Gaussian Integrals for Quantum States of well Defined Momentum

In summary, the Gaussian Integral (eqn 2.64) involves a constant of normalization that is rationalized by changing the sign of the exponent and squaring the quantity twice. This is done in order to satisfy the requirement that the integral of the probability density function from -inf to +inf equals 1. The momentum distribution, which corresponds to a normal distribution, is related to the position distribution by their standard deviations. In order to understand the process better, someone can show the steps to transform the equation and point out where any mistakes were made.
  • #1
24
0
Consider the Gaussian Integral (eqn 2.64).. is anyone able to explain how the constant of normalization is rationalised?
 

Attachments

  • Position Representation.jpg
    Position Representation.jpg
    48.6 KB · Views: 442
Last edited:
Physics news on Phys.org
  • #2
If by 'rationalized' you mean 'normalized' so that the integral of the pdf from -inf to + inf = 1, I would think in the usual fashion.
 
  • #3
Hey SteamKing, thanks for stopping by to help. I'm not referring to the normalisation process here, it's the constant on the exponential I'm having trouble with.. The equation shows a rationalisation of the numerator to give (2∏(hbar)^2 / 4σ^2)^-1/4.

From what i understand if one wants to move a quantity from the numerator under the fourth principal root sign in the denominator, i must change the sign of the exponent and square that quantity twice but it doesn't work out. Perhaps I'm doing something wrong here?
 
Last edited:
  • #4
It comes from normalizing ##\langle p | \psi \rangle##, that is, requiring that ##| \langle p | \psi \rangle |^2 = 1##.
 
  • #5
jtbell i apologise for any confusion here but I'm not referring to the normalisation process, it's the constant on the exponential. The equation shows a rationalisation of the numerator to give 1 /(2∏(hbar)^2 / 4σ^2)^1/4.

From what i understand if one wants to move a quantity from the numerator under the fourth principal root sign in the denominator, i must change the sign of the exponent and square that quantity twice but it doesn't work out. Perhaps I'm doing something wrong here?
 
Last edited:
  • #6
[itex] h = 2\pi \hbar[/itex]

[itex] \sqrt[4]{\frac{16\sigma^4\pi^2}{h^2 (2\pi \sigma^2)}} = \sqrt[4]{\frac{16\pi^2\sigma^2}{2\pi h^2}} = \sqrt[4]{\frac{16\sigma^2\pi^2}{2\pi\left(2\pi \hbar\right)^2}}[/itex]
[itex] = \sqrt[4]{\frac{16\sigma^2}{2\pi 4\hbar^2}} = \sqrt[4]{\frac{4\sigma^2}{2\pi \hbar^2}} = \frac{1}{\sqrt[4]{2\pi \left(\hbar^2/(4\sigma^2) \right)}}[/itex]

The momentum distribution does not describe a definite (well defined) momentum but rather corresponds to the case that both position and momentum are normal distributions, and this equation shows the relation between their standard deviations.
 
Last edited:
  • Like
Likes 1 person
  • #7
lukka said:
it doesn't work out.

If you show us explicitly what you thought it should work out to be, in stepwise fashion, someone can probably tell you what your mistake is.
 
  • #8
Equation 2.64 implies that the normalisation constants are equivalent. that is..

(2σ √∏) / √h(2∏σ^2)^1/4 = (2∏(hbar)^2 / 4σ^2)^-1/4.

here's the algebra i did..


(2σ √∏) / √h(2∏σ^2)^1/4 = √∏ / (2∏/16σ^2)^1/4 ..

..=√h (2∏/16σ^2 ^∏^2)^-1/4 = (2∏(hbar)^2 / 16σ^2)^-1/4.
 
  • #9
[itex]\frac{2\sigma \pi^{1/2}}{h^{1/2}(2\pi \sigma^2)^{1/4}}=\left [\frac{16\sigma^4\pi^2}{2\pi h^2\sigma^2}\right ]^{1/4}=\left [\frac{16\sigma^2\pi^2}{2\pi (2\pi\hbar)^2}\right ]^{1/4}=\left [\frac{16\sigma^2\pi^2}{8\pi^3 \hbar^2}\right ]^{1/4}=\left[\frac{4\sigma^2}{2\pi \hbar^2}\right ]^{1/4}[/itex]
 
  • Like
Likes 1 person
  • #10
Hey that's great! i never thought of transforming it that way.. i guess what i was attempting to do at first before was nonsensical although I'm not sure why, it seems they should have revealed the same? I'm sure I'm on the right track with this now. Thanks for going to the time to show me where i was going wrong..
 
  • #11
i can see where i going wrong with this now.. thanks Dauto!
 

1. What are Gaussian Integrals for Quantum States of well Defined Momentum?

Gaussian Integrals for Quantum States of well Defined Momentum are mathematical techniques used in quantum mechanics to calculate the probability of a particle being in a certain state with a defined momentum. They involve integrating over a Gaussian distribution, which represents the wave function of the particle.

2. How are Gaussian Integrals used in quantum mechanics?

Gaussian Integrals are used in quantum mechanics to calculate the probability amplitudes of quantum states with well-defined momentum. They are also used to analyze the behavior of particles in quantum systems and understand the dynamics of quantum systems.

3. What is the significance of well-defined momentum in quantum mechanics?

In quantum mechanics, the momentum of a particle is considered one of its fundamental properties. A well-defined momentum means that the particle has a definite momentum value and its wave function is sharply peaked at that value. This is important for understanding the behavior of particles in quantum systems.

4. Can Gaussian Integrals be used for particles with non-zero momentum?

Yes, Gaussian Integrals can be used for particles with non-zero momentum. In fact, they are commonly used to calculate the probability of a particle being in a state with a specific momentum value, regardless of whether it is zero or non-zero.

5. Are there any limitations to using Gaussian Integrals for Quantum States of well Defined Momentum?

While Gaussian Integrals are a powerful tool in quantum mechanics, they do have limitations. They may not accurately describe systems with strong interactions or high energies. Additionally, they are not suitable for systems with multiple particles or entangled states.

Suggested for: Gaussian Integrals for Quantum States of well Defined Momentum

Replies
4
Views
557
Replies
16
Views
1K
Replies
6
Views
735
Replies
9
Views
879
Replies
29
Views
1K
Replies
11
Views
997
Replies
1
Views
738

Back
Top