Find the normalization constant ##A##

In summary: And then I coulnd't get ##\psi## normalizated... I've calculated all of the integrals above and that's what I got $$ \left| A \right|^2 \sqrt {\frac {\pi} {2}} \left(2 - e^{i\alpha} + e^{-i\alpha} \right) = 1.$$ So, I don't know what to do next. What do think?In summary, the noralization constant of the function bellow is ##A##.
  • #1
Mutatis
42
0

Homework Statement



Find the noralization constant ##A## of the function bellow: $$ \psi(x) = A e^\left(i k x -x^2 \right) \left[ 1 + e^\left(-i \alpha \right) \right], $$ ##\alpha## is also a constant.

Homework Equations



##\int_{-\infty}^{\infty} e^\left(-\lambda x^2 \right) \, dx = \sqrt {\frac {\pi} {\lambda} }##

The Attempt at a Solution



Well, first I've tried to find the density of probability: $$ \left| \psi(x) \right|^2 = 2 \left| A \right|^2 e^\left( -2 x^2 \right) + e^\left( -2 i \alpha x^2 \right) +e^\left(2 i \alpha x^2 \right) .$$ Then I have no idea how to solve the integral to normalize ##\psi(x)##... The fisrt two terms I've solved using the gaussian integral above, but I can't do it to the third term: $$ 2 \left| A \right|^2 \left( \int_{-\infty}^{\infty} e^\left( -2 x^2 \right) dx + \int_{-\infty}^{\infty} e^\left( -2 i \alpha x^2 \right) dx + \int_{-\infty}^{\infty} e^\left(2 i \alpha x^2 \right) dx \right)= 1 \\ 2 \left| A \right|^2 \left( \frac {\sqrt {2 \pi}} {2} + \frac {\sqrt {2 i \alpha \pi}} {2 i \alpha} + \int_{-\infty}^{\infty} e^\left(2 i \alpha x^2 \right) dx \right)= 1$$.
I don't know if I'm doing it right, but I have no idea to get this solved... Help!
 
Physics news on Phys.org
  • #2
Mutatis said:
$$ \left| \psi(x) \right|^2 = 2 \left| A \right|^2 e^\left( -2 x^2 \right) + e^\left( -2 i \alpha x^2 \right) +e^\left(2 i \alpha x^2 \right) .$$
This is not correct. Check your work. If you don't see your mistake, then show us how you got this result. You should not get ##\alpha## multiplying ##x^2## in any of the exponents. ##|A|^2## should be an overall factor of the entire result.
 
  • Like
Likes Mutatis
  • #3
Mutatis said:

Homework Statement



Find the noralization constant ##A## of the function bellow: $$ \psi(x) = A e^\left(i k x -x^2 \right) \left[ 1 + e^\left(-i \alpha \right) \right], $$ ##\alpha## is also a constant.

Homework Equations



##\int_{-\infty}^{\infty} e^\left(-\lambda x^2 \right) \, dx = \sqrt {\frac {\pi} {\lambda} }##

The Attempt at a Solution



Well, first I've tried to find the density of probability: $$ \left| \psi(x) \right|^2 = 2 \left| A \right|^2 e^\left( -2 x^2 \right) + e^\left( -2 i \alpha x^2 \right) +e^\left(2 i \alpha x^2 \right) .$$ Then I have no idea how to solve the integral to normalize ##\psi(x)##... The fisrt two terms I've solved using the gaussian integral above, but I can't do it to the third term: $$ 2 \left| A \right|^2 \left( \int_{-\infty}^{\infty} e^\left( -2 x^2 \right) dx + \int_{-\infty}^{\infty} e^\left( -2 i \alpha x^2 \right) dx + \int_{-\infty}^{\infty} e^\left(2 i \alpha x^2 \right) dx \right)= 1 \\ 2 \left| A \right|^2 \left( \frac {\sqrt {2 \pi}} {2} + \frac {\sqrt {2 i \alpha \pi}} {2 i \alpha} + \int_{-\infty}^{\infty} e^\left(2 i \alpha x^2 \right) dx \right)= 1$$.
I don't know if I'm doing it right, but I have no idea to get this solved... Help!

