# Schrödinger Equation explained

**Schrödinger equation**- is a homogeneous, partial differential equation of second order which can be used to describe a quantum mechanical system (e.g. an electron or atom).

## Requirements

- Basics of differential and integral calculus
- Basics about the complex numbers
- Basics about the vectors
- Energy conservation

This is how the **time-independent Schrödinger equation** looks like:`\[ W \, \mathit{\Psi} ~=~ -\frac{\hbar^2}{2m} \, \frac{\partial^2 \mathit{\Psi}}{\partial x^2} ~+~ W_{\text{pot}} \, \mathit{\Psi} \]`

And this is how the **time-dependent Schrödinger equation** looks like:`\[ \mathrm{i} \, \hbar \, \frac{\partial \mathit{\Psi}}{\partial t} ~=~ -\frac{\hbar^2}{2m} \, \frac{\partial^2 \mathit{\Psi}}{\partial x^2} ~+~ W_{\text{pot}} \, \mathit{\Psi} \]`

We can already state that the Schrödinger equation is - mathematically speaking - a **partial differential equation of second order**.

- "
*differential equation*" means, that the searched quantity is not a variable, but a*function*and in the equation there are derivatives of this function. The function we are looking for in the Schrödinger equation is the so-called wave function. - "
*partial*" means that the equation contains derivatives with respect to multiple variables, such as derivative with respect to location \(x\) and with respect to time \(t\). - And "
*second order*" means that the highest derivative that occurs in the differential equation is of*second order*.

## Classical Mechanics vs. Quantum Mechanics

Most phenomena of our everyday life can be described by classical mechanics. The goal of classical mechanics is to determine how a body of mass \(m\) moves over time \(t\). We want to determine the trajectory, that is the path \(\boldsymbol{r}(t)\) of this body. In classical mechanics the trajectory allows us to predict where this body will be at any given time. For example we are able to describe...

- the movement of our earth around the sun
- the movement of a satellite around the earth
- the launch of a rocket
- the movement of a swinging pendulum
- the movement of a thrown stone.

These are all classical motions that can be calculated with the help of Newton’s second law of motion. So with the equation: \(F = m\,a\) or for the experts among you, with the differential equation: \(m \, \frac{\text{d}^2 \boldsymbol{r}}{\text{d}t^2} = - \nabla W_{\text{pot}}\). By solving this differential equation you can find the trajectory you are looking for for a specific problem. To solve this differential equation at all, the potential energy function \(W_{\text{pot}}\) must of course be given. The initial conditions characterizing the problem that you want to solve, must also be known. For example, if you describe the motion of a particle, then an initial condition could be the position and velocity of the particle at time zero: \(\boldsymbol{r}(0) = (0,0,0)\) und \(\boldsymbol{v}(0) = (0,0,0)\).

Once you have determined the trajectory \(\boldsymbol{r}(t)\) by solving the differential equation, you can then use it to find out all other relevant quantities, such as:

- the particle's velocity: \(\boldsymbol{v} = \frac{\text{d}\boldsymbol{r}}{\text{d}t}\)
- its momentum: \(\boldsymbol{p} = m \, \boldsymbol{v}\)
- its kinetic energy: \(W_{\text{kin}} = \frac{1}{2}\,m\,v^2\).

In the atomic world, however, classical mechanics does not work. The tiny particles here, like electrons, do not behave like *classical* point-like particles under all conditions, but they can also behave like waves. Because of this wave character, the location \(\boldsymbol{r}(t)\) of an electron cannot be determined precisely because a wave is not concentrated at a single location. And, if we try to squeeze it to a fixed location, the momentum can no longer be determined exactly. This behavour is described by the *uncertainty principle*; a fundamental principle of quantum mechanics, that cannot be bypassed. So we cannot determine a trajectory \(\boldsymbol{r}(t)\) of the electron as in classical mechanics and derive all other motion quantities from this trajectory. Instead we have to find another way to describe the quantum world. And this other way is the development of quantum mechanics and the Schrödinger equation.

