The Schrödinger Equation

Shaun Williams, PhD

Introduction

• A scientific postulate is a generally accepted statement that is consistent with experimental observations
• Postulates are also known as physical laws
• Postulates cannot be derived by any other fundamental considerations
• Newton's second law, \(F=ma\), is an example of a postulate
• One goal of science is to find the smallest and most general set of postulates that can explain all observations
• A whole new set of postulates was added with the invention of quantum mechanics
• The Schrödinger equation is a more general and fundamental postulate than both \(E=h\nu\) and \(p=\frac{h}{\lambda}\)

Differential Equations

• A differential equation is a mathematical equation involving one or more derivative
• The analytical solution to a differential equation is the expression or function for the dependent variable that gives an identity when substituted into the differential equation
• Numerical solutions to differential equations can also be obtained

Example 3.1

The differential equation \( \frac{dy(x)}{dx}=2 \) has the solution \( y(x)=2x+b \), where \(b\) is a constant. This function \( y(x) \) defines the family of straight lines on a graph with a slope of 2. Show that this function is a solution to the differential equation by substituting for \( y(x) \) in the differential equation. How many solutions are there to this differential equation? For one of these solutions, construct a table of data showing pairs of \( x \) and \( y \) values, and use the data to sketch a graph of the function. Describe this function in words.

An Interesting Case

• Some differential equations have the property that the derivative of the function gives the function back multiplied by a constant
• The differential equation for a first-order chemical reaction is one example
• This differential equation and the solution for the concentration of the reactant are $$ \begin{align*} \frac{dC(t)}{dt} &= -kC(t) \\ C(t) &= C_0 e^{-kt} \end{align*} $$

Example 3.2

Show that \( C(t) \) is a solution to the differential equation.

Another Interesting Case

• Another kind of differential equation has the property that the second derivative of the function yields the function multiplied by a constant $$ \frac{d^2 \psi(x)}{dx^2}=k\psi(x) $$
• Both kinds of these types of differential equations are found in Quantum Mechanics

Example 3.3

What is the value of the constant in the above differential equation when \( \psi(x)=\cos(3x)\)?

Example 3.4

What other functions, in addition to the cosine, have the property that the second derivative of the function yields the function multiplied by a constant?

Mathematical Functions and Waves

• Since some mathematical functions, such as sine and cosine, go through periodic maxima and minima, they produce graphs that look like wave
• Such functions can themselves be thought of as waves and can be called wavefunctions
• If electrons, atoms, and molecules have wave-like behavior, then there must be a mathematical function that is the solution to a differential equation that describes electrons, atoms, and molecules
• Thoughts such as this likely motivated Erwin Schrödinger to find (ie. create) the wave equation that is now accepted as the fundamental postulate of quantum mechanics

A Classical Wave Equation

• The easiest way to find a differential equation that will provide wavefunctions as solutions is to start with a wavefunction and work backwards
• We will consider a sine wave, take its first and second derivatives, and examine the results

A Sine Wave

• The amplitude of a sine wave can depend upon the position, \(x\), in space $$ A(x) = A_0 \sin \left(\frac{2\pi x}{\lambda}\right) $$ or upon time, \(t\) $$ A(t) = A_0 \sin \left( 2\pi \nu t \right) $$ or upon both space and time $$ A(x,t) = A_0 \sin \left( \frac{2\pi x}{\lambda} - 2\pi \nu t\right) $$
• We can simplify this by using the wave vector, \( k=\frac{2\pi}{\lambda}\) and the angular frequency, \(\omega = 2\pi \nu\) so that \( A(x,t) = A_0 \sin \left( kx - \omega t \right) \)

Partial Derivatives

$$ A(x,t) = A_0 \sin \left( kx - \omega t \right) $$

• Let's take the partial second derivatives of \(A(x,t)\) with respect to both \(x\) and \(t\) $$ \begin{align*} \frac{\partial^2 A(x,t)}{\partial x^2} &= -k^2 A_0 \sin \left( kx-\omega t\right) = -k^2 A(x,t) \\ \frac{\partial^2 A(x,t)}{\partial t^2} &= -\omega^2 A_0 \sin \left(kx-\omega t\right) = -\omega^2 A(x,t) \end{align*} $$
• Since both involve \( A(x,t) \) we find an equality $$ k^{-2} \frac{\partial^2 A(x,t)}{\partial x^2} = -A(x,t) = \omega^{-2} \frac{\partial^2 A(x,t)}{\partial t^2} $$

