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Schrödinger Equation and The Wave Function

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).
Level 3
Level 3 requires the basics of vector calculus, differential and integral calculus. Suitable for undergraduates and high school students.

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 \(\mathit{\Psi}\) (*saai*).
  • "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.
The goal is to solve the Schrödinger differential equation and to find a concrete wave function for a concrete quantum mechanical problem using given initial conditions.

Classical Mechanics vs. Quantum Mechanics

Trajectory of a particle from classical mechanics. Its position \(\boldsymbol{r}(t)\) at three different points in time.

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:

Klassisches Teilchen oder Materiewelle, je nach Situation.
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.

Plane wave at time \(t=0\) and at a later time \(t\).

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.

Plane wave as rotating vector in the complex plane.

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 clockwise. 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:

  1. First, use the rewritten de-Broglie relation for the momentum \( p = \hbar \, k \) and replace \(k^2\) in 11:12\[ \frac{\partial^2 \mathit{\Psi}}{\partial x^2} ~=~ -\frac{p^2}{\hbar^2} \, \mathit{\Psi} \]
  2. 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\[ \frac{\partial^2 \mathit{\Psi}}{\partial x^2} ~=~ -\frac{2m}{\hbar^2} \, W_{\text{kin}} \, \mathit{\Psi} \]
  3. If you now look at the law of conservation of energy 8 multiplied by the wave function, you will see that \( W_{\text{kin}} \, \mathit{\Psi} \) occurs there. So solve the equation 13 for \(W_{\text{kin}} \, \mathit{\Psi}\):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: 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 \(t\).

In general, the probability to find the particle at a certain location can change over time: \(P(t)\).

Example of the squared magnitude. The probability \(P\) to find the particle between \(a\) and \(b\) corresponds to the enclosed area between \(a\) and \(b\).

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.

Example of the squared magnitude of a wave function. The area under the curve must be 1 when integrating from \(x=-\infty\) to \(x=+\infty\).

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 condition17\[ \int_{-\infty}^{\infty} |\mathit{\Psi}|^2 \, \text{d}x ~=~ 1 \]

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 \(b\):18\[ \int_{a}^{b} |\mathit{\Psi}|^2 \, \text{d}x ~=~ 1 \]

Example: Using the normalization conditionLet'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 the negative electrode is at \(x=0\) and the positive electrode 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. 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 \( |e^{i\,(kx – omega t)}| = 1 \). 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.

Wave function in classically allowed and forbidden regions

We can learn something about the behavior of the wave function from the Schrödinger equation without having solved it already. In the Schrödinger equation, bring the term with the potential energy to the left hand side and bracket the wave function: 20\[ (W - W_{\text{pot}}) \, \, \mathit{\Psi} ~=~ -\frac{\hbar^2}{2m} \, \frac{\partial^2 \mathit{\Psi}}{\partial x^2} \]

The potential energy \(W_{\text{pot}}(x)\) generally depends on the location \(x\). One could also call it potential energy function (or ambiguously but briefly: potential). It indicates the potential energy of a particle at the location \(x\). This function could be for example quadratic in \(x\) - called harmonic potential. But the potential energy function could also have a completely different behavior.

Here we look at an example of a quadratic potential energy function. If a particle is in this potential, then it has greater potential energy when it is further away from the origin. The potential energy function should be given before you can solve the Schrödinger equation.

According to the law of conservation of energy, the total energy \(W\) is a certain constant value regardless of where the particle is in this potential. In a diagram (see Illustration 7) it is a horizontal line that intersects our one-dimensional potential energy function \(W_{\text{pot}}(x)\) in two points \(x_1\) and \(x_2\). A classical particle can under no circumstances exceed this total energy! Consequently, it can only move between the reversal points \(x_1\) and \(x_2\). Between these two points, it can completely convert its kinetic energy into potential energy and vice versa - without moving outside of \(x_1\) and \(x_2\). The particle is trapped in this region.

Classically forbidden region (red) with negative kinetic energy and allowed region (blue) with positive kinetic energy.

