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Lecture 6: Schrödinger Equation

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    38871
  • Recap of Lecture 5

    Schrödinger Equation is a wave equation that is used to describe quantum mechanical system and is akin to Newtonian mechanics in classical mechanics. The Schrödinger Equation is an eigenvalue/eigenvector problem. To use it we have to recognize that observables are associated with linear operators that "operate" on the wavefunction.

    The Schrödinger Equation: A Better Quantum Approach to Quantum

    Recall the wave equation

    \[\dfrac {\partial^2 u}{\partial x^2} = \dfrac {1}{\nu^2} \cdot \dfrac {\partial^2 u}{\partial t^2}.\]

    Through separation of variables, \(u(x,t) = \psi (x) cos(\omega t)\) where \(\psi (x)\) is the spatial amplitude. Replacing \(u(x,t)\) with \(\psi (x) \cos(\omega t)\), the classical wave equation becomes

    \[\dfrac {d^2 \psi (x)}{d x^2} + \dfrac {\omega^2}{\nu^2} \psi (x) = 0\]

    With manipulation, the equation can be rewritten to

    \[\dfrac {d^2 \psi (x)}{d x^2} + \dfrac {4\pi^2}{\lambda^2} \psi (x) = 0\]

    from the relation \(\omega = 2\pi \nu\).

    Using the de Broglie formula \(\lambda = h/p\), we solve for \(p\). The momentum \(p\) is given by \(p = mv\). Recalling that \(KE = 0.5\,mv^2\), \(KE\) can be written in terms of \(p\) by \(KE = p^2/2m\). \(E_{total}\) can now be solved to be \(p^2/2m + V(x)\). Solving for \(p\) results in

    \[p = \sqrt{2m(E_{total}-V(x))}\]

    Now that \(p\) is known, and \(h = 6.626 \times 10^{-34}\), \(\lambda\) can be solved to be equal to \(\dfrac {h}{\sqrt{2m(E-V(x))}}\). Through squaring that equation and taking the reciprocal of it, the result is

    \[\dfrac {1}{\lambda^2} = \dfrac {2m(E-V(x))}{h^2}\]

    This can be put into

    \[\dfrac {d^2 \psi (x)}{d x^2} + \dfrac {4\pi^2}{\lambda^2} \psi (x) = 0\]

    giving

    \[\dfrac {d^2 \psi (x)}{d x^2} + \dfrac {2m}{\hbar^2}[E-V(x)] \psi (x) = 0\]

    This can be rearranged to make the Schrödinger equation

    \[\dfrac {-\hbar^2}{2m} \dfrac {d^2 \psi (x)}{d x^2} + V(x)\psi (x) = E\psi (x)\]

    The kinetic operator in one dimension is

    \[KE_{1D}=\dfrac {-\hbar^2}{2m} \dfrac{d^2}{dx^2}\]

    In three dimensions, the operator is

    \[KE_{3D}=\dfrac {-\hbar^2}{2m} \nabla^2\]

    The Laplacian operator

    The three second derivatives in parentheses together are called the Laplacian operator, or del-squared,

    \[ \nabla^2 = \left ( \frac {\partial ^2}{\partial x^2} + \dfrac {\partial ^2}{\partial y^2} + \dfrac {\partial ^2}{\partial z^2} \right ) \label {3-20}\]

    with the del operator,

    \[\nabla = \left ( \vec {x} \frac {\partial}{\partial x} + \vec {y} \frac {\partial}{\partial y} + \vec {z} \frac {\partial }{\partial z} \right ) \label{3-21}\]

    also is used in Quantum Mechanics. The symbols with arrows over them are unit vectors.

    Two Flavors of Schrödinger Equations

    We have the time-dependent Schrödinger Equation, which is the complete and full description of the system (both spatial and temporal)

    \[ \textcolor{red} { \underbrace{ i\hbar\dfrac{\partial\Psi(x,t)}{\partial t}=\left[-\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2}+V(x)\right]\Psi(x,t) }_{\text{time-dependent Schrödinger equation in 1D}}}\label{3.1.16}\]

    For waves in three dimensions this is can be expanded to

    \[\textcolor{red}{ \underbrace{ i\hbar\dfrac{\partial}{\partial t}\Psi(\vec{r},t)=\left[-\dfrac{\hbar^2}{2m}\nabla^2+V(\vec{r})\right]\Psi(\vec{r},t)}_{\text{time-dependent Schrödinger equation in 3D}}}\label{3.1.17}\]

    The use of the time-dependent Schrödinger Equations will always be valid, albeit necessary in many situations.