Simplifying Our Equality

• Recall that \(\nu\) and \(\lambda\) are related in that their product gives the velocity of the wave, \(\nu \lambda = v \)
• Be careful to distinguish \(v\) (velocity) from \(\nu\) (frequency)
• If \(\omega = 2\pi \nu \) then \( \omega = 2\pi \nu = \frac{2\pi v}{\lambda} = vk \)
• Therefore we can simplify our equality earlier to the classical wave equation in one-dimension $$ \begin{align*} k^{-2} \frac{\partial^2 A(x,t)}{\partial x^2} &= \omega^{-2} \frac{\partial^2 A(x,t)}{\partial t^2} \\ \frac{1}{k^2} \frac{\partial^2 A(x,t)}{\partial x^2} &= \frac{1}{v^2k^2} \frac{\partial^2 A(x,t)}{\partial t^2} \end{align*} $$ $$ \begin{equation} \frac{\partial^2 A(x,t)}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 A(x,t)}{\partial t^2} \label{eq:1} \end{equation} $$

Invention of the Schrödinger Equation

• From the previous section, the classical wave equation in one-dimension has a solution that is a sine function
• Other functions, such as cosine, can also work
• Our goal is to seek a method of finding the wavefunctions that are appropriate for describing electrons, atoms, and molecules
• In order to get there, we need the appropriate wave equation.

Example 3.5

Show that the functions \( e^{i(kx+\omega t)} \) and \( \cos(kx−\omega t) \) also satisfy equation \eqref{eq:1} (the classical wave equation). Note that \(i\) is a constant equal to \( \sqrt{-1}\).

Separation of Variables

• A general method for finding solutions to differential equations that depend on more than one variable is to separate the variables into different terms.
• This technique is called the Method of Separation of Variables
• This separation makes is possible to write the solution as a product of functions that depend on the variables
• For the classical wave equation, separating variables is very easy because \(x\) and \(t\) do not appear together in the same term in the differential equation
• In other words, the variables are already separated

Beginning Quantum Mechanics (QM)

• It is common in QM to symbolize the functions that are solutions to the Schrödinger equation as \(\psi\), \(\Psi\), \(\phi\), or \(\Phi\)
• We could say, for instance, that a QM wavefunction is $$ \begin{equation} \Psi(x,t)=\psi(x)\cos(\omega t) \label{eq:3.3.2} \end{equation} $$
• This wavefunction can be substituted into the classical wave equation, which after simplifying yields $$ \cos (\omega t) \frac{\partial^2 \psi(x)}{\partial x^2} = -\frac{\omega^2}{v^2}\psi(x) \cos(\omega t) $$ which rearranges to $$ \frac{\partial^2 \psi(x)}{\partial x^2} + \frac{\omega^2}{v^2}\psi(x) = 0 $$

Moving Right Along

$$ \frac{\partial^2 \psi(x)}{\partial x^2} + \frac{\omega^2}{v^2}\psi(x) = 0 $$

• We now include the idea that we are trying to find a wave equation for a particle by introducing the particles momentum by using the de Broglie relation so that \( \frac{\omega^2}{v^2} = \frac{p^2}{\hbar^2} \) $$ \begin{equation} \frac{\partial^2 \psi(x)}{\partial x^2} + \frac{p^2}{\hbar^2}\psi(x) = 0 \label{eq:3.3.5} \end{equation} $$

Adding in Energy

• Now lets use the total energy of the particle as the sum of the kinetic energy and potential energy $$ \begin{equation} E=T+V(x) = \frac{p^2}{2m}+V(x) \label{eq:3.3.6} \end{equation} $$
• Notice that we have identified that the potential energy is a function of position
• Each atomic or molecular system will have a different potential energy function
• Let's solve equation \eqref{eq:3.3.6} for \(p^2\) $$ p^2 = 2m\left(E-V(x)\right) $$

