Calculate the normalization of a Particle in a box

In summary, the conversation discusses the calculation of the normalization factor for a particle described by the wave function \Psi(x)=Ae^{(-bx^2)}. The integral of the probability over the entire area is set equal to 1, and the discussion includes different methods for evaluating the integral. However, it is pointed out that the wave function given may not be appropriate for a particle in an infinite potential well, as it does not vanish at the boundaries. It is suggested that the question may be about a free particle, in which case the integral is easy to calculate.
  • #1
IHateMayonnaise
94
0
[SOLVED] Particle in a box

Homework Statement



A particle is described by the wave function [tex]\Psi(x)=Ae^{(-bx^2)}[/tex]. Calculate the normalization factor A.

Homework Equations

&

The Attempt at a Solution



So, to normalize [tex]\psi(x)[/tex], we integrate the probability ([tex]\Psi(x)^2[/tex]) over the entire area (this is in an infinite potential well, so let's say the particle is between x=0 and x=L) and set this equal to 1. Doing this, we have:

[tex]A^2\int_0^L{e^{-2bx^2}}dx=1}[/tex]

I feel pretty comfortable with calculus but I cannot figure out how to evaluate this integral. I was able to find this in an integral table, but it wasn't much help:

[tex]\int e^{ax^2} dx = -\frac{i\sqrt{\pi}}{2\sqrt{a}}\text{erf}\left(ix\sqrt{a}\right)[/tex]

I do not recall learning anything about the error function (erf), and google wasn't much help. How can I evaluate this integral? Should I expand it in series? Any help would be appreciated. Thanks

IHateMayonnaise
 
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  • #2
Is this the right equation or is there an 'i' on the exponential?

If not, try integrating a power series expansion of 'e^(x)'.

recall that if e^(-ix), then the normalization would be the integral of the wavefunction times its complex conjugate e^(ix), which would be much simpler.
 
  • #3
end3r7 said:
Is this the right equation or is there an 'i' on the exponential?

If not, try integrating a power series expansion of 'e^(x)'.

recall that if e^(-ix), then the normalization would be the integral of the wavefunction times its complex conjugate e^(ix), which would be much simpler.

it is not imaginary, unfortunately...I will try the power series expansion now though
 
  • #4
[itex](e^x)^2[/itex] is [itex]e^{2x}[/itex], not [itex]e^{x^2}[/itex]. And if you're in an infinite potential well, shouldn't the wavefunction vanish at the boundaries?
 
  • #5
StatusX said:
[itex](e^x)^2[/itex] is [itex]e^{2x}[/itex], not [itex]e^{x^2}[/itex]. And if you're in an infinite potential well, shouldn't the wavefunction vanish at the boundaries?

I apologize, that was a typo - it's fixed now. Yes the wavefunction will vanish at the boundaries...but since we know that our particle will be somewhere between x=0 and x=L, those are the limits when finding the normalization factor, A. Alternatively, the probability that we will find the particle anywhere should be 1...since our particle can only be between x=0 and x=L, these are the limits of the normalization integral.
 
  • #6
IHateMayonnaise said:
I apologize, that was a typo - it's fixed now. Yes the wavefunction will vanish at the boundaries...but since we know that our particle will be somewhere between x=0 and x=L, those are the limits when finding the normalization factor, A. Alternatively, the probability that we will find the particle anywhere should be 1...since our particle can only be between x=0 and x=L, these are the limits of the normalization integral.

The point is that the wavefunction you gave can not represent a particle in a square well because it does not vanish at the boundaries. My bet is that the question is about a free particle in which case you must integrate from minus infinity to plus infinity and the integral is easy to do (or look it up).
 
  • #7
I also say that the wave function that you posted in the original post [tex]\Psi(x)=Ae^{(-bx^2)}[/tex]
Can't represent a particle in a infinite deep box,
 

Related to Calculate the normalization of a Particle in a box

What is the formula for calculating the normalization of a Particle in a box?

The normalization of a Particle in a box is calculated using the formula ∫ τ*τ dx = 1, where τ is the wave function and dx represents the length of the box.

What is the significance of calculating the normalization of a Particle in a box?

Calculating the normalization of a Particle in a box allows us to determine the probability of finding the particle in a specific location within the box. It also ensures that the total probability of finding the particle within the box is equal to 1.

How does the size of the box affect the normalization of a Particle in a box?

The size of the box, represented by the value of dx, directly affects the normalization of a Particle in a box. As the size of the box decreases, the normalization value will increase, and vice versa.

What is the relationship between the normalization of a Particle in a box and the wave function?

The normalization of a Particle in a box is directly related to the wave function as it is used in the calculation of normalization. The wave function represents the probability amplitude of finding the particle at a specific location within the box.

Are there any limitations to calculating the normalization of a Particle in a box?

One limitation of calculating the normalization of a Particle in a box is that it only applies to one-dimensional systems and cannot be used for more complex systems. Additionally, it assumes that the particle is confined within the box and does not take into account any external forces or interactions.

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