    For conservative (stationary) systems, the energy is a constant, and the time-dependence can be separated from the space-only factor (via the Separation of Variables technique)

    \[ \textcolor{red}{ \underbrace{\left[-\dfrac{\hbar^2}{2m}\nabla^2+V(\vec{r})\right]\psi(\vec{r})=E\psi(\vec{r})} _{\text{time-independent Schrödinger equation}}} \label{3.1.19}\]

    The use of the time-independent Schrödinger Equations will valid when the potential \(V(x\) is NOT a function of time.

    Using the time-independent Schrödinger Equation does NOT mean that there is no time-dependence to the resulting wavefunction. It just means it is trivial

    \[\Psi(\vec{r},t)=\psi(\vec{r})e^{-iEt / \hbar}\label{3.1.18}\]

    where \(\psi(\vec{r})\) is a wavefunction dependent (or time-independent) wavefuction that only depends on space coordinates.

    Operators

    An operator is a generalization of the concept of a function applied to a function. Whereas a function is a rule for turning one number into another, an operator is a rule for turning one function into another.

    Presentation1.jpg

    For the time-independent Schrödinger Equation, the operator of relevance is the Hamiltonian operator (often just called the Hamiltonian) and is the most ubiquitous operator in quantum mechanics.

    \[ \hat{H} = -\dfrac{\hbar^2}{2m}\nabla^2+V(\vec{r})\]

    We often (but not always) indicate that an object is an operator by placing a `hat' over it, eg, \(\hat{H}\). So Equation \ref{3.1.19} can then be simplified to

    \[ \hat{H} \psi(\vec{r}) = E\psi(\vec{r}) \label{simple}\]

    Equation \ref{simple} says that the Hamiltonian operator operates on the wavefunction to produce the energy, which is a number, (a quantity or observable), times the wavefunction.

    Fundamental Properties of Operators

    Most properties of operators are straightforward, but they are summarized below for completeness.

    The sum and difference of two operators \(\hat{A} \) and \(\hat{b} \) are given by

    \[ (\hat{A} \pm \hat{B}) f = \displaystyle \hat{A} f \pm \hat{B} f \]

    The product of two operators is defined by

    \[ \hat{A} \hat{B} f \equiv \hat{A} [ \hat{B} f ] \]

    Two operators are equal if

    \[\hat{A} f = \hat{B} f \]

    for all functions \(f\).

    The identity operator \(\hat{1}\) does nothing (or multiplies by 1)

    \[ {\hat 1} f = f \]

    The associative law holds for operators

    \[ \hat{A}(\hat{B}\hat{C}) = (\hat{A}\hat{B})\hat{C} \]

    The commutative law does not generally hold for operators. In general,

    \[ \hat{A} \hat{B} \neq \hat{B}\hat{A}\]

    It is convenient to define the commutator of \(\hat{A} \) and \(\hat{b} \)

    \[ [\hat{A}, \hat{B}] \equiv \hat{A} \hat{B} - \hat{B} \hat{A} \]

    Note that the order matters, so that

    \[[\hat{A} ,\hat{B} ] = - [\hat{B} ,\hat{A} ] \].

    If \(\hat{A} \) and \(\hat{B} \) commute, then

    \[\hat{A} ,\hat{B} ] = 0\].

    The \(n\)-th power of an operator \(\hat{A}^n \) is defined as \(n\) successive applications of the operator, e.g.

    \[ \hat{A}^2 f = \hat{A} \hat{A} f \]

    Linearity of Operators

    Almost all operators encountered in quantum mechanics are linear operators.

    An operator \(\hat{A}\) is linear if

    \[ \hat{A}[c_1f(x)+c_2f_2(x)]= c_1\hat{A}f_1(x)+c_2\hat{A}f_2(x) \]

    and the operator is nonlinear if

    \[\hat{A}[c_1f_1(x)+c_2f_2(x)] \neq c_1\hat{A}f_1(x)+c_2\hat{A}f_2(x)\]

    Example \(\PageIndex{1}\): Linear or NOT?

