Orthogonality of momentum space wavefunctions

In summary, the conversation discusses using partial fractions to rewrite the product in the denominators of the even momentum space wavefunctions for the symmetric well, and using an integral table to show that the integral of the squared wavefunctions equals the Kronecker delta function. The conversation also includes a correction to the statement of the problem.
  • #1
ehrenfest
2,020
1
Page 152 Robinett:

Consider the (non-normalized) even momentum space wavefunctions for the symmetric well:,

[tex]\phi_n^+(p) = 2sin(w-m)/(w-m)+sin(w+m)/(w+m)[/tex] where
w = sin((n-1/2)pi) and
m = ap/hbar.

Show that

[tex] \int_{-\infty}^{\infty}\phi_n^+(p)^*\cdot \phi_n^+(p) dp = \delta_{n,m} [/tex]

The hint is to use partial fractions to rewrite the product found in the denominators and then use an integral table.

So, there are there are terms in the expansion of that integrand. Do I need to rewrite all of them in terms of partial fractions?

The first is 2sin(w-m)sin(w+m)/(w-m)(w+m), which I am having trouble with partial fractions. I get A=B=0 for the numerators?
 
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  • #2
ehrenfest said:
Page 152 Robinett:

Consider the (non-normalized) even momentum space wavefunctions for the symmetric well:,

[tex]\phi_n^+(p) = 2sin(w-m)/(w-m)+sin(w+m)/(w+m)[/tex] where
w = sin((n-1/2)pi) and
m = ap/hbar.

Show that

[tex] \int_{-\infty}^{\infty}\phi_n^+(p)^*\cdot \phi_n^+(p) dp = \delta_{n,m} [/tex]

The hint is to use partial fractions to rewrite the product found in the denominators and then use an integral table.

So, there are there are terms in the expansion of that integrand. Do I need to rewrite all of them in terms of partial fractions?

The first is 2sin(w-m)sin(w+m)/(w-m)(w+m)...

no, it's 2sin(w-m)(w+m)/(w-m)(w+m).
 
  • #3
So the integrand is

[tex] \frac{a}{2\pi\hbar} \left(sin^2(w-m)/(w-m) + 2 sin (m-w) sin (w+m)/(w-m)(w+m) + sin^2(w+n)/w+m\right) dp[/tex]

I think I can integrate the squared terms, but I am not sure how to do partial decomposition on the middle term to derive something useful from it.

When I try partial fraction decomposition on that term I get

2 sin(w-m) sin(w+m)/(2n- pi) (1/(w-m) +1/(w+m) )Sorry. The statement of the problem is wrong. (n-1/2)pi not the sine of that.
 
Last edited:
  • #4
Sorry. The statement of the problem is wrong. (n-1/2)pi not the sine of that.
 

Related to Orthogonality of momentum space wavefunctions

1. What is the concept of orthogonality in momentum space wavefunctions?

Orthogonality in momentum space wavefunctions refers to the property of two wavefunctions having a dot product of zero, indicating that they are perpendicular to each other. This means that their corresponding states have no overlap in momentum space.

2. How is orthogonality of momentum space wavefunctions related to the uncertainty principle?

The uncertainty principle states that there is a fundamental limit to the precision with which certain pairs of physical properties of a particle, such as position and momentum, can be known simultaneously. In momentum space, the wavefunctions of these properties are orthogonal, which reflects the fact that they cannot be precisely known at the same time.

3. Why is orthogonality important in quantum mechanics?

Orthogonality plays a crucial role in quantum mechanics because it allows for the calculation of probabilities in quantum systems. The squared magnitude of the dot product between two wavefunctions represents the probability of finding a particle in a particular state, making orthogonality a fundamental concept in understanding the behavior of particles at the quantum level.

4. How is orthogonality of momentum space wavefunctions used in quantum mechanics calculations?

In quantum mechanics calculations, the orthogonality of momentum space wavefunctions is used to determine the probability amplitudes of different states and the overall state of a system. It is also used in the calculation of expectation values and transition probabilities between states.

5. Can two non-orthogonal momentum space wavefunctions exist?

Yes, two non-orthogonal wavefunctions can exist in momentum space. This means that their corresponding states have some overlap, and the probability of finding a particle in either state is not zero. However, in quantum mechanics, it is often more useful to work with orthogonal wavefunctions as they simplify calculations and provide a clearer understanding of the system.

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