The Schrödinger Equation

• Plugging out equation for \(p^2\) into equation \eqref{eq:3.3.5} yields the Schrödinger Equation $$ \begin{equation} \frac{\partial^2 \psi(x)}{\partial x^2} + \frac{2m}{\hbar^2}\left(E-V(x)\right)\psi(x) = 0 \label{eq:3.3.7} \end{equation} $$
• The Schrödinger Equation is usually written in a rearranged form $$ \begin{equation} -\frac{\hbar^2}{2m} \frac{\partial^2 \psi(x)}{\partial x^2} + V(x)\psi(x) = E\psi(x) \label{eq:3.3.8} \end{equation} $$
• Notice that the left side consists of a kinetic energy term and a potential energy term and the right side consists of the total energy

Example 3.6

Show the steps that lead from equation \eqref{eq:1} and \eqref{eq:3.3.2} to equation \eqref{eq:3.3.8}.

Analysis of the Schrödinger Equation

• Equation \eqref{eq:3.3.8} concerns a particle of mass \(m\) moving in one dimension \(x\) in a potential field \(V(x)\)
• This equation does not contain time so it is called the Time-Independent Schrödinger Equation
• The Schrödinger equation for a particle moving in three dimensions (\(x\), \(y\), and \(z\)) is obtained by adding the other second derivative terms and by including the three-dimensional potential energy function

$$ \begin{equation} \frac{-\hbar^2}{2m}\left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \right) \psi(x,y,z) + V(x,y,z)\psi(x,y,z)=E\psi(x,y,z) \label{eq:3.3.9} \end{equation} $$

Example 3.7

Write the Schrödinger equation for a particle of mass \(m\) moving in a 2-dimensional space with the potential energy given by $$ V(x,y) = -\frac{\left(x^2+y^2\right)}{2} $$

New Symbols and Notations

• The three second derivatives in parentheses together are called the Laplacian operator, del-squared $$ \begin{equation} \nabla^2 = \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \right) \label{eq:3.3.11} \end{equation} $$
• The del operator by itself is $$ \begin{equation} \nabla = \left( \vec{x} \frac{\partial}{\partial x} + \vec{y} \frac{\partial}{\partial y} + \vec{z} \frac{\partial}{\partial z} \right) \label{eq:3.3.12} \end{equation} $$ where symbols with arrows over them are unit vectors

Example 3.8

Write the del-operator and the Laplacian operator for two dimensions and rewrite your answer to Example 3.7 in terms of the Laplacian operator.

Operators, Eigenfunctions, Eigenvalues, and Eigenstates

• The Laplacian operator is called an operator because it does something to the function that follows
• Symbols for operators are often denoted by a hat (^) over the symbol unless the symbol is used exclusively as an operator (\(\nabla\)) or does not involve differentiation (\(r\) for position)

The Hamiltonian

• Recall that we can identify the total energy operator, which is called the Hamiltonian operator, \(\hat{\mathcal{H}}\), as made from the kinetic energy operator and the potential energy operator $$ \begin{equation} \hat{\mathcal{H}} = -\frac{\hbar^2}{2m} \nabla^2 + \hat{V}(x,y,z) \label{eq:3.4.1} \end{equation} $$
• The term Hamiltonian is named after the Irish mathematician Hamilton
• It comes from the formulation of Classical Mechanics that is based on the total energy \( H=T+V \) rather than Newton's second law \(\vec{F}=m\vec{a}\)
• Using our Hamiltonian notation, we can write the Schrödinger Equation as $$ \begin{equation} \hat{\mathcal{H}}\psi(x,y,z)=E\psi(x,y,z) \label{eq:3.4.2} \end{equation} $$