    Determine from the following operators which are linear and nonlinear:

    1. \(\hat{A}f(x)= f(x)^2\) [square f(x)]
    2. \(\hat{A}f(x)= f^*(x)\) [form the complex conjugate of f(x)]
    3. \(\hat{A}f(x)= 0\) [multiply f(x) by zero]
    4. \(\hat{A}f(x)= [f(x)]^{-1}\) [take the reciprocal of f(x)]
    5. \(\hat{A}f(x)= f(0)\) [evaluate f(x) at x=0]
    6. \(\hat{A}f(x)= \ln f(x)\) [take the log of f(x)]
    Strategy:

    It is important to note that an operator \(\hat{A}\) is linear if

    \[ \underbrace{\hat{A}[c_1f(x)+c_2f_2(x)]}_{\text{left side}}= \underbrace{c_1\hat{A}f_1(x)+c_2\hat{A}f_2(x) }_{\text{right side}}\]

    and the operator is nonlinear if

    \[ \underbrace{ \hat{A}[c_1f_1(x)+c_2f_2(x)]}_{\text{left side}} \neq \underbrace{ c_1\hat{A}f_1(x)+c_2\hat{A}f_2(x) }_{\text{right side}}\]

    Answer a:

    Evaluate the left side

    \[\begin{align} \hat{A}[c_1f(x)+c_2f_2(x)] &= [c_1f_1(x)+c_2f_2(x)]^2 \\ &= c_1^2 f_1(x)^2+2c_1f_1(x) c_2f_2(x)+c_2^2f_2(x)^2 \end{align}\]

    Evaluate the right side

    \[c_1 \hat{A} f_1(x)+c_2\hat{A}f_2(x)=c_1[f_1(x)]^2+c_2[f_2(x)]^2 \neq \hat{A}[c_1f_1(x)+c_2f_2(x)] \]

    This operator is nonlinear

    Answer b:

    Evaluate the left side

    \[\hat{A}[c_1f_1(x)+c_2f_2(x)] = c_1^*f_1^*(x) + c_2^*f_2(x)\]

    Evaluate the right side

    \[c_1\hat{A}f_1(x) + c_2\hat{A}f_2(x) = c_1f_1^*(x) + c_2f_2^*(x)\]

    \[ = \hat{A}[c_1f_1(x) + c_2f_2(x)]\]

    This operator is linear

    Answer c:

    Evaluate the left side

    \[ \hat{A}[c_1f_1(x)+c_2f_2(x)] = 0\]

    Evaluate the right side

    \[c_1\hat{A}f_1(x) + c_2\hat{A}f_2(x) = c_1f_1(x) + c_2f_2(x) = 0\]

    \[ = \hat{A}[c_1f_1(x) + c_2f_2(x)]\]

    This operator is linear

    Answer d:

    Evaluate the left side

    \[\hat{A}[c_1f_1(x)+c_2f_2(x)] = \dfrac{1}{c_1f_1(x) + c_2f_2(x)}\]

    Evaluate the right side

    \[c_1\hat{A}f_1(x) + c_2\hat{A}f_2(x) = \dfrac{c_1}{f_1(x)} + \dfrac{c_2}{f_2(x)} \]

    \[ \neq \hat{A}[c_1f_1(x) + c_2f_2(x)]\]

    This operator is nonlinear

    Answer e:

    Evaluate the left side

    \[\hat{A}[c_1f_1(x)+c_2f_2(x)] = c_1f_1(0) + c_2f_2(0)\]

    Evaluate the right side

    \[ = c_1\hat{A}f_1(x) + c_2\hat{A}f_2(x)\]

    This operator is linear

    Answer f:

    Evaluate the left side

    \[\hat{A}[c_1f_1(x)+c_2f_2(x)] = \ln [c_1f_1(x) + c_2f_2(x)]\]

    Evaluate the right side

    \[c_1\hat{A}f_1(x) + c_2\hat{A}f_2(x) = c1 \ln f_1(x) + c_2 \ln f_2(x)\]

    \[ \neq \hat{A}[c_1f_1(x) + c_2f_2(x)]\]

    This operator is nonlinear