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Non normalisable wave function

  • #1

Homework Statement



At a given point, the wave function of a particle in a non normalisable state is 1+sin^2(kx). When you measure thekinetic energy, which values are expected and with which probabilities?

Homework Equations



K=<P2>/2m

The Attempt at a Solution



I guess I should apply the operator <P2> to the wave function, call it f(x), so

K=-h2d2f(x)

and I finally end up with -h2k2/m (cos2kx-sin2kx)

so? does that mean we have -h2k2/m and h2k2/m with 50% probability each?
Is this rigth if we can't normalise the wave function?

thanks

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
180
4
Express your wave function in terms of eigenfunctions of energy.
 
  • #3
that's

Hf(x) = Ef(x)

which gives E as a function of x ... and the probability?

thanks
 
  • #4
180
4
No, E should be a constant (in respect to x). Just express your wave function as a linear combination of functions that satisfy the above equation (Hint: eigenfunctions of the free hamiltonian are plain waves). You can get the probabilities from the coefficients of different eigenfuctions, just make sure that probabilities add up to 1.
 
  • #5
With sin2kx = 1/2-cos2kx/2 I get a cos2kx which represents a plain wave but what about the constant term the 1 + ... (I guess it's a silly question) ...

can I transform the wave function to the base of momentums with the Fourier transform (obtaining some Dirac deltas) and get the energies from there?

thank you for the patience
 
  • #6
180
4
sin2kx is not an eigenfunction of energy, eikx is however.

Hint 1: [tex]\sin(kx) = \frac{1}{2i}(e^{ikx} - e^{-ikx})[/tex].

Hint 2: [tex]1 = e^{i0x}[/tex]

Calculating the Fourier Transform should also work although I think it's kinda unnecessary because in order to calculate it you have to express sinkx in terms of exponentials at which point you will be able to see what the energies are anyway.
 
  • #7
right, I tried that before, but is eik0 an eigenfunction? then how this satisfy the schrodinger equation for a free particle? I tried but the second derivate of the constant it's 0 so:

[tex]\frac{P^{2}}{2m}\varphi = E\varphi[/tex]

[tex]\frac{2k^{2}}{m}(\frac{e^{2jkx}+e^{-2jkx}}{2}) = E (2 - \frac{e^{2jkx}+e^{-2jkx}}{2})[/tex]

that's why I asked for the representation in the space of momentums where I get 3 [tex]\delta[/tex] ... I don't know either how to get the possible energies from there and the probs, but i guess the possible values of p are the positions of the [tex]\delta[/tex] and the probabilities the square of the amplitud?.
 
  • #8
180
4
ei0x is an eigenfunction, it represents a particle at rest (k=0 => E=0)

[tex]-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} e^{i0x} = 0 e^{i0x}[/tex]

So just express your wave function as a linear combination of plain waves eikx (with different k). You know that the energy of a plain wave is

[tex]E_k = \frac{\hbar^2 k^2}{2m}[/tex]

[tex]\frac{P^{2}}{2m}\varphi = E\varphi[/tex]

[tex]\frac{2k^{2}}{m}(\frac{e^{2jkx}+e^{-2jkx}}{2}) = E (2 - \frac{e^{2jkx}+e^{-2jkx}}{2})[/tex]
This won't work because your wave function is not an eigenfunction of energy. It is a superposition of different energy eigenfunctions which satisfy the eigenvalue equation on their own with different energies.
 
  • #9
I think I get it now ...

[tex]H\varphi_{0}=E_{0}\varphi_{0}[/tex]

and

[tex]H\varphi_{1}=E_{1}\varphi_{1}[/tex]

and so on ...

therefore

[tex]\varphi=2\varphi_{0}-\frac{1}{2}\varphi_{1.1}+\frac{1}{2}\varphi_{1.2}[/tex]

having [tex] \varphi_{1.1} and \varphi_{1.2}[/tex] the same energy

[tex]E_{0}=0 and E_{1}=-\frac{2h^{2}k^{2}}{m}[/tex]

and the probabilities are the squares of coefficients in [tex]\varphi[/tex] normalized 8/9 and 1/9.

is it right?
 
  • #10
180
4
You're on the right track, only E1 is positive (kinetic energy can't be negative) since i squared cancels the minus sign in front of the derivative, and I'm not so sure about those probabilities.

[tex]1 + \sin^2(kx) = \frac{3}{2} - \frac{1}{4}e^{i2kx} - \frac{1}{4}e^{-i2kx}[/tex]
 
  • #11
You are right, there was a minus I forgot in the calculations.

The wave function was also wrong. It should be

[tex]
\varphi=2\varphi_{0}-\frac{1}{2}\varphi_{1.1}-\frac{1}{2}\varphi_{1.2}
[/tex]

What I used for the probabilities is that the square of the cofficients should sum up to 1.
So I calculate the resizing factor a

[tex]a\frac{1}{2^{2}}+a\frac{1}{2^{2}}+a2^{2}=1[/tex]

and end up with probability for E0 = 8/9
 
  • #12
180
4
That's right, but how did you arrive at those coefficients in the first place (I mean 2 and 1/2s)?
 
Last edited:
  • #13
I arrive easily to those coeficients making a stupid mistake in the calculation of the [tex]
1 + \sin^2(kx) [/tex]

The right coefficients are the ones you posted before. Then I get P(Eo)=18/19 and P(E1)=1/19

thanks for your help, I appreciate
 

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