Interpretation of this Schrödinger Equation

• Equation \eqref{eq:3.4.2} says that the Hamiltonian operator operates on the wavefunction to produce the energy times the wavefunction
• The energy is a number
• This means that the Hamiltonian operating on a function produces a constant number times the same function
• When this happens, the function is called an eigenfunction and the resulting numerical value (\(E\) in this case) is called the eigenvalue
• Eigen here is the German word meaning self or own
• It is a general principle of quantum mechanics that there is an operator for every physical observable
• If the wavefunction of the system is an eigenfunction of an operator then the value of the associated observable is extracted from the eigenfunction by operating on the eigenfunction with the appropriate operator

Momentum Operators

• One of the tasks we must be able to do is to replace classical variables in mathematical expression with corresponding quantum mechanical operators
• We have already seen the operators for the total energy and the kinetic energy
• We are now going to focus on momentum

Finding the Momentum Squared Operator

• The momentum operator can be obtained from the kinetic energy operator
• The classical expression for the kinetic energy of a particle moving in dimension (\(x\)) is \( T_x = \frac{P_x^2}{2m} \)
• Therefore we expect that $$ \begin{equation} \hat{T}_x = \frac{P_x^2}{2m} = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \label{eq:3.5.2} \end{equation} $$
• From this we can identify the operator for the square of the \(x\)-momentum $$ \begin{equation} \hat{P}_x^2 = -\hbar^2 \frac{\partial^2}{\partial x^2} \label{eq:3.5.3} \end{equation} $$

Finding the Momentum Operator

• Since we now know the operator for momentum squared, there are two possibilities for the momentum operator $$ \begin{equation} \hat{P}_x = i\hbar \frac{\partial}{\partial x} \label{eq:3.5.4} \end{equation} $$ and $$ \begin{equation} \hat{P}_x = -i\hbar \frac{\partial}{\partial x} \label{eq:3.5.5} \end{equation} $$
• Remember that \(i=\sqrt{-1}\)
• It turns out that the second option, \( \hat{P}_x=-i\hbar \frac{\partial}{\partial x} \), is the best option as we shall see

Which \(\hat{P}_x\) is better?

Let's start with \( \hat{P}_x \psi(x)=P_x \psi(x) \) where \(\psi(x)=e^{ikx}\) $$ \hat{P}_x \psi(x)=P_x \psi(x) $$ $$ \pm i\hbar \frac{\partial}{\partial x} e^{ikx} = P_x e^{ikx} $$ $$ \pm i\hbar (ik) e^{ikx} = P_x e^{ikx} $$ $$ \mp \hbar k e^{ikx} = P_x e^{ikx} $$ $$ P_x = \mp \hbar k $$ So, if we use the momentum operator that has the - sign, we get the momentum and wave vector pointing in the same direction which is the preferred result corresponding to the de Broglie relation

Example 3.9

Show the following unit vector relations: \( \vec{x}\cdot \vec{x} =1\) and \(\vec{x}\cdot\vec{y}=0\)

Example 3.10

Consider a particle moving in three dimensions. The total momentum, which is a vector, is \( p=\vec{x}P_x + \vec{y}P_y + \vec{z}P_z \) where \(\vec{x}\), \(\vec{y}\), and \(\vec{z}\) are unit vectors pointing in the \(x\), \(y\), and \(z\) directions, respectively. Write the operators for the momentum of this particle in the \(x\), \(y\), and \(z\) directions, and show that the total momentum operator is \( -i\hbar\nabla = -i\hbar \left( \vec{x}\frac{\partial}{\partial x}+\vec{y}\frac{\partial}{\partial y}+\vec{z}\frac{\partial}{\partial z} \right) \) is the vector operator called del. Show that the scalar product \(\nabla \cdot \nabla\) produces the Laplacian operator.

