## Schrödinger Equation

Time-independent Schrödinger equation is$$\left[-\frac{{\hslash}^{2}}{2m}{\nabla}_{x}^{2}+V\left(x\right)\right]\Psi \left(x\right)=E\Psi \left(x\right)$$ | 1 |

*i*is the √(-1), ℏ is Planck's constant divided by 2π,

*m*is the mass, ∇

^{2}is the second derivative operator with respect to coordinate (

*d*in this case), and

^{2}/dx^{2}*V*(

*x*) is the potential energy operator.

*Ψ*(

*x*) and

*E*are, respectively, wave function and energy of the system.

We are going to look at details of how we solve Schrödinger equation for certain systems, including chemical systems that we are famililar with, such as reaction energies and transition state structure and activation energy.

The form of Schrödinger equation does not change, but different system
has different potential energy (or shape of potential energy) for which we
model the physical system. So, in one sense, we seek solutions, meaning
that we look for *Ψ*(x) that satisfy the equation, then at the same
time, we can learn about the energy of the state of the system.

For a larger system, multiparticle cases, we must think of additional strategy, but it would be simple enough for relatively simple systems. For electronic structure of molecules, we must go beyond the simple approach and incurr another level of strategy.

Here is what Schrödinger might have done for derivation of the equation.

*λ*,

$$\lambda =\frac{h}{p}$$ | C1 |

*h*is Planck's constant and

*p*is the particle momentum. From classical mechanics of wave, the wave amplitude,

*Ψ*is given by following equation,

$${\nabla}^{2}\Psi =-{k}^{2}\Psi $$ | C2 |

*d/dx*and

*k = 2π/λ*is the wave vector. By substituting above into Eqn. C1,

$${\nabla}^{2}\Psi =-{\left(\frac{2\pi p}{h}\right)}^{2}\Psi =-\frac{4{\pi}^{2}{p}^{2}}{{h}^{2}}\Psi $$ | C3 |

*E = T + V*, and

$$E=\frac{1}{2}m{v}^{2}+V$$ | C4 |

$$m{v}^{2}=2\left(E-V\right)$$ | C5 |

*p*,

^{2}= m(mv^{2})$${\nabla}^{2}\Psi =-\frac{4{\pi}^{2}m\left(2\left(E-V\right)\right)}{{h}^{2}}\Psi $$ | C6 |

$$-\frac{{\hslash}^{2}}{2m}{\nabla}^{2}\Psi +V\Psi =E\Psi $$ | C7 |

*ℏ = h/2π*.

## Quantum Mechanical Postulates

Before we move on to solving Schrödinger equation, we need to understand that quantum mechanics is based upon several postulates. Even though, these are postulates, for application to chemistry these have not deviated from the experimental findings.

Here we list six postulates:

1) For any state of a system, there is a wave function *Ψ*,
that depends on the coordinates of the system and time,
describes the system completely.

$$\Psi =\Psi \left({x}_{1},{x}_{2},{x}_{3},\cdots ,{x}_{N},t\right)$$ | 2 |

*N*is the maximum number of coordinates in the system.

In general the wave function can be complex numbers. The wave function is the key in understanding the quantum system. Much of studies applies quantum mechanics devoted to solve the wave function. By itself, the wave function does not have physical meaning.

2) For every observable quantity, there exists a corresponding linear Hermitian operator. We will look at this more closely in the next section.

3) The state function *Ψ* is given as a solution to the
Schrödinger
equation

$$\widehat{H}\Psi =E\Psi $$ | 3 |

4) The state function *Ψ* also satisfies

$$i\hslash \frac{\partial \Psi}{\partial t}=\widehat{H}\Psi $$ | 4 |

5) The square of the state function,
|*Ψ*(*x,t*)|^{2}*dτ*,
gives probability density of the particle in the system, with
the *dτ* describes the volume element. This postulate is due to Max Born.

6) The average quantity in quantum mechanics for a given operator is called
*expectation value*, which is

$$\langle \widehat{O}\rangle =\frac{{\displaystyle \int {\Psi}^{*}\widehat{O}\Psi d\tau}}{{\displaystyle \int {\Psi}^{*}\Psi d\tau}}=\frac{\langle \Psi |\widehat{O}|\Psi \rangle}{\langle \Psi |\Psi \rangle}$$ | 5 |

