Quantum mechanics: Normalization and the width of distribution

In summary, the conversation is discussing the normalization of a wave function and finding the corresponding function in position-space. The attempt at finding the width of the distribution in the x variable leads to a complicated equation and the suggestion to use the standard deviation as a different approach. The ultimate goal is to show the uncertainty principle using the relationship between the wave function in position-space and momentum-space.
  • #1
Kentaxel
14
0
1. Homework Statement [/b]

Normalize the following wave function, obtain the corresponding function in position-space (fourier transform) and find the width of the distribution in the x variable.


Homework Equations



[itex]\phi(p_x) =
\begin{cases}
0, & \;\; |p_x-p_0| > \gamma \\
C, & \;\; |p_x-p_0| \leq \gamma
\end{cases}[/itex]

3. The Attempt at a Solution

Now the normalization of the initial wave function pops out the relation*

[itex]C=\frac{1}{\sqrt{2\gamma}}[/itex]

So for so good. Using this result to obtain a corresponding wave function in position space (1dim x-space) i am able to obtain the function

[itex]\Psi (x) = \sqrt{\frac{\hbar}{\pi \gamma}} e^{i x p_0 / \hbar} \frac{\sin(\gamma x / \hbar)}{x}[/itex]

Which in fact normalizises to unity as expected, suggesting I'm on the right track. Now however i encounter problems. In trying to find the width of the distribution in the variable x (i.e Δx), I'm not sure if my calculatins are incorrect or if my interpretation of the problem is incorrect or if my solution is actually fine but the problem it self is incredibly complicated. My approach is as follows:

I want to know for what value of x the amplitude of the probability density i.e |ψ(x)|^2 has droped to e^(-1) times its maximum value right? Thus i simply calculate where the function obtains it's maximum value using the derivative with respect to x and setting it equal to zero, or so i thought. It turns out however that this yields a slightly uncomfortable equation To calculate (though by no means impossible) this fact however leads me to believe its not the right way to go about solving the problem. I push on, and find the maximum to be at x=0 and |ψ(0)|^2 = γ/(∏*h-bar). Thus i want to know for what x |ψ(x)|^2 = γ/(e*∏*h-bar) right? This expression however seems to yield a result only solvable with numerical methods which is not what i expect and does not give the nice representation of the width Δx that i want but rather*

[itex]\frac{\hbar}{\pi \gamma} \frac{\sin^2(\gamma x / \hbar)}{x^2} = \frac{\gamma}{e \pi \hbar}[/itex]

The problem is that I am supposed to show that the result leads to the uncertanity relation between the different distrubutions, ΔxΔp ≥ h-bar, which with this result seems impossible to me.
 
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  • #2
Why not use the most common form of the standard deviation?

[itex]\sigma _x = \sqrt{\langle x^2 \rangle - \langle x \rangle ^2}[/itex]

(I'm assuming what you mean by width of distrobution is the same thing I mean by standard deviation)
 
  • #3
genericusrnme said:
Why not use the most common form of the standard deviation?

[itex]\sigma _x = \sqrt{\langle x^2 \rangle - \langle x \rangle ^2}[/itex]

(I'm assuming what you mean by width of distrobution is the same thing I mean by standard deviation)

So how would i actually apply this to my problem? The initial task was to show that the uncertainity principle comes out of the relation between the wavefunction in position-space and momentum-space. This seems to me like using another definition to obtain the desired result.

By width of distribution i mean the value of x where the probability density has dropped to e^(-1) of it's maximum amplitude as discribed in the text.
 

1. What is normalization in quantum mechanics?

Normalization in quantum mechanics refers to the process of ensuring that the total probability of all possible outcomes of a quantum system is equal to 1. This is done by dividing the wave function by a constant known as the normalization constant.

2. Why is normalization important in quantum mechanics?

Normalization is important in quantum mechanics because it ensures that the probabilities of all possible outcomes of a measurement add up to 1. This is necessary for the mathematical consistency of the theory and allows for accurate predictions of the behavior of quantum systems.

3. How is the width of a distribution related to normalization in quantum mechanics?

The width of a distribution in quantum mechanics is related to normalization through the uncertainty principle. The more spread out a distribution is, the less certain we are about the value of the physical quantity being measured. Normalization helps to ensure that the distribution is not too spread out, thus reducing the uncertainty in the measurement.

4. Can normalization affect the shape of a probability distribution in quantum mechanics?

Yes, normalization can affect the shape of a probability distribution in quantum mechanics. The normalization constant can cause the amplitude of the wave function to change, which in turn can alter the shape of the distribution. This is why normalization is an important step in calculating probabilities in quantum mechanics.

5. How is normalization different from normalization by integration in quantum mechanics?

Normalization by integration in quantum mechanics refers to the process of using mathematical integration to determine the normalization constant and ensure that the wave function is normalized. This is typically done for continuous distributions. Normalization, on the other hand, is a more general concept and can also refer to the process of ensuring that the total probability is equal to 1 for discrete distributions. Both methods achieve the same goal of ensuring the mathematical consistency of the theory.

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