# Normalize the Gaussian wave packet

tanaygupta2000
Homework Statement:
To normalize the wave equation given
Relevant Equations:
A^2 (psi)*(psi) = 1 For normalizing this wave function, I began by finding the complex conjugate of psi and then multiplied it with the original psi.
Now what I am getting is A^2 integral exp(2cx^2-4ax) dx = 1
Now I am not getting how to solve this exponential term. I tried by completing the square method but it is not giving a good result. Please help!

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I would try completing the square of the exponential argument before multiplying by the complex conjugate and then do a change of variables. Why is it not a good result?

On edit: I take it back. Completing the square after finding ##\psi^*\psi## works. Note that given real constant ##c## must be negative for the normalization integral to converge and needs a little bit of extra care.

Last edited:
• tanaygupta2000
tanaygupta2000
Yes sir, integral will only converge if c is replaced by -c.
On doing so, and making some simple substitution like (x + a/c) = t and dx = dt,
I am successfully getting value of normalization constant A = (2c/pi)^(1/4) exp(-a^2 /c)

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I would write the coefficient up front as ##\left[\frac{2(-c)}{\pi}\right]^{\frac{1}{4}}.## That's because ##c## is a negative number. Also, the argument of the exponential is not what I got. What does your completed square for ##\psi^*\psi## look like?

tanaygupta2000
It's 2c[(x - a/c)^2 - (a/c)^2]

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It's 2c[(x - a/c)^2 - (a/c)^2]
Yes, I agree. Your normalization is good. Sorry for the confusion, I had an extra factor of two. Can you finish the problem now?

• tanaygupta2000
tanaygupta2000
Yes, I agree. Your normalization is good. Sorry for the confusion, I had an extra factor of two. Can you finish the problem now?
Sure, sir. Thank you very much for the help!
So I get this: solve by using c --> -c so that the integral can be evaluated and in the final answer revert it to +c back (using iota wherever necessary).

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Sure, sir. Thank you very much for the help!
So I get this: solve by using c --> -c so that the integral can be evaluated and in the final answer revert it to +c back (using iota wherever necessary).

I think you missed my point in post #4. You do not use ##i=\sqrt{-1}## wherever necessary. To normalize, you do the integral $$1=\int (N\psi)^*(N\psi) dx=|N|^2 \int \psi^*\psi~dx$$ and solve for ##|N|.## You can drop the absolute signs, but you must keep in mind that the normalization constant ##N## is real. To remember that in this case, it helps to write the constant up front as ##k=\left[\frac{2(-c)}{\pi}\right]^{\frac{1}{4}}.## If someone gives you a value ##c=-8~\text{m}^{-2}##, then you can substitute and say ##k=\frac{2}{\pi^{1/4}}~\text{m}^{-1/2}.## If someone gives you a value ##c=+8~\text{m}^{-2}##, then you cannot substitute and must say "this wavefunction cannot be normalized"; you don't say ##k=\frac{\sqrt{2}+i\sqrt{2}}{\pi^{1/4}}~\text{m}^{-1/2}.##

What did you get for the constant exponential term that multiplies ##k## when you write ##N=k \exp(?)##. I don't think it is ##\exp(-a^2/c^2)## as you have in post #3.

Last edited:
• tanaygupta2000
tanaygupta2000
Okay, so we write N = [-2c/pi]^(1/4) with c strictly negative as the normalization constant.
So for calculating the expectation values, can I use
psi(x) = [2c/pi]^(1/4) exp(-cx^2 - 2(a+ib)x - a^2 /c) ?

because for this integral to convege, there should be a - sign before the c.
So I replaced c with -c to evaluate it and hence there is no - sign in N.
Or should I adopt some other method?

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If you replace ##c## with ##-c##, you are likely to be confused in the end. What I would do is replace ##c=-\gamma^2## where ##\gamma## is real. Do your calculations and integrals with that and then in the very end replace ##\gamma^2## with ##(-c)## (keep the parentheses) and ##\gamma^4## with ##c^2##. That will automatically sort out the negative signs.

• tanaygupta2000
tanaygupta2000
In the end, I am getting a factor of sqrt(gamma) in the normalization constant.
How to justify it?

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In the end, I am getting a factor of sqrt(gamma) in the normalization constant.
How to justify it?
I don't get a factor of gamma. Please show me how it comes about in your calculation.

Last edited:
tanaygupta2000 Homework Helper
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You have ##A^2e^{2a^2/\gamma^2}\sqrt{\frac{\pi}{2\gamma^2}}=1##.
If you follow the algebra, it should be ##A^2e^{2a^2/\gamma^4}\sqrt{\frac{\pi}{2\gamma^2}}=1##.

Then ##A^2=e^{-2a^2/\gamma^4}\left[\frac{2\gamma^2}{\pi}\right]^{1/2}=e^{-2a^2/c^2}\left[\frac{2(-c)}{\pi}\right]^{1/2}.##

• tanaygupta2000
tanaygupta2000 Sir, am I correct upto this portion?
After this step, I am getting only (gamma)^2 below 2a^2 as (gamma)^2/ (gamma)^4 = (gamma)^(-2). isn't it?

