Normalize the Gaussian wave packet

• tanaygupta2000
In summary, the conversation discussed normalizing a wave function by finding the complex conjugate of psi and then multiplying it with the original psi. There was also a discussion about using completing the square method and making a change of variables to solve the exponential term. The conversation also clarified that the normalization constant must have a negative value for the integral to converge. It was also mentioned that the normalization constant should be written as k = [2(-c)/pi]^1/4, and that using c = -γ^2 and then replacing it with -c in the end will automatically sort out the negative signs. The conversation ended with a discussion about the normalization constant and the justification of a factor of gamma in the final calculation, which was resolved by
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!

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.

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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)

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?

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

tanaygupta2000 said:
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
kuruman said:
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).

tanaygupta2000 said:
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.

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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?

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

tanaygupta2000 said:
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.

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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

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?

tanaygupta2000 said:
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.

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
kuruman said:
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!

PeroK said:
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.

I'm getting confused.

tanaygupta2000 said:
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
PeroK said:
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.

tanaygupta2000 said:
But my value of A is coming to be complex. It must be real.
Why is ##A## complex?

PeroK said:
Why is ##A## complex?
It is involving sqrt of iota.

after making substitution c = -(gamma)^2

tanaygupta2000 said:
It is involving sqrt of iota.
Not in post #15, where you have the right answer.

PeroK said:
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?

tanaygupta2000 said:
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
PeroK said:
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
PeroK said:
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.
PeroK said:
##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

1. What is a Gaussian wave packet?

A Gaussian wave packet is a mathematical function that describes the probability distribution of a quantum particle in space and time. It is a solution to the Schrödinger equation and is commonly used to model the behavior of particles in quantum mechanics.

2. Why is it important to normalize a Gaussian wave packet?

Normalizing a Gaussian wave packet ensures that the total probability of finding the particle in any location is equal to 1. This is necessary because the wave function represents the probability amplitude of the particle, and the total probability must always be conserved in quantum mechanics.

3. How do you normalize a Gaussian wave packet?

To normalize a Gaussian wave packet, you need to find the normalization constant, which is the square root of the inverse of the integral of the absolute square of the wave function. This constant is then multiplied to the wave function to ensure that the total probability is equal to 1.

4. What is the relationship between the width of a Gaussian wave packet and its normalization?

The width of a Gaussian wave packet is inversely proportional to its normalization constant. This means that as the width of the wave packet decreases, the normalization constant increases, and vice versa. This is because a narrower wave packet has a higher probability density and therefore requires a larger normalization constant to maintain a total probability of 1.

5. Can a Gaussian wave packet be normalized if it has complex values?

Yes, a Gaussian wave packet can still be normalized even if it has complex values. The normalization constant will also have a complex value, and it will be multiplied to both the real and imaginary parts of the wave function to ensure that the total probability is equal to 1.

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