If we find the particle outside of \(x_1\) and \(x_2\), its potential energy would be greater than its total energy. So the kinetic energy \(W - W_{\text{pot}}\) would be negative. A negative kinetic energy could have the particle only with an imaginary velocity. But imaginary velocity is not measurable, not physical. Therefore a negative kinetic energy is also not physical. This is exactly why we can expect that a classical particle can never be outside of \(x_1\) and \(x_2\). We call the region outside \(x_1\) and \(x_2\) the classically forbidden region. And the region within \(x_1\) and \(x_2\) as the classically allowed region.

In quantum mechanics, however, you often have the case that the wave function \(\mathit{\Psi} \) in the classically forbidden region is not zero. But if the wave function is not zero, the probability of finding the particle in the classically forbidden region is not zero either. This property of the wave function allows the particle to pass through regions that are classically forbidden. This behavior of the wave function is the basis for the quantum tunneling.

At first glance, this seems to be a serious contradiction, because if the wave function enters the forbidden region, the quantum mechanical particle can be found there with a certain probability. But there its potential energy is greater than its total energy. Consequently, the quantum mechanical particle would have to have a negative kinetic energy.

But this contradiction is resolved by the Heisenberg’s uncertainty principle: According to this principle, the potential and kinetic energy of a particle cannot be determined simultaneously with arbitrary accuracy. If the particle had a potential energy greater than its total energy, it can be calculated that the uncertainty in the measurement of kinetic energy is always at least as large as the energy difference \(W - W_{\text{pot}}\). This energy difference is the kinetic energy of a classical particle, but not of a quantum mechanical particle. In quantum mechanics you have to get rid of the idea that a quantum mechanical particle has an exact potential and exact kinetic energy simultaneously. Because of the uncertainty principle you cannot claim that the kinetic energy in the forbidden region becomes negative because \(W - W_{\text{pot}}\) IS NOT a kinetic energy. Therefore, a quantum mechanical particle can with a low probability be in the classically forbidden region without violating the principles of physics.

From the Schrödinger equation you can extract interesting information about the behavior of the wave function. This can be seen when you look at the signs of the energy difference \(W - W_{\text{pot}}\) and the wave function \(\mathit{\Psi} \) (see left hand side of Eq. 20).

Their two signs, together with the minus sign on the other side of the equation, determine the sign of the second spatial derivative of the wave function. The second spatial derivative \(\frac{\partial^2 \mathit{\Psi}}{\partial x^2}\) is called curvature.

You can visualize the curvature as follows: Imagine the wave function is a road that you want to ride along with a bicycle. A negative curvature means that the wave function bends to the right. You would have to steer your bicycle to the right. A positive curvature, on the other hand, means that the wave function curves to the left. You would therefore have to steer your bicycle to the left.

Let us first look at the two cases where the energy difference \(W - W_{\text{pot}}\) is positive. Here the total energy is greater than the potential energy. So we are in the classically allowed region.

  1. The wave function at location \(x\) is positive:20.1\[ \mathit{\Psi} ~>~ 0 \] Then the right hand side of Eq. 20 with the minus sign in front of it must also be positive to satisfy the equation. Consequently, the curvature at this location must be negative:20.2\[ \frac{\partial^2 \mathit{\Psi}}{\partial x^2} ~<~ 0 \] The curvature causes the positive wave function to always bend towards the \(x\)-axis. Even if it rises a little initially, this rise will become smaller and smaller until the wave function inevitably falls.
  2. The wave function at the location \(x\) is negative: 20.3\[ \mathit{\Psi} ~<~ 0 \]Then the curvature at this location must be positive: 20.4\[ \frac{\partial^2 \mathit{\Psi}}{\partial x^2} ~>~ 0 \]Again, the curvature causes the negative wave function to bend from below towards the \(x\)-axis.

If you compare the sign of the curvature with the sign of the wave function in these two classically allowed cases, you will see that they always have an opposite sign. In summary, this behavior results in an oscillation of the wave function around the \(x\)-axis.

