Find the normalization constant ##A##

  • #1
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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!!!
 

Answers and Replies

  • #2
TSny
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$$ \left| \psi(x) \right|^2 = 2 \left| A \right|^2 e^\left( -2 x^2 \right) + e^\left( -2 i \alpha x^2 \right) +e^\left(2 i \alpha x^2 \right) .$$
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.
 
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  • #3

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}##?...
 
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  • #4
You made mistake in multiplying the exponential. The normalization factor should come as an overall factor.

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  • #5
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Ok guys, thank you. I'm going to review my calculations and then I get back here.
 
  • #6
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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 thats 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
TSny
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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.)
 
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  • #8
PeroK
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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 thats 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
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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
PeroK
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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
TSny
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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
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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.

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
TSny
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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.
 
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  • #14
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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
TSny
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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?
 

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