In quantum mechanics you do not calculate a trajectory \(\boldsymbol{r}(t)\), but a so-called wave function \(\Psi\). This is a function that generally depends on the location \(\boldsymbol{r}\) and the time \(t\). Where now the location \(\boldsymbol{r}\) is a space coordinate (and not an unknown trajectory). The tool with which we can find the wave function is the Schrödinger equation.

## Applications

It is only through this novel approach to nature using the Schrödinger equation that humans have succeeded in making part of the microcosm controllable. As a result, humans are now able to build *lasers* that are indispensable in medicine and research today. Or scanning *tunneling microscopes*, which significantly exceed the resolution of conventional light microscopes. It was only through the Schrödinger equation that we were able to fully understand the *periodic table* and *nuclear fusion in our sun*. This is only a small fraction of the applications that the Schrödinger equation has given us. So let us first find out, where this powerful equation comes from.

## Derivation of the time-independent Schrödinger equation (1d)

Unfortunately it is not possible to derive the Schrödinger equation from classical mechanics alone. But it can be derived, for example, by including the *wave-particle duality*, which does not occur in classical mechanics. However, experiments and modern technical society show that the Schrödinger equation works perfectly and is applicable to most quantum mechanical problems. Let us try to understand the fundamental principles of the Schrödinger equation and how it can be derived from a simple special case.

We make our lives easier by first looking at a one-dimensional movement. In one dimension a particle can only move along a straight line, for example along the spatial axis \(x\).

Now consider a particle of mass \(m\) flying with velocity \(v\) in \(x\)-direction. Because the particle moves, it has a kinetic energy \( W_{\text{kin}} \). It is also located in a *conservative * field, for example in a gravitational field or in the electric field of a plate capacitor. Conservative means: When the particle moves through the field, the total energy \(W\) of the particle does not change over time. Consequently, the **energy conservation law ** applies and a potential energy, lets call it \( W_{\text{pot}} \), can be assigned to the particle.

The total energy \(W\) of the particle is then the sum of the kinetic and potential energy:`1\[ W ~=~ W_{\text{kin}} ~+~ W_{\text{pot}} \]`

This is nothing new, you already know this from classical mechanics. The energy conservation law is a fundamental principle of physics, which is also fulfilled in quantum mechanics in modified form. The weirdness of quantum mechanics is added by the wave-particle duality. This allows us to regard the particle as a **matter wave**. A matter wave characterized by the **de-Broglie wavelength**:

`2\[ \lambda ~=~ \frac{h}{p} \]`

\(p\) is the *momentum* of the particle, which is the product of the velocity \(v\) and the mass \(m\) of the particle. And \(h\) is the *Planck constant*, a natural constant that appears in many quantum mechanical equations. By the way: Because of its tiny value of only \( 6.626 \cdot 10^{-34} \, \text{Js} \) it is understandable why we do not observe quantum mechanical effects in our macroscopic everyday life.

According to the wave-particle duality, we can regard a particle as a wave and assign physical quantities to this particle that are actually only intended for waves, such as the wavelength in this case.

In quantum mechanics it is common practice to express the momentum \(p = \frac{h}{\lambda} \) not with the de-Broglie wavelength, but with the wavenumber \( k = \frac{2\pi}{\lambda} \). Thus the momentum becomes: \(p = \frac{h \, k}{2\pi} \). And \( \frac{h}{2\pi} \) is defined as a reduced planck constant \(\hbar\) („h bar“). So we can write the momentum more compact as:`3\[ p ~=~ \hbar \, k \]`

We will need this equation later.

