Orthogonality of wave functions

In summary, the conversation is about proving that the integral of psi1*psi0 is 0, with the use of integration by parts and considering boundary conditions. The poster is struggling to show that the expression is equal to 0 and is unsure about the necessary boundary conditions. The discussion continues with suggestions for possible boundary conditions and how to evaluate the integral at limits. Ultimately, the issue seems to be with the wording of the question and the lack of given information.
  • #1
Plaetean
36
0

Homework Statement


aaBkgpy.png


Homework Equations

The Attempt at a Solution



http://i.imgur.com/tktQBsp.jpg
[/B]
I assume that you need to prove that the integral of psi1*psi0 is 0, so I have written out the integral and attempted to solve using integration by parts, but whichever way I write out the integration by parts it doesn't seem to cancel down. Do I need to write the integral as a definite integral instead and evaluate it at limits? Or perhaps invoke some properties of waves instead that will allow me to say =0?
 
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  • #2
I would imagine you should have some boundary conditions at infinity due to physical conservation laws. What do you get if you assume ## \psi_0 (\pm \infty)=0##?
 
  • #3
What boundary conditions must ##\psi## fulfil in order for it to be normalisable? (It really does not matter if these boundary conditions are in infinity or not.)
 
  • #4
From the previous questions, we are told that psi0 is normalised, real and even. There's nothing in terms of boundary conditions given, which is why I'm having trouble getting any further than just writing out the integration by parts.
 
  • #5
Please show your actual work.
 
  • #7
Move the integral from the righthand side to the lefthand side.
 
  • #8
vela said:
Move the integral from the righthand side to the lefthand side.
That just gives me 2A(integral)=APsi02, which is a circular argument and just proves that the integral = the integral. I feel like I must be missing something really obvious and its driving me a little crazy.
 
  • #9
I'm not sure why you think it's showing the integral equals the integral. You have an equation that says the integral is equal to some other expression. You need to show or argue that that expression is equal to 0.
 
  • #11
What boundary conditions do you usually put on psi?
 
  • #12
Orodruin said:
What boundary conditions do you usually put on psi?
We've done infinite square well, square well, free particle and harmonic oscillator. However there's no info given here and I don't know how to prove it 'generally', so I think it just may be an issue with the way the question is worded. I'm finding the questions worth far more marks a lot easier than this one, so I don't think its intended to be this confusing. I'll speak to the tutor tomorrow.

edit: thank you everyone for the replies though, its very much appreciated
 
  • #13
Plaetean said:
However there's no info given here and I don't know how to prove it 'generally', so I think it just may be an issue with the way the question is worded.

There is no issue. The boundary conditions are general if you want to have a wave function that is possible to normalize (which it has to be in order to be normalized).
 
  • #14
Plaetean said:
http://imgur.com/T4fQLqt Proving Psi02=0 seems even more difficult surely?

You didn't write down limits in that image. Suggest you do so. You don't have to prove ##{\Psi_0}^2=0##.
 
  • #15
I don't know what the limits would be, -infinity to +infinity? 0 to +infinity? How you would even evaluate the integral at any limits considering we aren't actually given the function?
 
  • #16
What conditions would you get if the potential becomes infinite at some point? What conditions would you get if it does not? I suggest starting with the second case as it is more general - the former can be seen as a trivial extension.
 
  • #17
Plaetean said:
I don't know what the limits would be, -infinity to +infinity? 0 to +infinity? How you would even evaluate the integral at any limits considering we aren't actually given the function?

You are not given the region in this problem, so you have to allow a general region (within physical reason, so you probably don't have to consider, e.g. disconnected sets). You should put in the limits:
$$2\int_a^b \psi_1(x)\psi_0(x) dx = A\left.\psi_0(x)^2\right|_a^b$$ Now you just have to figure out what ##\left.\psi_0(x)^2\right|_a^b## has to be in order for the wave function to be normalizable and continuous and differentiable. So you have to know what ##\psi_0(a)## and ##\psi_0(b)## are, these are the boundary conditions that people keep talking about.

This problem seems to be a bit hard if you have to also prove that all physical boundary conditions lead to the conclusion that ##\left.\psi_0(x)^2\right|_a^b=0##, but at least for all the boundary conditions I can think to impose, this condition is met.
 

Related to Orthogonality of wave functions

What is the concept of orthogonality in wave functions?

The concept of orthogonality in wave functions refers to the mathematical property where two functions are perpendicular or at right angles to each other. In the context of quantum mechanics, this means that the overlap or inner product of two wave functions is equal to zero, indicating that they are independent of each other.

Why is orthogonality important in quantum mechanics?

Orthogonality is a fundamental concept in quantum mechanics because it allows us to describe and analyze the behavior of particles in a system. It plays a crucial role in determining the probability of a particle being in a particular state, and it also helps us to understand the symmetries and properties of wave functions.

How is orthogonality of wave functions related to the uncertainty principle?

The uncertainty principle, a key principle in quantum mechanics, states that it is impossible to know the exact position and momentum of a particle simultaneously. The concept of orthogonality of wave functions is closely related to this principle, as it is through the superposition of orthogonal wave functions that we can describe the position and momentum of a particle with a certain degree of uncertainty.

Can wave functions that are not orthogonal exist?

Yes, it is possible for wave functions that are not orthogonal to exist. However, in the context of quantum mechanics, these wave functions would not be considered as valid solutions to the Schrödinger equation, which describes the behavior of particles in a quantum system. Orthogonality is a crucial criterion for determining the validity of wave functions in quantum mechanics.

How do we determine the orthogonality of two wave functions?

The orthogonality of two wave functions can be determined by calculating their inner product, which is the integral of the product of the two wave functions over a given range. If the inner product is equal to zero, the wave functions are orthogonal. If it is non-zero, the wave functions are not orthogonal. In some cases, the orthogonality of wave functions can also be determined by examining their functional forms and boundary conditions.

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