The Time-Dependent Schrödinger Equation

• The time-dependent Schrödinger equation is used to find the time dependence of the wavefunction
• The equation relates the energy to the first time derivative analogous to the classical wave equation that involves the second time derivative $$ \begin{equation} \hat{\mathcal{H}}(r,t)\Psi(r,t)=i\hbar \frac{\partial}{\partial t}\Psi(r,t) \label{eq:3.6.1} \end{equation} $$ where \(r\) represents the spatial coordinates \((x,y,z)\)
• Even if the Hamiltonian does not depend on time, we can use this equation to find the time dependence \(\phi(t)\) of the eigenfunctions of \(\vec{\mathcal{H}}\)

Starting to Derive the Time-Dependent Schrödinger Equation

• We write the wavefunction \(\Psi(r,t)\) as a product of a space function \(\psi(r)\) and a time function \(\phi(t)\) $$ \Psi(r,t)=\psi(r)\phi(t) $$
• We use this because the Hamiltonian does not depend on time
• Since \(\psi(r)\) is an eigenfunction of \(\vec{\mathcal{H}}\) with eigenvalue \(E\) $$ \begin{align} \hat{\mathcal{H}}(r)\psi(r)\phi(t) &= i\hbar \frac{\partial}{\partial t}\psi(r)\phi(t) \\ E\psi(r)\phi(t) &= i\hbar\psi(r)\frac{\partial}{\partial t}\psi(t) \\ \frac{d\phi(t)}{\phi(t)} &= \frac{-iE}{\hbar}dt \label{eq:3.6.5} \end{align} $$

Integrating Towards a Solution

• Integrating equation \eqref{eq:3.6.5} yields $$ \begin{equation} \phi(t)=e^{-i\omega t} \label{eq:3.6.6} \end{equation} $$ after we set the integration constant to 0 and using the definition that \(\omega = \frac{E}{\hbar}\)
• Thus we see that the time dependent Schrödinger equation contains the condition \(E=\hbar\omega\) proposed by Planck and Einstein
• The eigenfunction of a time-independent Hamiltonian therefore have an oscillatory time dependence given by the complex function $$ \begin{equation} \Psi(r,t) = \psi(r)e^{-i\omega t} \label{eq:3.6.7} \end{equation} $$

Analysis of Eigenfunctions

• When molecules are described by these sorts of eigenfunctions, they are said to be in an eigenstate of the time-independent Hamiltonian operator
• We will see that all observable properties of a molecule in an eigenstate are independent of time
• A wavefunction with this oscillatory time dependence \(e^{-i\omega t}\) therefore is called a stationary-state function
• When a system is not in a stationary state, wavefunction can be represented by a sum of eigenfunction like those above
• In non-stationary state calculations, the time dependence does not cancel out

Example 3.11

Complete the steps leading from Equation \eqref{eq:3.6.1} to Equation \eqref{eq:3.6.7}.

Example 3.12

Show that Equation \eqref{eq:3.6.7} is a solution to Equation \eqref{eq:3.6.1} when the Hamiltonian operator does not depend on time and \(\psi(r)\) is an eigenfunction of the Hamiltonian operator.

Meaning of the Wavefunction

• Since wavefunctions can be complex functions, the physical significance cannot be found from the function itself because \(\sqrt{-1}\) is not physical
• The physical significance is found in the product of the wavefunction and its complex conjugate
  • the absolute square of the wavefunction
  $$ \begin{equation} \psi^* (r,t) \psi(r,t) = \left| \psi(r,t) \right|^2 \label{eq:3.7.1} \end{equation} $$
• The square is used, rather than the modulus, just like the intensity of a light wave depends on the square of the electric field

Meaning of the Square of the Wavefunction

• At one time, it was thought that for an electron described by the wavefunction \(\psi(r)\), the quantity \(e \psi^*(r_i)\psi(r_i)d\tau \) was the amount of charge to be found in the volume \(d\tau\) located at \(r_i\)
• However, Max Born found this was inconsistent with scattering experiments
• The Born interpretation, which is generally accepted, is that \(\psi^*(r_i)\psi(r_i)d\tau\) is the probability that the electron in in the volume \(d\tau\) located at \(r_i\)
• The Born interpretation called the wavefunction the probability amplitude, the absolute square of the wavefunction is called the probability density, and the probability density times a volume element in 3D (\(d\tau\)) is the probability

Example 3.13

Show that the square of the modulus of \( \Psi(r,t)=\psi(r)e^{-i\omega t} \) is time independent. What insight regarding stationary states do you gain from this proof?

Example 3.14

According to the Born interpretation, what is the physical significance of \( e \psi^*(r_0)\psi(r_0)d\tau \)?