The de-Broglie wavelength 2

is also a measure of whether the object behaves more like a *particle* or a *wave*. Particle-like behaviour can be described by classical mechanics. More exciting is the case, when the particle behaves like a wave. To distinguish it from classical, point-like particles, such an object is called a **quantum mechanical particle**. The larger the de-Broglie wavelength 2

, the more likely the object behaves quantum mechanically. A particle has a larger de Broglie wavelength if it has a smaller momentum \(p\). In other words, smaller mass and velocity. Perfect candidates for such quantum mechanical particles are electrons. They have a tiny mass\( m_{\text e} = 9.1 \cdot 10^{-31} \, \text{kg} \) and their velocity can be greatly reduced by means of electric voltage or cooling in liquid hydrogen.

Thus the classical particle behaves more like an extended matter wave, which can be described mathematically with a **plane wave**. We call it by the capital Greek letter \(\mathit{\Psi}\). A plane matter wave generally depends on the location \(x\) and the time \(t\): \(\mathit{\Psi}(x,t)\).

You can describe a plane wave, which has the wave number \(k\), frequency \(\omega\) and amplitude \(A\), by a cosine function:`4\[ \mathit{\Psi}(x,t) ~=~ A \, \cos(k\,x - \omega \, t) \]`

It does not matter whether you express the plane wave with sine or cosine function. You might as well have used sine. When time \(t\) advances, the wave moves in the positive \(x\)-direction, just like our considered particle.

In order to do math with such waves without using any addition theorems, we transform the plane wave 3

into a complex exponential function. Add to the cosine function the imaginary sine function:`5\[ \mathit{\Psi}(x,t) ~=~ A \, \left[ \cos(k\,x - \omega \, t) + \mathrm{i} \,\sin(k\,x - \omega \, t)\right] \]`

You have thus transformed a real function 4

into a complex function 5

. Where \( \mathrm{i} \) is the imaginary unit, \(\text{Re}(\mathit{\Psi}) = \cos(k\,x - \omega \, t) \) is the *real part* and \(\text{Im}(\mathit{\Psi}) = \sin(k\,x - \omega \, t) \) is the *imaginary part* of the complex function \( \mathit{\Psi}\).

The good thing is that we can take advantage of the enormous benefits of complex notation and then declare that in the experiment we are only interested in the real part 3

(cosine function). In our case, we can simply label the imaginary part as non-physical and just ignore it. However, keep in mind that a complex plane wave 5

can also be a possible solution to the Schrödinger equation. But we will deal with this later.

In the next step we use the *Euler relationship* from mathematics:`6\[ A \, e^{\mathrm{i}\,\varphi} ~=~ A \, \left[ \cos(\varphi) + \mathrm{i}\,\sin(\varphi)\right] \]`It connects the complex exponential function \(e^{\mathrm{i}\,\varphi}\) with Cosine and Sine. So that's exactly what you need right now. Because, with it you can convert the complex plane wave to an exponential function:`7\[ \mathit{\Psi}(x,t) ~=~ A \, e^{\mathrm{i}\,(k\,x - \omega\,t)} \]`

And you have already represented a plane wave in a complex exponential notation. Whenever you see a term like 7

, you know it's a plane wave. Remember: Our original plane wave 4

as a cosine function is contained in the complex function as information, namely as the real part of this function.

You can easily illustrate the complex exponential function 7

(see Illustration 3). It is a *vector* in the complex plane: And

- The amplitude \(A\) corresponds to the magnitude of the vector (that is its length).
- the argument \( k\,x - \omega\,t \) corresponds to the
*angle*\(\varphi\) (also called*phase*) enclosed between the real axis and the \(\mathit{\Psi}\) vector.

When the time \(t\) passes, the angle \(\varphi = k\,x - \omega\,t \) changes and the vector rotates in the complex plane - in our case counterclockwise. This rotation represents the propagation of the plane wave in the positive \(x\)-direction.

The complex exponential function 7

is a *function* that describes a plane *wave*. Therefore it is also called **wave function**, especially in connection with quantum mechanics. Often the wave function \(\mathit{\Psi }\) is also called the state of the particle. The particle is in the state \(\mathit{\Psi}\). So always remember: When we talk about state in quantum mechanics, we mean the wave function. Of course, there are different states that different particles can take under different conditions. The plane wave is just one simple example of a possible state.

