretation
In quantum mechanics, an operator is interpreted as the act of observing a physical quantity. Therefore, the eigenvalue equation below is a mathematical expression of the act of observing a physical quantity. It can be interpreted that when the physical quantity $A$ of the wave function $\psi$ is observed, its value comes out to be $a$.
$$ A \psi = a \psi $$
To use a simple analogy of measuring a person’s weight, $A$ is the act of stepping onto a scale, $\psi$ is the person, and $a$ is the weight of $\psi$. What this implies is extremely important, because it shows that in quantum mechanics the physical quantities of a particle (wave) are quantized. The Hamiltonian operator $H$ is the operator that observes the energy of a particle.
$$ H \psi = E \psi $$
Now, if we solve the above eigenvalue equation for a given potential $V$ and the eigenvalues do not appear continuously but instead appear discretely as $E_{1}$, $E_{2}$, $E_{3}$, $\dots$, it means that under the given conditions the only energies the particle can have are $E_{1}$, $E_{2}$, $E_{3}$, $\dots$. That is to say, the energy is quantized. So solving an eigenvalue problem in quantum mechanics amounts to finding the states (eigenfunctions) the particle can have and the physical quantities (eigenvalues) it can have in those states.
Ground State and Excited State
Among the energy eigenvalues obtained by solving the eigenvalue problem of the Hamiltonian as above, the eigenstate corresponding to the lowest energy is called the ground state. An eigenstate that is not the ground state is called an excited state. The subscript of the ground state sometimes starts from $n=0$ and sometimes from $n=1$, depending on the convention of the article or textbook.
The ground state is the lowest-energy state a particle can have in a given system, so unless energy is supplied from the outside, the particle cannot descend to a lower energy state than this. A point worth noting is that this minimum energy is generally greater than the minimum value of the potential. For example, the ground-state energy of the infinite potential well is $E_{1} = \dfrac{\pi^{2}\hbar^{2}}{2ma^{2}}$, and the ground-state energy of the quantum harmonic oscillator is $E_{0} = \dfrac{1}{2}\hbar\omega$, both of which are greater than $0$, the minimum value of the potential. This is a quantum-mechanical feature that contrasts with classical mechanics, where a particle can sit still at the bottom of the potential and make its energy $0$. Meanwhile, the ground-state energy (minimum energy) of the hydrogen atom is about $-13.6 \mathrm{eV}$, and when an electron in an excited state falls to the ground state, it emits light equal to the difference between those energy levels.
Solving the Eigenvalue Equation
In $(1)$, if we denote the eigenvalue as $\lambda$ and rearrange so that the right-hand side becomes $0$,
$$ \begin{align*} && A\mathbf{x} &= \lambda \mathbf{x} \\ \implies && A\mathbf{x}-\lambda \mathbf{x} &=0 \\ \implies && (A-\lambda I)\mathbf{x} &=0 \end{align*} $$
Here, let us assume that the inverse of the matrix $(A-\lambda I)$ exists. Multiplying both sides by the inverse matrix, we obtain the following.
$$ \mathbf{x}=(A-\lambda I)^{-1} \cdot 0 = 0 $$
However, the solution $\mathbf{x} = 0$ is meaningless, so let us focus on finding a nonzero $\mathbf{x}$. To do so, the inverse of $(A-\lambda I)$ must not exist. The condition for an arbitrary matrix not to have an inverse is that its determinant is $0$. Therefore, the condition for finding a nonzero $\mathbf{x}$ is as follows.
$$ |A-\lambda I |=0 $$
At this point, the above equation is called the characteristic equation, and the solutions of this characteristic equation are precisely the eigenvalues.
Finding the Eigenvalues
Suppose the matrix $A$ is given as $A=\begin{pmatrix} 6 & 2 \\ 2 & 3 \end{pmatrix}$. Then $A-\lambda I$ is as follows.
$$ \begin{pmatrix} 6 & 2 \\ 2 & 3 \end{pmatrix}-\lambda I = \begin{pmatrix} 6 -\lambda& 2 \\ 2 & 3-\lambda \end{pmatrix} $$
Solving the characteristic equation,
$$ \begin{align*} \begin{vmatrix} 6 -\lambda & 2 \\ 2 & 3-\lambda \end{vmatrix} &= (\lambda -6)(\lambda -3)-4 \\ &= \lambda^2-9\lambda+14 \\ &=(\lambda-7)(\lambda-2)=0 \end{align*} $$
Therefore, we can see that the two eigenvalues are $\lambda=7$ and $\lambda=2$.
Finding the Eigenfunctions
Once the eigenvalues are found, we can find the eigenfunction corresponding to each eigenvalue. Let $\mathbf{x}=\begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix}$.
Case 1 $\lambda = 7$
$$ \begin{align*} && (A-\lambda I) \mathbf{x}&=0 \\ \implies && \begin{pmatrix} 6-7 & 2 \\ 2 & 3-7 \end{pmatrix} \begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix}&= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ \implies && \begin{pmatrix} -1 & 2 \\ 2 & -4 \end{pmatrix} \begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix}&= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ \implies && \begin{pmatrix} -1x_{1} + 2x_{2} \\ 2x_{1} -4x_{2} \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \end{align*} $$
Here, any $x_{1}$, $x_{2}$ satisfying $x_{1}=2x_{2}$ satisfies the equation above, so there are infinitely many solutions $\mathbf{x}$ to the equation above. However, in quantum mechanics $\mathbf{x}$ is precisely the wave function, and a wave function is meaningless unless it is normalized. Therefore, since we will normalize it anyway, we may just pick any pair $x_{1}$, $x_{2}$ satisfying the equation above. Whichever pair we choose, after going through the normalization process it ends up being the same value. The important point here is to choose the simplest possible form. That is to say, there is no need to needlessly choose a solution where $x_{1}=2000$ when $x_{2}=1000$. Since $x_{1}=2$ when $x_{2}=1$, the eigenfunction is as follows.
$$ \mathbf{x}_{1}=\begin{pmatrix} 2 \\ 1 \end{pmatrix} $$
Another important point here is that normalization must be performed. The normalized eigenfunction is as follows.
$$ \mathbf{x}_{1}=\frac{1}{\sqrt{5}}\begin{pmatrix} 2 \\ 1 \end{pmatrix} $$
Case 2 $\lambda = 2$
Finding the normalized eigenfunction $\mathbf{x}_{2}$ by the same method as above gives the following.
$$ \mathbf{x}_{2}= \frac{1}{\sqrt{5}}\begin{pmatrix} 1 \\ -2 \end{pmatrix} $$
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