A particle is described by the wave function ψ ( x ) = { c e x / L x ≤ 0 mm c e − x / L x ≥ 0 mm \psi (x)=\begin{cases} ce^{x/L} & x\leq 0\text{ mm} \\ ce^{-x/L} & x\geq 0\text{ mm} \end{cases} mm where L = 2.0 mm. Determine the normalization constant c.

Hi everyone. Let's take a look at this practice problem dealing with the normalization constant of a wave function. So in this problem, we have a particle and it's confined to the region of X is equal to negative 10 millimeters to X equals positive 10 millimeters. And it's given by the wave function um I is equal to sin not E to the negative absolute value of X divided by D where sin not here is the normalization constant and D is equal to four millimeters. And the question wants us to determine sin not. So we're given four choices for our answers. Voice A is 0.25 millimeters to the negative one half. Choice B is 0.33 millimeters to the negative one half oc is 0.5 millimeters to the negative one half and choice D 0.66 millimeters to the negative one half. To begin with, we need to recall um our condition for normalization. And so basically, we take over all space going from minus infinity to infinity the modules squared of our wavefunction. So I have the modulus squared of P si and that needs to be multiplied by DX because the um sites, the module squared of S ID X is our probability. And if we integrate over all space, that means that we have to find a, our particle at some point. And so that means that this integral has to equal one. Now we need to do is um plug in our function for Psy. And we can also change our limits of integration as well since the wave function is only defined between negative 10 millimeters and 10 millimeters. And so outside of that, our wave function is zero. And if you integrate zero, you will always get zero. So we'll have the limited integration going from negative 10 millimeters you 10 millimeters. And for I will plug in our actual function. Um I here is a real value function. So we don't need to worry about the complex conjugate. So I'm just gonna be squaring what I have here for Psy. So I have the quantity of thy knot E to the minus absolute value of X divided by D. That whole quantity is squared multiplied by DX and this equals one. Now, I'd like to get rid of that absolute value um sign in the exponent there. And I noticed that I have an even function here. So that means I can change my limits of integration if I multiply this integral by two. And so I can now integrate from 0 to 10 millimeters and I'll have the quantity of I not E to the minus X over D. So I can get rid of the absolute value there because my X values are all positive at this point. And that whole quantity gets squared multiplied by DX equals one. So now I can actually um look at possibly performing this integration. So I'm going to pull out this constant term of sin not. So my normalization constant is just a number doesn't depend on X. So it can move outside the integral, I'll have two thy not squared multiplying the integral of zero uh 10 millimeters of E to the minus two X divided by D multiplied by DX. And this all has the 01. So let me perform the integration here and I'm actually gonna plug in the value for D in the next step. So I'll have two thy not squared. And so when I integrate E to the negative two X divided by D I get a factor of negative D over two. So I'll have negative four millimeters divided by two. And that's gonna be multiplying the quantity. So now I need to plug in my upper limit. So that's gonna be E to the minus two multiplying 10 millimeters divided by the four millimeters for D minus the lower limit. So E to the zero is just one and so all that has to equal one. So at this point, I just use a software cy not so I'll have day night is equal to some things cancel out here. So this will cancel out with this factor of one half. And so what I'm going to be left with is the square root Versailles of one divided by the quantity of four millimeters multiplied by the quantity of one minus E to the negative five. So all the units here cancel out in the exponent and I'm left with just a pure number which is five. And so when I actually calculate this in my calculator, that turns out to have a sin not to be equal to 0.5 millimeters to the negative one half power. So that's the answer that I'm looking for. And that corresponds to answer C So just to recap real quick, we use um the fact that the probability of finding a particle anywhere has to equal one. And we use that to calculate our normalization constant for our wave function. So I hope that this has been useful. And I'll see you in the next video.

Table of contents

Skip topic navigation

0. Math Review31m
- Math Review
  31m
1. Intro to Physics Units1h 29m
2. 1D Motion / Kinematics3h 56m
3. Vectors2h 43m
4. 2D Kinematics1h 42m
5. Projectile Motion3h 6m
6. Intro to Forces (Dynamics)3h 22m
7. Friction, Inclines, Systems2h 44m
8. Centripetal Forces & Gravitation7h 26m
9. Work & Energy1h 59m
10. Conservation of Energy2h 54m
11. Momentum & Impulse3h 40m
12. Rotational Kinematics2h 59m
13. Rotational Inertia & Energy7h 4m
14. Torque & Rotational Dynamics2h 5m
15. Rotational Equilibrium3h 39m
16. Angular Momentum3h 6m
17. Periodic Motion2h 9m
18. Waves & Sound3h 40m
19. Fluid Mechanics4h 27m
20. Heat and Temperature3h 7m
21. Kinetic Theory of Ideal Gases1h 50m
22. The First Law of Thermodynamics1h 26m
23. The Second Law of Thermodynamics3h 11m
24. Electric Force & Field; Gauss' Law3h 42m
25. Electric Potential1h 51m
26. Capacitors & Dielectrics2h 2m
27. Resistors & DC Circuits3h 8m
28. Magnetic Fields and Forces2h 23m
29. Sources of Magnetic Field2h 30m
30. Induction and Inductance3h 38m
31. Alternating Current2h 37m
32. Electromagnetic Waves2h 14m
33. Geometric Optics2h 57m
34. Wave Optics1h 15m
35. Special Relativity2h 10m

35. Special Relativity

Inertial Reference Frames

Physics

35. Special Relativity

Inertial Reference Frames

Previous problemNext problem

5:53 minutes

Problem 38b

Textbook Question

A particle is described by the wave function $ψ (x) = {\begin{matrix} c e^{x / L} & x \leq 0 mm \\ c e^{- x / L} & x \geq 0 mm \end{matrix}$ mm where L = 2.0 mm. Determine the normalization constant c.