In the classically allowed region between the locations \(x_1\) and \(x_2\) each wave function oscillates.
The wave function oscillates in the classically allowed region. In the forbidden region, it approaches zero for certain energies.

The higher the total energy \(W\) of the particle, the more the wave function oscillates. The matter wave then has a smaller de-Broglie wavelength. If the total energy is lower, the wave function oscillates less. The matter wave then has a larger de-Broglie wavelength.

Now let's look at the forbidden region and see how the wave function must behave there. In the forbidden region the potential energy is greater than the total energy. Its energy difference \(W - W_{\text{pot}}\) is therefore always negative.

In the classically forbidden region, the wave function and the curvature do not always have the opposite sign, but the same sign. Therefore the wave function is no longer forced to bend towards the \(x\)-axis. Instead, it can show two other behaviors. On the one hand it could grow into positive or negative infinity. But this behavior is not physical, because it violates the normalization condition 17. On the other hand, it can drop exponentially. This behavior is compatible with the normalization condition and therefore physically possible.

Energy quantization in harmonic potential \(W_{\text{pot}}(x)\).

If you take a closer look, you will notice that this behavior is only achieved for certain values of the total energy. This is called quantization, which means the fact, that the allowed total energies can only take discrete values. If you trap a quantum mechanical particle somewhere, as in our case between \(x_1\) and \(x_2\), the total energy of this particle is always quantized. It can then accept values \(W_0\), \(W_1\), \(W_2\), \(W_3\) and so on, but no energy values in between. For each of these allowed energies there is a corresponding wave function \(\mathit{\Psi}_0\), \(\mathit{\Psi}_1\), \(\mathit{\Psi}_2\), \(\mathit{\Psi}_3\) and so on.

The different possible wave functions and the corresponding allowed energies are numbered with an integer \(n\): \(\mathit{\Psi}_n\) and \(W_n\). Here \(n\) is a so-called quantum number.

We say: A particle with the smallest possible energy \(W_0\) is in the ground state \(\mathit{\Psi}_0\). And a particle that has an energy greater than \(W_0\) is in the excited state.In the classically allowed region the wave function oscillates and in the classically forbidden region the wave function drops exponentially. And the total energy of the trapped particle described by this wave function is quantized.

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 operator26\[ \hat{H} \, \mathit{\Psi} ~=~ W \, \mathit{\Psi} \]

Using the Hamilton operator, you formulated the Schrödinger equation as an eigenvalue equation, which you probably know from linear algebra. You apply the Hamilton 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 hamilton operator. These eigenvalues are discrete for most hamilton operators that you will encounter.

Time-dependent Schrödinger equation (1d and 3d)

What if the total energy \(W\) of the quantum mechanical particle is not constant in time? This can happen, for example, if the particle interacts with its environment and thus its total energy changes. For such problems, the time-independent Schrödinger equation 15 or 24 is not applicable. For this you need a more general form of the Schrödinger equation, the time-dependent Schrödinger equation

Now we assume a time-dependent total energy \(W(t)\). For simplicity we assume that the particle is not in an external field and therefore has no potential energy: \(W_{\text{pot}} = 0\). The total energy \(W\) of the particle is therefore only the time-dependent kinetic energy:27\[ W(t) ~=~ W_{\text{kin}}(t) \]

Multiply Eq. 27 by the wave function \(\mathit{\Psi}\):28\[ W \, \mathit{\Psi} ~=~ W_{\text{kin}} \, \mathit{\Psi} \]

Does the expression\(W_{\text{kin}} \, \mathit{\Psi}\) look familiar to you? You've already seen it in the derivation of the time-independent Schrödinger equation when we were looking at the second spatial derivative of the plane wave. Let’s use the resulting Eq. 14:29\[ W \, \mathit{\Psi} ~=~ -\frac{\hbar^2}{2m} \, \frac{\partial^2 \mathit{\Psi}}{\partial x^2} \]

Now we do some magic. Nobody has yet succeeded in deriving the time-dependent Schrödinger equation from fundamental principles. But the important thing is that it still works perfectly in experiments.