As has been pointed out by TSny, the ##A^*A## should be an overall factor of the probability density. You made another mistake multiplying the exponentials when finding the probability density: don't you know that ##e^a e^b = e^{a+b}## and not ##e^{ab}##?...
 
  • Like
Likes Mutatis
  • #4
You made mistake in multiplying the exponential. The normalization factor should come as an overall factor.

Embedded video removed by mentor. Posting complete solutions is not permitted at this site.
 
Last edited by a moderator:
  • Like
Likes Mutatis
  • #5
Ok guys, thank you. I'm going to review my calculations and then I get back here.
 
  • #6
So, I'm here again... I've done my calculations right this time, but I still can't get ##A## though... I got the probability density: $$ \left| \psi(x) \right|^2 = \left| A \right|^2 \left[ 2 e^\left(-2 x^2\right) +e^\left(-2x^2 +i\alpha \right) + e^\left(-2x^2 -i\alpha \right) \right] . $$
And then I coulnd't get ##\psi## normalizated... I've calculated all of the integrals above and that's what I got $$ \left| A \right|^2 \sqrt {\frac {\pi} {2}} \left(2 - e^{i\alpha} + e^{-i\alpha} \right) = 1.$$ So, I don't know what to do next. What do think?
 
  • #7
Mutatis said:
I got $$ \left| A \right|^2 \sqrt {\frac {\pi} {2}} \left(2 - e^{i\alpha} + e^{-i\alpha} \right) = 1.$$ So, I don't know what to do next. What do think?
Check whether or not the minus sign is correct for the middle term. Note that with this minus sign, your expression will not be real and therefore cannot equal 1. Otherwise, I think your expression is right.

I think you can assume that A is a real, positive number. Otherwise, there will not be a unique answer. So, once you fix the sign error, you can just solve the expression for A. (You can combine the two complex exponentials into a trig function.)
 
Last edited:
  • #8
Mutatis said:
So, I'm here again... I've done my calculations right this time, but I still can't get ##A## though... I got the probability density: $$ \left| \psi(x) \right|^2 = \left| A \right|^2 \left[ 2 e^\left(-2 x^2\right) +e^\left(-2x^2 +i\alpha \right) + e^\left(-2x^2 -i\alpha \right) \right] . $$
And then I coulnd't get ##\psi## normalizated... I've calculated all of the integrals above and that's what I got $$ \left| A \right|^2 \sqrt {\frac {\pi} {2}} \left(2 - e^{i\alpha} + e^{-i\alpha} \right) = 1.$$ So, I don't know what to do next. What do think?

Note that the expression in ##\alpha## is just a constant factor, so it didn't need to get tangled up in your integral. Or your expression for ##|\psi(x)|^2##.
 
  • #9
This question is from the master degree qualification test that I'm intended to do here in my city (northeast of Brazil). I've obtained ##A## doing what you've told me to. First I got the exponentials in terms of ##cosh##, then I've turned it into a constant ##w## so I got: $$ \frac {\left| A \right|^2} {2} \sqrt \frac {\pi} {2} \left( 1 + \cosh(i \alpha) \right) = 1 \\ A = \sqrt \frac {2} {w} \left( \frac {2} {\pi} \right) ^{1/4}. $$
I don't know if this is right, because It is a very ugly equation.
 
  • #10
Mutatis said:
This question is from the master degree qualification test that I'm intended to do here in my city (northeast of Brazil). I've obtained ##A## doing what you've told me to. First I got the exponentials in terms of ##cosh##, then I've turned it into a constant ##w## so I got: $$ \frac {\left| A \right|^2} {2} \sqrt \frac {\pi} {2} \left( 1 + \cosh(i \alpha) \right) = 1 \\ A = \sqrt \frac {2} {w} \left( \frac {2} {\pi} \right) ^{1/4}. $$
I don't know if this is right, because It is a very ugly equation.

You should have ##cos## not ##cosh## in that equation.
 