Next, multiply the equation 1

for the total energy by the wave function 7

. In this way, you combine the law of conservation of energy and the wave-particle duality inherent in the wave function:`8\[ W \, \mathit{\Psi} ~=~ W_{\text{kin}} \, \mathit{\Psi} ~+~ W_{\text{pot}} \, \mathit{\Psi} \]`

But this equation does not help you much yet. You still have to find a way to convert it into a differential equation.

A plane wave is a typical wave that appears in optics and electrodynamics when describing electromagnetic waves. And from there we know that a plane wave is a possible solution of the wave equation. A one-dimensional plane wave, as in our case, solves the one-dimensional wave equation:`9\[ \frac{\partial^2 \mathit{\Psi}}{\partial x^2} ~=~ \frac{1}{c^2} \, \frac{\partial^2 \mathit{\Psi}}{\partial t^2} \]`

\(c\) is the phase velocity of the wave. In the case of electromagnetic waves it is the speed of light. In the case of matter waves it is the phase velocity \( c = \frac{\omega}{k}\). But for us this is not important for the time being. I quickly want to show you the wave equation to motivate our next step. On the left side of the wave equation is the second derivative of the wave function with respect to \(x\). So let's calculate the second derivative of our plane wave:`10\[ \frac{\partial^2 \mathit{\Psi}}{\partial x^2} ~=~ -k^2 \, A \, e^{\mathrm{i}\,(k\,x - \omega\,t)} \]`

The second derivative adds \(k^2\) and a minus sign. The minus sign because \(i^2 = -1\). The wave function as an exponential function remains unchanged with derivation - as you hopefully know:`11\[ \frac{\partial^2 \mathit{\Psi}}{\partial x^2} ~=~ -k^2 \, \mathit{\Psi} \]`

Next, we again make steps, which at first sight appear to be arbitrary, but in the end they will lead us to the Schrödinger equation. We will somehow try to connect the second derivative 11

of the wave function with the conserved total energy 8

:

- First, use the rewritten de-Broglie relation for the momentum \( p = \hbar \, k \) and replace \(k^2\) in
11

:`12\[ \frac{\partial^2 \mathit{\Psi}}{\partial x^2} ~=~ -\frac{p^2}{\hbar^2} \, \mathit{\Psi} \]` - Now, to bring the kinetic energy \(W_{\text{kin}}\) into play, replace the momentum \(p^2\) with the help of the relation: \( W_{\text{kin}} = \frac{p^2}{2m} \). You can easily obtain this form from \( W_{\text{kin}} = \frac{1}{2}\,m\,v^2 \) by rearranging the momentum \( p = m\,v\) and inserting it into velocity \(v\):
`13\[ \frac{\partial^2 \mathit{\Psi}}{\partial x^2} ~=~ -\frac{2m}{\hbar^2} \, W_{\text{kin}} \, \mathit{\Psi} \]` - If you now look at the law of conservation of energy
8

multiplied by the wave function, you will see that \( W_{\text{kin}} \times \mathit{\Psi} \) occurs there. So solve the equation13

for \(W_{\text{kin}} \times \mathit{\Psi}\):`14\[ W_{\text{kin}} \, \mathit{\Psi} ~=~ -\frac{\hbar^2}{2m} \, \frac{\partial^2 \mathit{\Psi}}{\partial x^2} \]`

Now if you just insert 14

into the law of conservation of energy 8

, you get the Schrödinger equation:

**Zeitunabhängige Schrödinger-Gleichung (1d)**

`15\[ W \, \mathit{\Psi} ~=~ -\frac{\hbar^2}{2m} \, \frac{\partial^2 \mathit{\Psi}}{\partial x^2} ~+~ W_{\text{pot}} \, \mathit{\Psi} \]`

This Schrödinger equation is one-dimensional and time-independent. You can recognize the one-dimensionality immediately by the fact that only the derivative with respect to a single space coordinate occurs. In this case with respect to \(x\). And you can recognize the time independence of the Schrödinger equation by the fact that a constant total energy \(W\) occurs. In general, however, the wave function \(\mathit{\Psi}\) may be time-dependent: \(\mathit{\Psi}(x,t)\).