## Eigenvalue Equation: Schrödinger Equation

The time-independent Schrödinger equation, is given as eigenvalue equation. You notice that Hamiltonian operator operated on the wave function gives energy E and the same wave function back. One way to understand what Schrödinger equation tells is to consider physical meaning of the equation. Regardless of where the equation comes from (in fact, one can not derive Schrödinger equation--Schrödinger fudged it somehow during the Christmas holiday in the Tyrollean Mountains), it tells you that Hamiltonian operator describes the physics of the system. It contains operators for kinetic and potential energy. As was the case in the classical mechanics, the kinetic energy plus the potential energy in the phase space gives total energy of the system. Also as in the classical mechanics, the total energy a constant of motion, meaning that the total energy of the system is a conserved quantity.Another way to look at this equation is that the total energy is given by transforming the operator into coordinate independent quantity, such as energy via the wave function. The transformation of the kind, $\widehat{H}\to E$, is for Schrödinger equation. Therefore, you should ask yourself a question, "Given the kinetic energy and the between the particles, what wave function gives me the energy?" The answer to this question gives the solution to Schrödinger equation. By the way, as we mentioned before, the force between particles is given by $F=-\frac{\partial V\left(x\right)}{\partial x}$. In atoms and molecules, the forces between electrons and nuclei are based on electromagnetic forces, in particular Coulomb attractions and repulsions.

Let's look at properties necessary to construct mathematical structure of quantum mechanics next.

## Conditions on Ψ to be well-behaved

1) Single-valued so that the probability is unambiguously normalizable.2) Continuous

3) Nowhere infinite

4) Piecewise continuous of the first derivatives

5) Commutation with constant

6) Square-integrable

**Figure 1. a) Examples of bad wave function, b) good wave function.**

Examples of bad and good wave functions look like, are shown in Figure 1. The first panel of a) is not single-valued, thus it is not even a function. The second panel of a) has one side tended toward infinity. The third has a break 2) and the first derivatives also don't seem to match 4). Figure 1b is an example of good wave function that satisfies all conditions.

The last one, square integrability, hits the core of quantum mechanics. First, it means that the integral of the probability density must be finite. This is so that to the wave function is normalizable. It makes perfect sense when you consider a physical system where the particle in the system must exist somewhere in the system! This is expressed mathematically, in the denominator of Eqn. 5,

$$\int {\Psi}^{*}\Psi d\tau}=c$$ | 4 |

*Ψ*is the complex conjugate of

^{*}*Ψ*.

The wave function can be complex,

$$\Psi ={\Psi}_{re}+i{\Psi}_{im}$$ | 5 |

*Ψ*is the real part of the total wave function,

_{re}*i*is square-root of negative 1, and

*Ψ*is the imaginary part. The complex conjugate is constructed by changing the sign on the imaginary part,

_{im}$$\Psi ={\Psi}_{re}-i{\Psi}_{im}$$ | 6 |

$$\langle \Psi |\Psi \rangle ={\displaystyle \int \left({\Psi}_{re}-i{\Psi}_{im}\right)}\left({\Psi}_{re}+i{\Psi}_{im}\right)d\tau ={\displaystyle \int \left({\Psi}_{re}^{2}+{\Psi}_{im}^{2}\right)}d\tau $$ | 7 |

*Ψ*is taken as unnormalized wave function we can obtain normalized function,

*Ψ'*=

*AΨ*, where

*A*is the normalization factor. The square integrating the normalized wave function is forced to be one.

$$\int \Psi {\text{'}}^{*}\Psi \text{'}d\tau}=1$$ | 8 |

$${A}^{2}{\displaystyle \int {\Psi}^{*}\Psi}d\tau =1$$ | 9 |

$${A}^{2}c=1$$ | 10 |

$$A={\left(\frac{1}{c}\right)}^{1/2}$$ | 11 |

$$\Psi \text{'}={\left(\frac{1}{c}\right)}^{1/2}\Psi $$ | 12 |

## Operators

The observable quantities in quantum mechanics is expressed in terms of*operators*. Operator is used in

*eigenvalue equation*(Eqn. 3). When operated on

*eigenfunction, Ψ*, the operator is transformed into

*eigenvalue*. In Eqn. 3, Hamiltonian operator, $\widehat{H}$ , is transformed into energy,

*E*.

For us, we need mostly two operators that are important in understanding
chemistry, and these are: *momentum operator* and *coordinate
operators*. These constitute the component of Hamiltonian operator as well
as some of the spectroscopic quantities, such as *dipole moment* operator.

### Momentum operator

The momentum operator is given in one dimension (*x*coordinate) is

$${\widehat{p}}_{x}=-i\hslash \frac{\partial}{\partial x}$$ | 13 |

*ℏ*is Planck's constant divided by 2

*π*. So, if you have $\Psi =A{e}^{ikx}$ as a wave function, its momentum is, then

$$\widehat{p}\Psi =\hslash kA{e}^{ikx}=\hslash k\Psi $$ | 14 |

*p*=

*ℏ k*.

