Normalisation of a Wavefunction

[misslondonuk]

What condition must a 1D wavefuntion satisfy to be normalised?

Is the fact that it the wavefuntion squared has to equal the probability of finding a particle or that the wavefuntion has to be finite or something totally different??

please help,
thanks

 

[dextercioby]

<To be an element of a preBanach space, or, in particular, an element of L^2 (R, dx)> is a valid answer to your bolded question.

 

[Schrodinator]

The location x,t and hence degrees of freedom must be finite and countable yes to render the probability realistic or reality based. In 3+1 dimensions there are limits on how and where something can be and in what time and the likelihood of it being so is related to this. Hence renormalisation. Obviously probabilities like - (1:6) don't make much sense hence the squaring to remove them, there is not a -1 in -6 chance of anything clearly. It's just a practical consideration to render results meaningful given experiment and the induction hence of the Schrödinger equation there from. As to whether it actually should be used or it reflects what's really going on, well people would probably debate that until the cows come home.

 

[Matterwave]

<To be an element of a preBanach space, or, in particular, an element of L^2 (R, dx)> is a valid answer to your bolded question.

lol, as wave-functions are vectors in the Hilbert Space, which is automatically a Banach space...I don't think this condition has any real significance to the question as presented other than perhaps as a tautology.

Anyways,
The wave-function must be normalizable. This means that you must be able to find some constant A such A times the integral of the absolute square of the wave-function over all of space is 1.

An example of a wave-function which could not be normalized would be psi(x)=x (extending over all space). In this case, the integral over all space of the square of the wave function diverges and you cannot find such an A.

A sufficient condition for the wave function to be normalizable is that it decays to 0 at plus or minus infinity. This is not technically a necessary condition, as some pathological functions can have finite integrals but doesn't decay to 0 at plus or minus infinity. Such pathological functions don't appear in physics.

 

[tom.stoer]

A sufficient condition for the wave function to be normalizable is that it decays to 0 at plus or minus infinity.

This is not sufficient as can be seen by looking at

$$\psi(x) \sim 1/\sqrt{x}$$

This function is not normalizable in L₂. Just try to calculate the integral

$$\int_a^\infty dx\,|\psi(x)|^2 \sim \int_a^\infty dx\,\frac{1}{x}$$

which diverges logarithmically. You can construct similar pathological functions for other L_p spaces.

Given the L₂ Norm on L₂[a,b] the sufficient condition is simply

$$\int_a^b dx\,|\psi(x)|^2 < \infty$$

 

[Matterwave]

This is not sufficient as can be seen by looking at

$$\psi(x) \sim 1/\sqrt{x}$$

This function is not normalizable in L₂. Just try to calculate the integral

$$\int_a^\infty dx\,|\psi(x)|^2 \sim \int_a^\infty dx\,\frac{1}{x}$$

which diverges logarithmically. You can construct similar pathological functions for other L_p spaces.

Given the L₂ Norm on L₂[a,b] the sufficient condition is simply

$$\int_a^b dx\,|\psi(x)|^2 < \infty$$

Ah yes, I forgot about that example. This is what I get for stating things off the top of my head...

I suppose then the wave function must go to zero faster than $$\frac{1}{\sqrt{\abs{x}}}$$ as x goes to plus or minus infinity. This should be a sufficient condition.

 

[tom.stoer]

I suppose then the wave function must go to zero faster than $$\frac{1}{\sqrt{\abs{x}}}$$ as x goes to plus or minus infinity. This should be a sufficient condition.

It should be OK physically but is too restrictive mathematically.

Think about a function which is peaked at the integers n. For large n width and height of the peaks are adjusted such that the integral still converges. You can construct such functions by playing around with repesentations of delta functions quite easily. Of course this is physically nonsense but perfectly acceptable mathematically.

So I would conclude that the only sufficient condition is the integral condition itself.

 

[dextercioby]

lol, as wave-functions are vectors in the Hilbert Space, which is automatically a Banach space...I don't think this condition has any real significance to the question as presented other than perhaps as a tautology. [...]

You find it funny, but the finite integral condition Tom wrote is a particular case of my statement.

 

[Matterwave]

It indeed is a particular case of your statement as a L^2 (I think more commonly l^2 for continuous functions?) Banach space (Lebesgue space?) does not admit objects with no norm, but what I was saying is, I don't think your statement is going to help the OP all that much...

What I mean is, you went overboard with your generality. For example if I asked "what's a vector?", you could very well answer "It's an object with both a magnitude and a direction", or you could answer "It's a geometrical object defined on the tangent space to a manifold, the components of which transform as $$A^\mu'=\frac{\partial x^\mu'}{\partial x^\nu}}A^\nu$$"

Sure the second answer is correct, but it doesn't necessarily help...

@Tom: Ok, you mathematicians got me...I'm out of ideas...I am no good as a mathematician! >.>

 

[tom.stoer]

I don't really see the problem: the OP's question was What condition must a 1D wavefuntion satisfy to be normalised? Assuming that this refers to the L₂ norm my answer was that the integral must be finite. I think we should now wait for his response, further questions etc. and pause this discussion for a while ...

 

[frustr8photon]

Strictly speaking, the OP asked "what makes a wavefunction NORMALIZED," not normalizable.

For a wavefunction to be normalized, the integral the wavefunction from -infinity to infinity must equal 1 (as suggested in Post #4).

 

[Pengwuino]

Christ, you guys sure know how to go overboard with an explanation.

Something being normalized (to 1) simply means that given the function $$\psi$$,

$$\int_{-\infty)^{\infty} |\psi|dx = 1$$.

In other words, the total probability of finding a particle if you look at the entire x-axis sums up to 1, or 100% probability.

EDIT: Whoa what is up with the latex?