  1. Take the time derivative of the plane wave 7:30\[ \frac{\partial \mathit{\Psi}}{\partial t} ~=~ -\mathrm{i} \, \omega \, \mathit{\Psi} \]
  2. The total energy in our case corresponds to the kinetic energy and this can be written using the frequency \(\omega\) because of the wave-particle duality (analogous to the de-Broglie wavelength): \( W = \hbar \omega\). Use this equation to express the frequency \(\omega\) in 30 with the total energy:31\[ \frac{\partial \mathit{\Psi}}{\partial t} ~=~ -\frac{\mathrm{i}}{\hbar} \, W \, \mathit{\Psi} \]
  3. Rearrange Eq. 31 for \(W \, \mathit{\Psi}\):32\[ W \, \mathit{\Psi} ~=~ - \frac{\hbar}{\mathrm{i}} \, \frac{\partial \mathit{\Psi}}{\partial t} \]
  4. To beautify Eq. 32 a little, extend the fraction \(\frac{\hbar}{\mathrm{i}}\) with \(\mathrm{i}\). So \(\mathrm{i}^2\) becomes -1. This eliminates the fraction and the minus sign: 33\[ W \, \mathit{\Psi} ~=~ \mathrm{i} \, \hbar \, \frac{\partial \mathit{\Psi}}{\partial t} \]

Now our modified equation 33 is ready for insertion into \(W \, \mathit{\Psi}\) in 29:34\[ \mathrm{i} \, \hbar \, \frac{\partial \mathit{\Psi}}{\partial t} ~=~ -\frac{\hbar^2}{2m} \, \frac{\partial^2 \mathit{\Psi}}{\partial x^2} \]

And you have already obtained the time-dependent Schrödinger equation for a special case - for a particle without potential energy.

Erwin Schrödinger now assumed that the Eq. 34 is also fulfilled if the time-dependent potential energy \( W_{\text{pot}}(x,t)\), multiplied by the wave function, is added to the kinetic term in 34:

Time-dependent Schrödinger equation (1d)35\[ \mathrm{i} \, \hbar \, \frac{\partial \mathit{\Psi}}{\partial t} ~=~ -\frac{\hbar^2}{2m} \, \frac{\partial^2 \mathit{\Psi}}{\partial x^2} ~+~ W_{\text{pot}} \, \mathit{\Psi} \]

One-dimensional, time-dependent, Schrödinger equation has a similar form as the time-independent Schrödinger equation 15, with the only difference that the term for the total energy has changed.

Analogous to the one-dimensional time-independent equation 15, the time-dependent Schrödinger equation 35 can be extended to three dimensions. Just replace the second spatial derivative with the Laplace operator \(\nabla^2\):

Time-dependent Schrödinger equation (3d)36\[ \mathrm{i} \, \hbar \, \frac{\partial \mathit{\Psi}}{\partial t} ~=~ -\frac{\hbar^2}{2m} \, \nabla^2 \, \mathit{\Psi} ~+~ W_{\text{pot}} \, \mathit{\Psi} \]

Separation of variables and stationary states

Solving the time-dependent Schrödinger equation 35 is not that easy. But you can simplify the solving of this partial differential equation considerably if you convert it into two ordinary differential equations. One differential equation then depends only on time and the other only on space. The trick is called separation of variables, because you separate the space and time dependenies from each other. This is a very important approach in physics to simplify and solve differential equations.

The only requirement for variable separation is that the potential energy \( W_{\text{pot}}(x) \) does not depend on time \(t\) (but it may well depend on location \(x\)). The wave function itself, of course, can still depend on both location and time: \( \mathit{\Psi}(x,t)\).

Make the following variable separation. Divide the wave function \( \mathit{\Psi}(x,t) \) into two parts:

  • Into a part that depends only on the location \(x\). Let's denote this function as the small Greek letter \( \psi(x) \) (*saai*).
  • Into a part that depends only on time \(t\). Let's denote this function as the small Greek letter \( \phi(t) \) (*faai*).