Expectation Values

• An important deduction can be made if we multiply the left-hand side of the Schrödinger equation by \(\psi^*(x)\) and integrate over all values of \(x\) and examine the potential energy term that arises
• We can deduce that the potential energy integral provides the average value for the potential energy
• Likewise, the kinetic energy integral provides the average value for the kinetic energy

$$ \begin{equation} \begin{split} \expect{\mathcal{H}} = \int_{-\infty}^\infty \psi^*(x)\hat{\mathcal{H}}\psi(x)dx= &\int_{-\infty}^\infty \psi^*(x) \left( \frac{-\hbar^2}{2m} \right) \frac{\partial^2}{\partial x^2} \psi(x)dx \\ &+ \int_{-\infty}^\infty \psi^*(x) V(x) \psi(x) dx \end{split} \label{eq:3.8.1} \end{equation} $$

Further Analysis of Expectation Values

• The kinetic energy operator involves the differentiation of the wavefunction to the right of it
• This step must be completed before multiplying by the complex conjugate of the wavefunction
• The potential energy usually depends only on the position and not momentum - therefore no differentiation and no hat over the \(V\)
• For example, the harmonic potential in one dimension is \(\frac{1}{2}kx^2\) where, here, \(k\) is the force constant
• Since the potential energy integral only involves the products of functions and therefore the order of multiplication does not effect the result - the commutative property

$$ \begin{equation} \expect{V} = \int_{-\infty}^\infty \psi^*(x)V(x)\psi(x)dx = \int_{-\infty}^\infty V(x)\psi^*(x)\psi(x)dx \label{eq:3.8.2} \end{equation} $$

More Analysis

• Integral \eqref{eq:3.8.2} is telling us to take the probability that the particle is in the interval \(dx\) at \(x\) and multiplying it by the potential energy at \(x\)... then integrate over all possible values of \(x\)
• This procedure is just the way to calculate the average potential energy \(\expect{V}\) of the particle
• This integral therefore is called the average-value integral or the expectation-value integral because it gives the average result of a large number of measurements of the particle's potential energy
• When the operator involves differentiation, it does not commute with the wavefunctions $$ \begin{equation} \psi^*(x) \frac{\partial^2}{\partial x^2} \psi(x) \ne \psi^*(x) \psi(x) \frac{\partial^2}{\partial x^2} \ne \frac{\partial^2}{\partial x^2} \left( \psi^*(x) \psi(x) \right) \label{3.8.3} \end{equation} $$

Importance of Expectation Values

• Expectation values are very important
• They provide us with the average values of physical properties like energy, momentum, and position
• This is important because in many cases, precise values cannot be determined, even in principle
• If we know the average of some quantity, it is also important to know whether the distribution is narrow or broad
• The width of the distribution is characterized by its variance

Postulates of Quantum Mechanics

Let's summarize the postulates of Quantum Mechanics that we have introduced

Postulate 1

• The properties of a quantum mechanical system are determined by a wavefunction \(\psi(r,t)\) that depends upon the spatial coordinates of the system and time
• For more than one particle, \(r\) is used to represent the complete set of coordinates \(r=(x_1,y_1,z_1,x_2,y_2,z_2,\dots ,x_n,y_n,z_n)\)
• Since the state of a system is defined by its properties, \(\psi\) specifies or identifies the state and sometimes is called the state function

Postulate 2

• The wavefunction is interpreted as probability amplitude with the absolute square of the wavefunction, \( \Psi^*(r,t)\Psi(r,t) \), interpreted as the probability density at time \(t\)
• The probability density times a volume is probability $$ \begin{equation} \psi^*(x_1,y_1,z_1)\psi(x_1,y_1,z_1)dx_1dy_1dz_1 \label{eq:3.9.1} \end{equation} $$ is the probability that the particle is in the volume \(dxdydz\) located at \(x_1,y_1,z_1\) at time \(t\)
• For a particle system, we write the volume element as \(d\tau = dx_1 dy_1 dz_1 \dots dx_n dy_n dz_n\) and \(\Psi^*(r,t)\Psi(r,t)d\tau\) is the probability that particle 1 is in the volume \(dx_1dy_1dz_1\) at \(x_1y_1z_1\) and particle 2 is in the volume \(dx_2dy_2dz_2\) at \(x_2y_2z_2\), etc.