  • #11
Mutatis said:
First I got the exponentials in terms of ##cosh##,
Since the exponentials have imaginary arguments, it will be nicer to express things in terms of the cosine function with real argument rather than in terms of the cosh function with imaginary argument. @PeroK has pointed this out.
https://webhome.phy.duke.edu/~rgb/Class/phy51/phy51/node15.html
then I've turned it into a constant ##w## so I got: $$ \frac {\left| A \right|^2} {2} \sqrt \frac {\pi} {2} \left( 1 + \cosh(i \alpha) \right)$$
I don't see where the denominator of 2 (under |A|2) is coming from.
 
  • #12
PeroK said:
You should have ##cos## not ##cosh## in that equation.
Now I've changed it to ##cos(\alpha)##. But it doesn't change the final result in terms of ##w## like I wrote above.

TSny said:
Since the exponentials have imaginary arguments, it will be nicer to express things in terms of the cosine function with real argument rather than in terms of the cosh function with imaginary argument. @PeroK has pointed this out.
https://webhome.phy.duke.edu/~rgb/Class/phy51/phy51/node15.html
I don't see where the denominator of 2 (under |A|2) is coming from.

To write the exponentials in terms of cossine I'd divided ## \left( 2 + e^{i \alpha} + e^{-i\alpha} \right) ## for ##2##. Is this an aceptable answer? Because what I've got to do next is to find the current of probability ##j(x)##.
 
  • #13
Mutatis said:
To write the exponentials in terms of cossine I'd divided ## \left( 2 + e^{i \alpha} + e^{-i\alpha} \right) ## for ##2##.

Note that ## \left( 2 + e^{i \alpha} + e^{-i\alpha} \right) = \left(2 + 2\cos\alpha \right) = 2\left(1 + \cos\alpha \right) ##

If you want, you can use another identity to write this as ##2\left(2\cos^2\frac{\alpha}{2} \right) = 4 \cos^2 \frac{\alpha}{2}##
So, this is essentially your ##w##, and ##\sqrt{w}## has a nice simple form.
 
  • Like
Likes PeroK
  • #14
Thank you guys! I got this solved. My problem this time is to find ##<x^2>##. I did some calculation and it leads me to ##<x^2> = \frac {1} {8} ## and it doesn't seems the right answer.
 
  • #15
Mutatis said:
Thank you guys! I got this solved. My problem this time is to find ##<x^2>##. I did some calculation and it leads me to ##<x^2> = \frac {1} {8} ## and it doesn't seems the right answer.
I get a larger value for ##\langle x^2 \rangle##. What expression do you get for ##|\psi (x)|^2## if you simplify it as much as possible?
 

Related to Find the normalization constant ##A##

1. What is the purpose of finding the normalization constant?

The normalization constant, denoted by ##A##, is used to scale a mathematical function so that its integral over its entire domain is equal to 1. This is important in probability and statistics as it allows us to interpret the function as a probability density function.

2. How do you find the normalization constant?

To find the normalization constant, you need to integrate the function over its entire domain and then solve for ##A## such that the integral is equal to 1. This involves solving a mathematical equation and may require the use of techniques such as integration by parts or substitution.

3. Can the normalization constant be negative?

No, the normalization constant must be a positive value in order for the function to be a valid probability density function. This ensures that the total probability of all possible outcomes is equal to 1.

4. What happens if the normalization constant cannot be found?

If the normalization constant cannot be found, it may indicate that the function is not a valid probability density function. This could be due to the function having a negative value or the integral not converging. In such cases, the function may need to be reevaluated or a different approach may need to be taken.

5. Is the normalization constant unique?

Yes, the normalization constant is unique for a given function. This means that no matter how the function is scaled, the integral over its entire domain will always be equal to 1. However, different functions may have different normalization constants.

Similar threads

  • Advanced Physics Homework Help
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
24
Views
1K
  • Advanced Physics Homework Help
Replies
19
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
925
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
3
Views
2K
  • Advanced Physics Homework Help
Replies
10
Views
1K
  • Advanced Physics Homework Help
Replies
8
Views
1K
Replies
1
Views
1K
Back
Top