### Coordinate operator

The coordinate operator is a multiplicative operator of coordinate, say*x*. So,

$$\widehat{x}\Psi =x\Psi $$ | 15 |

$$\widehat{d}=q\widehat{x}$$ | 16 |

$$\langle \Psi |\widehat{d}|\Psi \rangle =q\langle \Psi |\widehat{x}|\Psi \rangle $$ | 17 |

### Function operator

A function*f(x)*is operated on

*Ψ*. Again, we can express in terms of expectation value of, for example, of potential energy of diatomic molecule,

$$V(x)={D}_{e}{\left(1-{e}^{-a\left(x-{x}_{eq}\right)}\right)}^{2}$$ | 18 |

*D*is the dissociation energy and

_{e}*a*is a constant related to the force constant.

$$\langle \Psi |\widehat{V}\left(x\right)|\Psi \rangle $$ | 19 |

### Hamiltonian operator

Hamiltonian operator is to calculate the energy of the system. Since the total energy is expressed classically as $H=T+V$ where*T*is the kinetic energy and

*V*is the potential energy. The quantum mechanical expression in terms of operator is

*Hamiltonian operator*. We saw this already in Eqn. 1, but let me reiterate here, in terms of operator.

$$\widehat{H}=\widehat{T}+\widehat{V}=-\frac{{\hslash}^{2}}{2m}\frac{{\partial}^{2}}{\partial {w}^{2}}+V\left(w\right)$$ | 20 |

*m*is the mass. In general, the number of coordinates,

*w*, is 3 (or 4, if one includes spin) for a given particle, therefore, the partial derivative is used in the kinetic energy operator.

### Commutation relation

When two coordinate operators, say*x*and

*y*, are used to operate on a wave function, the order of the operator doesn't matter in obtaining eigenvalues.

$$\widehat{x}\widehat{y}\Psi =xy\Psi $$ | 21 |

$$\widehat{y}\widehat{x}\Psi =yx\Psi =xy\Psi $$ | 22 |

*x*and

*y*are just numbers. But when derivative operator is used, the order matters.

$$\widehat{d}\widehat{x}\Psi \ne \widehat{x}\widehat{d}\Psi $$ | 23 |

$$\widehat{d}\widehat{x}\Psi =x{\Psi}^{\prime}+\Psi \ne \widehat{x}\widehat{d}\Psi =x{\Psi}^{\prime}$$ | 24 |

*not commutative*. And if they are equal, then they are said to be

*commutative*.

$$\widehat{A}\widehat{B}-\widehat{B}\widehat{A}=0$$ | Commuting Operators | 25 |

$$\widehat{A}\widehat{B}-\widehat{B}\widehat{A}\ne 0$$ | Noncommuting Operators | 26 |

When two operators commute, simultaneous measurements corresponding to the eigenvalues of the two operators are possible. On the other hand, the two operators don't commute, one can not measure eigenvalues simultaneously.

The commutation relation is expressed with a square bracket, as

$$\left[\widehat{A},\widehat{B}\right]=\widehat{A}\widehat{B}-\widehat{B}\widehat{A}$$ | 27 |

$$\left[\widehat{x},\widehat{p}\right]\Psi =i\hslash \left(-x\frac{d}{dx}+\frac{d}{dx}x\right)\Psi =i\hslash \left(-x\frac{d\Psi}{dx}+x\frac{d\Psi}{dx}+\Psi \right)=i\hslash \Psi $$ | 28 |

$$\left[\widehat{x},\widehat{p}\right]=i\hslash $$ | 29 |

## Uncertainty principle

Using the result, Eqn. 29, Heisenberg was able to deduce the one of the most important concepts in quantum mechanics,*uncertainty principle*, which states that the product of the errors associated with the measurements of coordinate and speed must be greater than or equal to

*ℏ*/2.

$$\delta x\delta p\ge \frac{\hslash}{2}$$ | 30 |

The variance, square of error, in *x* and *p* are

$${\left(\delta x\right)}^{2}=\langle {\left(\widehat{x}-\langle x\rangle \right)}^{2}\rangle ={\displaystyle \int {\Psi}^{*}}{\left(\widehat{x}-\langle x\rangle \right)}^{2}\Psi dx$$ | C1 |

$${\left(\delta p\right)}^{2}=\langle {\left(\widehat{p}-\langle p\rangle \right)}^{2}\rangle ={\displaystyle \int {\Psi}^{*}}{\left(\widehat{p}-\langle x\rangle \right)}^{2}\Psi dx$$ | C2 |

*δx*is root-mean-square deviation in

*x*from the average value of

*x*, i.e. <

*x*>, and similar language can be used for

*p*. So, the deviation is a spread of the quantity under consideration. The relation tells us that when the two systems are prepared identically, and try to measure position and speed simultaneously in each system, you would not obtain the same measurement twice.