Postulate 2 Continued

• Because of this probabilistic interpretation, the wavefunction must be normalized $$ \begin{equation} \int\limits_{allspace} \psi^*(r,t)\psi(r,t)d\tau=1 \label{eq:3.9.2} \end{equation} $$
• The integral sign in \eqref{eq:3.9.2} represents a multi-dimensional integral involving all coordinates
• Integration in three-dimensional space will be integration over \(dV\), which can be expanded:
  • Cartesian coordinates: \(dV=dx\, dy\, dz\)
  • Spherical coordinates: \(dV=r^2 \sin \phi\, dr\, d\theta\, d\phi\)
  • Cylindrical coordinates: \(dV=r\, dr\, d\theta\, dz\)

Postulate 3

• For every observable property there is a quantum mechanical operator
• The operator for position of a particle in three dimensions is just the set of coordinates \(x\), \(y\), and \(z\), which is written as a vector $$ r=(x,y,z)=\vec{xi}+\vec{yj}+\vec{zk} $$
• The operator for a component of momentum is $$ \hat{P}_x = -i\hbar \frac{\partial}{\partial x} $$
• The operator for kinetic energy in one dimension is $$ \hat{T}_x = -\left( \frac{\hbar^2}{2m} \right) \frac{\partial^2}{\partial x^2} $$

Postulate 3 Continued

• The operator for momentum in three dimensions $$ \hat{p}=-i\hbar \nabla $$
• The operator for kinetic energy in three dimensions $$ \hat{T} = \left( -\frac{\hbar^2}{2m} \right) \nabla^2 $$
• The Hamiltonian operator \(\hat{\mathcal{H}}\) is the operator for the total energy
• In many cases only the kinetic energy of the particles and the electrostatic or Coulomb potential energy due to their charges are considered
• Other terms are possible in the Hamiltonian

Postulate 4

• The time-independent wavefunctions of a time-independent Hamiltonian are are found by solving the time-independent Schrödinger equation $$ \hat{\mathcal{H}} \psi(r) = E\psi(r) $$
• These wavefunctions are called stationary-state functions because the properties of a system in such a state are time independent

Postulate 5

• The time evolution or time dependence of a state is found by solving the time-dependent Schrödinger equation $$ \hat{\mathcal{H}}(r,t)\psi(r,t)=i\hbar \frac{\partial}{\partial t}\psi(r,t) $$
• For the case where \(\hat{\mathcal{H}}\) is independent of time, the time dependent part of the wavefunction is \(e^{-i\omega t}\) where \(\omega=\frac{E}{\hbar}\) or equivalently \(\nu=\frac{E}{h}\)
• This shows that the energy-frequency relation used by Planck, Einstein, and Bohr results from the time-dependent Schrödinger equation
• The oscillatory time dependence of the probability amplitude does not affect the probability density or the observable properties because in the calculation of these quantities
• The imaginary part cancels in multiplication by the complex conjugate

Postulate 6

• If a system is described by the eigenfunction \(\psi\) of an operator \(\hat{A}\) then the value measured for the observable property corresponding to \(\hat{A}\) will always be the eigenvalue \(a\)
• This can be calculated from the eigenvalue equation $$ \hat{A}\psi = a\psi $$

Postulate 7

• If a system is described by a wavefunction \(\psi\), which is not an eigenfunction of an operator \(\hat{A}\), then a distribution of measured values will be obtained, and the average value of the observable property is given by the expectation value integral, $$ \begin{equation} \expect{A} = \frac{\int \psi^* \hat{A} \psi d\tau}{\int \psi^* \psi d\tau} \label{eq:3.9.11} \end{equation} $$ where the integration is over all coordinates involved in the problem
• The average value \(\expect{A}\), also called the expectation value, is the average of many measurements
• If the wavefunction is normalized, then the normalization integral is the denominator of equation \eqref{eq:3.9.11} equals 1