Let's recap for a moment. To derive the Schrödinger equation 15

, we have combined the *law of conservation of energy* and the *wave-particle duality*; introducing the wave-particle duality by assuming a plane matter wave.

So you could say that the time-independent Schrödinger equation is the **energy conservation law of quantum mechanics**. This term stands for total energy, this one for kinetic energy and this one for potential energy.

## Squared magnitude, probability and normalization

Let's assume that you have solved the Schrödinger equation and found a specific wave function. It doesn't matter how exactly you did it. Of course, depending on the problem, you will generally not get a plane wave. The found wave function can also be a complex function. So you can not just neglect the imaginary part of it, as we agreed on in the beginning with our plane wave. By omitting the imaginary part, the result of the Schrödinger equation would no longer agree with the results of experiments. For an experimentalist, however, such complex functions are quite bad because they cannot be measured. But how can you check your calculation in an experiment if the complex wave function cannot be measured at all? What does the wave function actually mean?

Here the predominant statistical interpretation of quantum mechanics comes into play, the so-called Copenhagen interpretation. It does not say what the wave function \(\mathit{\mathit{\Psi}}\) means, but it interprets its **square of the magnitude**: `\[ |\mathit{\mathit{\Psi}}|^2 \]`

By forming the square of the magnitude \(|\mathit{\mathit{\Psi}}|^2\) you get a *real-valued* function. That is a function measurable for the experimentalist.

In addition, the square of the magnitude is always positive, so there is no reason why it should not be interpreted as **probability density**. Because as you know: Probabilities are always positive, never negative. In the one-dimensional case, the square of magnitude would then be a *probability per length* and in the three-dimensional case a *probability per volume*.

Let's stay with the one-dimensional case and the wave function of a particle: If you integrate the probability density, that is the squared magnitude of the wave function, over the location \(x\) within the length between \(x = a\) and \( x = b\), then you get a *probability* \(P(t)\):`16\[ P(t) ~=~ \int_{a}^{b} |\mathit{\Psi}(x,t)|^2 \, \text{d}x \]`

The integral of the squared magnitude \(|\mathit{\Psi}(x,t)|^2\) indicates the probability \(P(t)\) that the particle is in the region between \(a\) and \(b\) at the time \(t\).

If you plot the squared magnitude \(|\mathit{\Psi}(x,t)|^2\) against \(x\), you can read out two pieces of information from it:

- The probability \(P\) is the area under the \(|\mathit{\Psi}(x,t)|^2\)-curve.
- The most likely way to find the particle is to find it at the maxima. Most unlikely at the minima.

Note, however, that it is not possible to specify the probability of the particle being at a particular location \(x = a\), but only for a *space region* (here between \(a\) and \(b\)), because otherwise the integral would be zero. Obviously! Because there are * infinitely many* space points on the distance between \(a\) and \(b\). If each of these space points had a finite probability, then the sum (that is the integral) of all the probabilities would be infinite, which would make no sense at all. Therefore, we always calculate the probability to find the particle in a specific *region* of space.

In order for the statistical interpretation to be compatible with the Schrödinger equation, the solution of the Schrödinger equation, that is the wave function \( \mathit{\Psi} \) must satisfy the so-called **normalization condition**. This means that the particle must exist somewhere in space. In one-dimensional case, it must therefore be found one hundred percent somewhere between \(x = -\infty\) and \(x = +\infty\). In other words, the integral for the probability, integrated over the entire space, must be 1:

