# Calculate the normalization of a Particle in a box

[SOLVED] Particle in a box

## Homework Statement

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

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## The Attempt at a Solution

So, to normalize $$\psi(x)$$, we integrate the probability ($$\Psi(x)^2$$) 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:

$$A^2\int_0^L{e^{-2bx^2}}dx=1}$$

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:

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

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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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.

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

StatusX
Homework Helper
$(e^x)^2$ is $e^{2x}$, not $e^{x^2}$. And if you're in an infinite potential well, shouldn't the wavefunction vanish at the boundaries?

$(e^x)^2$ is $e^{2x}$, not $e^{x^2}$. 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.

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).

malawi_glenn
I also say that the wave function that you posted in the original post $$\Psi(x)=Ae^{(-bx^2)}$$