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View attachment 278261

Sir, am I correct upto this portion?
After this step, I am getting only (gamma)^2 below 2a^2 as (gamma)^2/ (gamma)^4 = (gamma)^(-2). isn't it?
That looks right to me.

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Here's an alternative approach I would take to break the problem down into easier steps:

First, evaluate: $$\int_{-\infty}^{+\infty} e^{-x^2 -\beta x}dx$$ And write down this answer and keep it somewhere safe.

Then, using this result, evaluate $$\int_{-\infty}^{+\infty} e^{-\alpha x^2 -\beta x}dx$$ using the substitution ##u = \sqrt{\alpha} x##. And keep that safe too.

(I have six pages of standard integrals like this! I find it faster and more reliable than typing things into Wolfram.)

Finally, evaluate your integral with ##\alpha = -2c, \ \beta = 4a##.

You can figure out ##A## at the end.

• tanaygupta2000
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You have ##A^2e^{2a^2/\gamma^2}\sqrt{\frac{\pi}{2\gamma^2}}=1##.
If you follow the algebra, it should be ##A^2e^{2a^2/\gamma^4}\sqrt{\frac{\pi}{2\gamma^2}}=1##.

Then ##A^2=e^{-2a^2/\gamma^4}\left[\frac{2\gamma^2}{\pi}\right]^{1/2}=e^{-2a^2/c^2}\left[\frac{2(-c)}{\pi}\right]^{1/2}.##
I get no more than ##\gamma^2## in there. And I'm just looking up my standard integrals!

tanaygupta2000
Here's an alternative approach I would take to break the problem down into easier steps:

First, evaluate: $$\int_{-\infty}^{+\infty} e^{-x^2 -\beta x}dx$$ And write down this answer and keep it somewhere safe.

Then, using this result, evaluate $$\int_{-\infty}^{+\infty} e^{-\alpha x^2 -\beta x}dx$$ using the substitution ##u = \sqrt{\alpha} x##. And keep that safe too.

(I have six pages of standard integrals like this! I find it faster and more reliable than typing things into Wolfram.)

Finally, evaluate your integral with ##\alpha = -2c, \ \beta = 4a##.

You can figure out ##A## at the end.
Thank You sir! But unfortunately the normalization constant is still coming imaginary.

tanaygupta2000
I'm getting confused.

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Thank You sir! But unfortunately the normalization constant is still coming imaginary.
Your answer for ##A^2## is real and positive in post #15. That is correct.

All these integrals are real and positive. They can't possibly have a negative value. The only confusion is that they gave you ##c## (which forces ##c## to be a negative number). Or, let ##c = -\gamma^2##, where ##\gamma## is positive, as you've done.

• tanaygupta2000
tanaygupta2000
Your answer for ##A^2## is real and positive in post #15. That is correct.

All these integrals are real and positive. They can't possibly have a negative value. The only confusion is that they gave you ##c## (which forces ##c## to be a negative number). Or, let ##c = -\gamma^2##, where ##\gamma## is positive, as you've done.
But my value of A is coming to be complex. It must be real.

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But my value of A is coming to be complex. It must be real.
Why is ##A## complex?

tanaygupta2000
Why is ##A## complex?
It is involving sqrt of iota.

tanaygupta2000
after making substitution c = -(gamma)^2

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It is involving sqrt of iota.
Not in post #15, where you have the right answer.

tanaygupta2000
Not in post #15, where you have the right answer.
Gamma = iota times c
So already-complex A will be more complex as sq root of i comes into picture.
I am wondering is the question incorrect?

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Gamma = iota times c
So already-complex A will be more complex as sq root of i comes into picture.
I am wondering is the question incorrect?
No, no, no!

##c## is negative, ##-c## is positive, ##\gamma## is positive by definition.

There are no complex numbers involved.

• tanaygupta2000
tanaygupta2000
No, no, no!

##c## is negative, ##-c## is positive, ##\gamma## is positive by definition.

There are no complex numbers involved.
Excellent sir! Now I understood everything !
Thank You so so much for yours and Kuruman sir's precious efforts in helping me.

• kuruman and PeroK
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I get no more than ##\gamma^2## in there. And I'm just looking up my standard integrals!
You are absolutely correct. I forgot the ##\gamma^2## up front when I separated the exponentials. This is especially infuriating to me because I squared ##\gamma## in the first place so that its dimensions match ##a## and make troubleshooting easier by dimensional analysis.
##c## is negative,##-c## is positive, ##\gamma## is positive by definition
Actually the definition is ##c=-\gamma^2## where ##\gamma## is real. ##\gamma## doesn't have to be positive.

To @tanaygupta2000 :
Your expression in post #15 ##A^2e^{2a^2/\gamma^2}\sqrt{\frac{\pi}{2\gamma^2}}=1## is correct. Sorry for the confusion.

• tanaygupta2000