Next, write the original wave function \(\mathit{\Psi}\), as a product of the two separated wave functions:37\[ \mathit{\Psi}(x,t) ~=~ \psi(x) \, \phi(t) \]

Not all wave functions can be separated in this way. But since the Schrödinger equation is linear, you can form a linear combination of such solutions and thus obtain all wavefunctions (even those that cannot be separated).

As you can see from the time-dependent Schrödinger equation 35, the time derivative \( \frac{\partial \mathit{\Psi}}{\partial t} \) and the second spatial derivative \( \frac{\partial^2 \mathit{\Psi}}{\partial x^2} \) occur there. Execute these two derivatives independently from each other using the product rule:

  1. Take the derivative of the separated wave function 37 with respect to time \(t\):38\[ \frac{\partial \mathit{\Psi}(x,t)}{\partial t} ~=~ \psi(x) \, \frac{\partial \phi(t)}{\partial t} \]
  2. Take the second derivative of the separated wave function 37 with respect to \(x\):39\[ \frac{\partial^2 \mathit{\Psi}(x,t)}{\partial x^2} ~=~ \phi(t) \, \frac{\partial^2 \psi(x)}{\partial x^2} \]

Now you have two equations: 38 and 39.

You can now insert the time derivative 38 and the space derivative 39 into the Schrödinger equation 35. Addionally insert the separated wave function 37 in the term with the potential energy in Eq. 35:40\[ \mathrm{i} \, \hbar \, \psi(x) \, \frac{\partial \phi(t)}{\partial t} ~=~ - \frac{\hbar^2}{2m} \, \phi(t) \, \frac{\partial^2 \psi(x)}{\partial x^2} ~+~ W_{\text{pot}}(x) \, \psi(x)\, \phi(t) \]

You can make a little plastic surgery here. We know that \(\phi(t)\) only depends on time and that \(\psi(x)\) only depends on space. You can imply this information by replacing the partial derivatives (noted with curved \( \partial \)) (*del* like in delta) with so-called total derivatives (noted with a regular \( \text{d} \)). This way you don't have to write \(\phi(t)\) or \(\psi(x)\) all the time, but can simply write \phi and \psi. From the notation of the derivative it is then clear that the function depends only on one variable, the one that‘s in the derivative. It is not important if you do not know what a total or partial derivative is. Just replace the \( \partial \)) symbols with regular \(d\) symbols: 41\[ \mathrm{i} \, \hbar \, \psi \, \frac{\text{d} \phi}{\text{d} t} ~=~ - \frac{\hbar^2}{2m} \, \phi \, \frac{\text{d}^2 \psi}{\text{d} x^2} ~+~ W_{\text{pot}} \, \psi \, \phi \]

Now you have to reformulate differential equation 41 so that its left hand side depends only on time \(t \) and its right hand side only on location \( x \). This is achieved by dividing the equation by the product \( \psi \, \phi \):42\[ \mathrm{i} \, \hbar \, \frac{1}{\phi} \, \frac{\text{d} \phi}{\text{d} t} ~=~ - \frac{\hbar^2}{2m} \, \frac{1}{\psi} \, \frac{\text{d}^2 \psi}{\text{d} x^2} ~+~ W_{\text{pot}} \]

What does that do for you? If you vary the time \( t \) (which only occurs on the left hand side), only the left hand side of the equation will change, while the right hand side remains unchanged. But if the right hand side does not change with time, it is constant. It is a real-valued constant, otherwise it would violate the normalization condition. This constant corresponds to the total energy \(W\) which is constant in time. Denote the right hand side as a constant \(W\):43\[ \mathrm{i} \, \hbar \, \frac{1}{\phi} \, \frac{\text{d} \phi}{\text{d} t} ~=~ W \]

Bring \( \mathrm{i} \, \hbar \) to the other side. Here \( \frac{1}{\mathrm{i}} \) becomes \( -\mathrm{i} \):