**Normalization condition**

`17\[ \int_{-\infty}^{\infty} |\mathit{\Psi}|^2 \, \text{d}x ~=~ 1 \]`

The **normalization condition** is a necessary condition that every physically possible wave function must fulfill. After solving the Schrödinger equation, the found wave function \(\mathit{\Psi}\) must be normalized using the normalization condition 17

. Normalizing means that you must calculate the integral 17

and then determine the amplitude of the wave function so that the normalization condition is satisfied. The normalized wave function then remains normalized for all times \(t\). If this were(konjunktiv was vs were?) not the case, the Schrödinger equation and the statistical interpretation would be incompatible. There are of course solutions to the Schrödinger equation, for example \(\mathit{\Psi} = 0 \), which cannot be normalized. Such solutions are unphysical. They do not exist in reality. By the way: Wave functions that can be normalized are called **square-ingrable functions** in mathematics.

If you know with one hundred percent that the particle is located between \(a\) and \(b\), then you must reduce the normalization condition accordingly to the region between \(a\) and \(b\) (the general normalization condition 17

of course still applies):`18\[ \int_{a}^{b} |\mathit{\Psi}|^2 \, \text{d}x ~=~ 1 \]`

**Example: Using the normalization condition**Let's take a specific example of how a wave function is normalized. Let us consider a simple one-dimensional case. An electron moves straight from the negative electrode to the positive electrode of a plate capacitor. The two electrodes have the distance \(d\) to each other. You have determined the wave function by solving the Schrödinger equation:

`18.1\[ \mathit{\Psi}(x,t) ~=~ A \, e^{\mathrm{i}\,(k\,x - \omega\,t)} \]`

The amplitude \(A\) is unknown. Therefore you use the normalization condition to normalize the wave function and determine \(A\) at the same time. You know that with one hundred percent probability the electron must be between the two electrodes. If one electrode is at \(x=0\) and the other at \(x=d\), then the electron is somewhere between these two points:`18.2\[ \int_{0}^{d} |\mathit{\Psi}(x)|^2 \, \text{d}x ~=~ 1 \]`

First you have to determine the squared magnitude \(|\mathit{\Psi}(x)|^2\). The magnitude of the wave function is formed in the same way as the magnitude of a vector. This is the first time the usefulness of the complex exponential function comes into play. It is always true that \( |e^{i\,\varphi}| = 1 \) is equal to 1. Perfect. You don't have to do complicated math. So the squared magnitude of the wave function 18.1

is:`18.3\[ |\mathit{\Psi}|^2 ~=~ A^2 \]`

Insert the squared magnitude 18.3

into the normalization condition 18.2

:`18.4\[ \int_{0}^{d} A^2 \, \text{d}x ~=~ 1 \]`

The amplitude \(A\) is independent of \(x\), so it is a constant and you can put it before the integral. Integrate. Insert die integration limits. Rearrange for amplitude and you get:`18.5\[ A ~=~ \frac{1}{\sqrt{d}} \]`

In this way you normalize the wave function and determine the amplitude for a given problem.

In the example of the normalization condition, you can see from the amplitude 18.5

that it has the unit "one over square root of meter". Because the exponential function is dimensionless the wave function has the same unit as the amplitude:`19\[ [\mathit{\Psi}] ~=~ \frac{1}{\sqrt{\text m}} \]`In three dimensions the unit of the wavefunction is \(\frac{1}{\sqrt{\text{m}^3}}\).

Once you have determined the wave function by solving the Schrödinger equation, you can use it to find out not only the probability of the particle's location, but also the mean value of the location \(\langle x\rangle\) and that of all other physical quantities. For example the mean value of the momentum \(\langle p \rangle\), the velocity \(\langle v\rangle\) or kinetic energy \(\langle W_{\text{kin}} \rangle\). In quantum mechanics, the mean value is written in angle brackets. Why only a mean value and not an exact value and how this can be determined you will learn in detail in another video. Important for you is to know that you can describe a quantum mechanical particle with the wave function as well as you can describe a classical particle with the trajectory.

