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!

 

[kuruman]

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.

 

[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)

 

[kuruman]

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]

 

[kuruman]

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]

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).

 

[kuruman]

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.

 

[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?

 

[kuruman]

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?

 

[kuruman]

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.

 

[tanaygupta2000]

 

[kuruman]

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?
Please help me where I am wrong.

 

[PeroK]

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?
Please help me where I am wrong.

That looks right to me.

 

[PeroK]

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.

 

[PeroK]

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]

Can somebody please upload a picture with the solution?
I'm getting confused.

 

[PeroK]

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]

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.

 

[PeroK]

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

 

[PeroK]

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?

 

[PeroK]

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]

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]

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.