Verified step by step guidance

To normalize the wave function, we use the condition that the total probability of finding the particle over all space is equal to 1. Mathematically, this is expressed as: ∫|ψ(x)|² dx = 1, where the integral is taken over all space.

Substitute the given wave function into the normalization condition. Since the wave function is piecewise, split the integral into two parts: ∫|ψ(x)|² dx = ∫[c²e²ˣ/ᴸ] dx (for x ≤ 0) + ∫[c²e⁻²ˣ/ᴸ] dx (for x ≥ 0).

Evaluate each integral separately. For the first integral (x ≤ 0), calculate ∫[c²e²ˣ/ᴸ] dx from -∞ to 0. For the second integral (x ≥ 0), calculate ∫[c²e⁻²ˣ/ᴸ] dx from 0 to ∞. Use the standard integral formula for exponential functions: ∫eᵃˣ dx = (1/a)eᵃˣ + C, where a ≠ 0.

Combine the results of the two integrals and set the total equal to 1. This will give you an equation involving the normalization constant c. Solve for c by isolating it on one side of the equation.

Substitute the given value of L = 2.0 mm into the equation to express c in terms of numerical values. This will provide the normalized constant c, ensuring the wave function satisfies the normalization condition.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above

Video duration:

Play a video:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Wave Function

The wave function, denoted as ψ(x), is a mathematical description of the quantum state of a particle. It contains all the information about the system and is used to calculate probabilities of finding a particle in a given position. The wave function must be normalized, meaning the total probability of finding the particle in all space must equal one.

Normalization

Normalization is the process of adjusting the wave function so that the integral of the absolute square of the wave function over all space equals one. This ensures that the total probability of finding the particle in any location is 100%. For a piecewise wave function, normalization involves integrating each segment and setting the sum equal to one to solve for the normalization constant.

Integration

Integration is a fundamental mathematical operation used to calculate areas under curves, which in quantum mechanics is essential for determining probabilities. In the context of normalization, it involves integrating the square of the wave function over the entire space to find the normalization constant. The integration limits will depend on the defined regions of the wave function.

Watch next

Master Inertial Reference Frames with a bite sized video explanation from Patrick

Start learning

Related Videos

Related Practice

Textbook Question

Consider the electron wave function $ψ (x) = {\begin{matrix} c \sqrt{1 - x^{2}} & ∣ x ∣ \leq 1 cm \\ 0 & ∣ x ∣ \geq 1 cm \end{matrix}$ where x is in cm. Determine the normalization constant c.

views

Textbook Question

Consider the electron wave function $ψ (x) = {\begin{matrix} c \sqrt{1 - x^{2}} & ∣ x ∣ \leq 1 cm \\ 0 & ∣ x ∣ \geq 1 cm \end{matrix}$ where x is in cm. Draw a graph of ψ(x) over the interval −2 cm ≤ x ≤ 2 cm. Provide numerical scales on both axes.

views

Textbook Question

Consider the electron wave function $ψ (x) = {\begin{matrix} c \sqrt{1 - x^{2}} & ∣ x ∣ \leq 1 cm \\ 0 & ∣ x ∣ \geq 1 cm \end{matrix}$ where x is in cm. Draw a graph of |ψ(x)|² over the interval −2 cm ≤ x ≤ 2 cm. Provide numerical scales.

views

Textbook Question

A particle is described by the wave function $ψ (x) = {\begin{matrix} c e^{x / L} & x \leq 0 mm \\ c e^{- x / L} & x \geq 0 mm \end{matrix}$ where L = 2.0 mm. Sketch graphs of both the wave function and the probability density as functions of x.

views

Textbook Question

Heavy nuclei often undergo alpha decay in which they emit an alpha particle (i.e., a helium nucleus). Alpha particles are so tightly bound together that it’s reasonable to think of an alpha particle as a single unit within the nucleus from which it is emitted. A ²³⁸U nucleus, which decays by alpha emission, is 15 fm in diameter. Model an alpha particle within a ²³⁸U nucleus as being in a one-dimensional box. What is the maximum speed an alpha particle is likely to have?

views

Textbook Question

Heavy nuclei often undergo alpha decay in which they emit an alpha particle (i.e., a helium nucleus). Alpha particles are so tightly bound together that it’s reasonable to think of an alpha particle as a single unit within the nucleus from which it is emitted. The probability that a nucleus will undergo alpha decay is proportional to the frequency with which the alpha particle reflects from the walls of the nucleus. What is that frequency (reflections/s) for a maximum-speed alpha particle within a ²³⁸U nucleus?

views

Textbook Question

The electrons in a rigid box emit photons of wavelength 1484 nm during the 3→2 transition. What kind of photons are they—infrared, visible, or ultraviolet?

views

Textbook Question

The electrons in a rigid box emit photons of wavelength 1484 nm during the 3→2 transition. How long is the box in which the electrons are confined?

views

Show more practices