If we define Ψ_{x} and Ψ_{p} to be

$${\Psi}_{x}=\left(\widehat{x}-\langle x\rangle \right)\Psi $$ | C3 |

$${\Psi}_{p}=\left(\widehat{p}-\langle p\rangle \right)\Psi $$ | C4 |

*δp*, can be calculated as

$$\langle {\Psi}_{p}|{\Psi}_{p}\rangle ={\displaystyle \int {\Psi}_{p}^{*}{\Psi}_{p}dx={\displaystyle \int {\left[\left(\widehat{p}-\langle p\rangle \right)\Psi \right]}^{*}}}\left(\widehat{p}-\langle p\rangle \right)\Psi dx$$ | C5 |

$$={\displaystyle \int {\Psi}^{*}}{\left(\widehat{p}-\langle p\rangle \right)}^{2}\Psi dx={\left(\delta p\right)}^{2}$$ | C6 |

$$\langle {\Psi}_{x}|{\Psi}_{x}\rangle ={\left(\delta x\right)}^{2}$$ | C7 |

Now, what we want to do is to create a condition for inequality of certain type to exist.
Let us first make a linear combination of *Ψ _{x}* and

*Ψ*,

_{p}$$\Psi ={\Psi}_{x}-ik{\Psi}_{p}$$ | C8 |

$$\langle \Psi |\Psi \rangle \ge 0$$ | C9 |

Let us expand on this.

$$\langle \Psi |\Psi \rangle ={{\displaystyle \int \left({\Psi}_{x}-ik{\Psi}_{p}\right)}}^{*}\left({\Psi}_{x}-ik{\Psi}_{p}\right)dx\ge 0$$ | C10 |

$$={\displaystyle \int \left({\Psi}_{x}+ik{\Psi}_{p}\right)}\left({\Psi}_{x}-ik{\Psi}_{p}\right)dx$$ | C11 |

$$={\displaystyle \int \left[{\Psi}_{x}^{2}-ik\left({\Psi}_{x}^{*}{\Psi}_{p}-{\Psi}_{p}^{*}{\Psi}_{x}\right)+{k}^{2}{\Psi}_{p}^{2}\right]}dx$$ | C12 |

*x*and

*p*due to the fact that <

*x*> and <

*p*> are just numbers, they can be taken out of the integral, leaving operators of

*x*and

*p*. Hence,

$$\langle \Psi |\Psi \rangle ={\displaystyle \int \left[{\Psi}_{x}^{2}-ik\left[\widehat{x},\widehat{p}\right]+{k}^{2}{\Psi}_{p}^{2}\right]}dx={\displaystyle \int \left[{\left(\delta x\right)}^{2}+k\hslash +{k}^{2}{\left(\delta p\right)}^{2}\right]}dx\ge 0$$ | C13 |

*k*must also be positive.

$${\left(\delta x\right)}^{2}+k\hslash +{k}^{2}{\left(\delta p\right)}^{2}\ge 0$$ | C14 |

*k*to be applicable, the solution to have unique positive real solution in the following.

$$\frac{-\hslash \pm {\left({\hslash}^{2}-4{\left(\delta p\right)}^{2}{\left(\delta x\right)}^{2}\right)}^{1/2}}{2{\Psi}_{p}^{2}}$$ | C15 |

*D*, to have negative value, that is

$$D={\hslash}^{2}-4{\left(\delta p\right)}^{2}{\left(\delta x\right)}^{2}\le 0$$ | C16 |

$$\frac{{\hslash}^{2}}{4}\le {\left(\delta p\right)}^{2}{\left(\delta x\right)}^{2}$$ | C17 |

Coordinate and momentum are *conjugate variables* and they do not commute.
Energy and time are also conjugate variables, thus life time of the state and
energy are not simultaneously measurable. Fourier transformation connects
one variable to the other. The time domain signals in NMR can be Fourier
transformed into frequency domain to obtain NMR spectra.

In explaining the relation, Heisenberg used one of the greatest gedanken experiment. In this thought experiment is to show how simple act of observation can change the trajectory of quantum particle because of photon can interact with the quatnum particle.

**Figure 2. Heisenberg's gedanken experiment on trajectory of electron**

Act of looking requires turning a light on. In order to make sharper image of the quantum particle, one must send higher frequency light to the quantum particle. The image (or position) of the particle is now sharp, but because high frequency light would interact with the particle strongly and change the momentum of the particle enough to blurr the velocity of the particle. Opposite can also happen where in order to sharpen the velocity, the position is blurred with low frequency light.

Fundamentally, in the quantum regime, particle trajectory in the sense of classical mechanics lose its meaning. Act of merely looking at particles influence the behavior of quantum particles.