## Time-independent Schrödinger equation (3d)

You can generalize the *one-dimensional* Schrödinger equation 15

to a *three-dimensional* Schrödinger equation. We assume that the wave function \(\Psi(x,t)\) depends not only on one spatial coordinate \(x\) but on three spatial coordinates \(x,y,z\): \(\Psi(x,y,z,t)\). You can also combine the three space coordinates more compactly to a vector \(\boldsymbol{r}\) (vectors are shown in bold here): \(\Psi(\boldsymbol{r},t)\).

In the one-dimensional Schrödinger equation 15

, you have to add the second derivative with respect to \(y\) and \(z\) to the second derivative with respect to \(x\), so that all three spatial coordinates occur in the Schrödinger equation. This is what the multidimensional analysis tells you to do, if you want to convert the one-dimensional Schrödinger-equation into the three-dimensional Schrödinger-equation:`21\[ W \, \mathit{\Psi} ~=~ -\frac{\hbar^2}{2m} \, \left( \frac{\partial^2 \mathit{\Psi}}{\partial x^2} + \frac{\partial^2 \mathit{\Psi}}{\partial y^2} + \frac{\partial^2 \mathit{\Psi}}{\partial z^2} \right) ~+~ W_{\text{pot}} \, \mathit{\Psi} \]`

This is our time-independent three-dimensional Schrödinger equation. You can write it more compactly. Bracket the wave function:`22\[ W \, \mathit{\Psi} ~=~ -\frac{\hbar^2}{2m} \, \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \right)\,\mathit{\Psi} ~+~ W_{\text{pot}} \, \mathit{\Psi} \]`

The sum of the spatial derivatives in the brackets form a so-called *Laplace operator* \(\nabla^2\) (Nabla squared. Sometimes also noted as \(\Delta\)):`23\[ \nabla^2 ~=~ \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \]`

An *operator*, like the Laplace operator, only has an impact when applied to a function. Because an isolated spatial derivate 23

makes no sense. Here you apply the Laplace operator to the wave function \(\mathit{\Psi}\):

**Time-independent Schrödinger equation (3d)**

`24\[ W \, \mathit{\Psi} ~=~ -\frac{\hbar^2}{2m} \, \nabla^2 \, \mathit{\Psi} ~+~ W_{\text{pot}} \, \mathit{\Psi} \]`

The result \(\nabla^2 \, \mathit{\Psi}\) gives the second spatial derivative of the wave function, that is exactly what we had before in 21

. With the three-dimensional time-independent Schrödinger equation 24

, many time-independent problems of quantum mechanics can be solved, be it a particle in the potential well, quantum mechanical harmonic oscillator, the description of a helium atom and many other problems.

Do you see another possible operator on the right hand side in 24

? Bracket the wave functions:`25\[ W \, \mathit{\Psi} ~=~ \left( -\frac{\hbar^2}{2m} \, \nabla^2 ~+~ W_{\text{pot}} \right) \, \mathit{\Psi} \]`

The operator in the brackets on the right hand side is called **Hamiltonian operator** \(\hat{H}\) or just Hamiltonian. You can use it to write the Schrödinger equation very compactly:

**Time-independent Schrödinger equation using Hamilton operator**

`26\[ W \, \mathit{\Psi} ~=~ \hat{H} \, \mathit{\Psi} \]`

Using the Hamiltonian operator, you formulated the Schrödinger equation as an eigenvalue equation, which you probably know from linear algebra. You apply the Hamiltonian operator (imagine it as a matrix) to the eigenfunction \(\mathit{\Psi}\) (imagine it as an eigenvector). Then you get the eigenvector \(\mathit{\Psi}\) again unchanged, scaled with the corresponding energy eigenvalue \(W\). With this eigenvalue problem you can mathematically see why the energy \(W\) can be quantized in quantum mechanics. The energy eigenvalues depend on the hamiltonian operator. These eigenvalues are discrete for most hamiltonian operators that you will encounter. As you can see, the energy \(W\) can generally not be arbitrary!