DEQ for time-dependent part of the total wave function 44\[ \frac{1}{\phi} \, \frac{\text{d} \phi}{\text{d} t} ~=~ -\frac{\mathrm{i} \, W}{\hbar}\]

The same applies for the space coordinate. If you vary \(x\) on the right hand side in 42, the left hand side remains constant because it is independent of \(x\). Because of the equality, the left hand side in 42 must correspond to the same constant \(W\). Denote the left hand side with the constant \(W\):45\[ W ~=~ - \frac{\hbar^2}{2m} \, \frac{1}{\psi} \, \frac{\text{d}^2 \psi}{\text{d} x^2} ~+~ W_{\text{pot}} \]

If you now multiply the differential equation 45 by \(\psi\), you get the time independent Schrödinger equation. In this version, the wave function \(\psi(x)\) depends only on \(x\) and not on time, because of the variable separation:

DEQ for space-dependent part of the total wave function 46\[ W \, \psi ~=~ - \frac{\hbar^2}{2m} \, \frac{\text{d}^2 \psi}{\text{d} x^2} ~+~ W_{\text{pot}} \, \psi \]

What did you achieve overall? As already mentioned: With this you have obtained two less complicated, ordinary differential equations 44 and 46 from a more complicated partial differential equation 35.

You can even immediately specify the solution for the temporal differential equation 44. Multiply Eq. 44 by \( \text{d}t \): 47\[ \frac{1}{\phi} \, \text{d} \phi ~=~ -\frac{\mathrm{i} \, W}{\hbar} \, \text{d} t \]

Now you just have to integrate both sides (we can omit the integration constants and include them in \(\psi(x)\)):48\[ \int \frac{1}{\phi} \, \text{d} \phi ~=~ -\frac{\mathrm{i} \, W}{\hbar} \int \, \text{d} t \]

On the left hand side the integration of \(\frac{1}{\phi}\) yields the natural logarithm and on the right hand side the integration yields\(t\):49\[ \ln(\phi) ~=~ -\frac{\mathrm{i} \, W}{\hbar} \, t \]

Rearranging for the searched function \(\phi\) gives the solution of the differential equation 44:

Solution for the time-dependent part 50\[ \phi(t) ~=~ \mathrm{e}^{-\frac{\mathrm{i} \, W}{\hbar}t}\]

You cannot solve the second, space-dependent differential equation 46, that is the stationary Schrödinger equation, without a given potential energy function \( W_{\text{pot}} \). But it’s ok. With the separation of variables we have simplified the solving process a lot.

The great thing is now: Instead of solving a more complicated time-dependent Schrödinger equation 35, you can solve the stationary Schrödinger equation 46. This way you only get the space-dependent part \(\psi(x)\) of the whole wave function.

But, if you look at the separation ansatz 37, you just have to multiply the space-dependent part \( \psi(x)\) with the time-dependent part \(\phi(t)\) to get the total wave function \(\mathit{\Psi}(x,t)\). And for \(\phi(t)\) you have found that it is an exponential function 50. (This time dependent part is the same for all seperable wave functions you will encounter):

Time-dependent total wave function 51\[ \mathit{\Psi}(x,t) ~=~ \psi(x) \, \mathrm{e}^{-\frac{\mathrm{i} \, W}{\hbar}t}\]

A wave function, which can be separated into a space and time dependent functions, describes a stationary state. By stationary we mean that the wave function \(\mathit{\Psi} \) itself is time-dependent, but its squared magnitude \(|\mathit{\Psi}|^2\) is not! All other physical quantities describing the particle are also time-independent. For example, a particle whose wave function is a stationary state has a constant mean value of energy \(\langle W\rangle\), constant mean value of momentum \(\langle p\rangle\), and so on.

The Schrödinger equation is not generally applicable. It is a non-relativistic equation. This means that it fails for quantum mechanical particles that move almost at the speed of light.

Furthermore, it does not naturally take into account the spin of a particle. All these problems are only solved by the more general equation of quantum mechanics, by the Dirac equation.

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