Normalizing a wave function, what does it mean?

So if you normalize a wavefunction so that the probability is always 1, you are saying that the wavefunction represents a state that is always possible. But what does that mean practically?f
  • #1
I can normalize wave functions all day, but I still don't know what I mean when I normalize them! So my question, what does it mean to normalize a wave function?
  • #2
Normalizing a wave function means finding the form of the wave function that makes the statement

[tex]\int^\infty_{-\infty} \psi^* \psi dx = 1 [/tex]

true. Essentially, normalizing the wave function means you find the exact form of [tex] \psi [/tex] that ensure the probability that the particle is found somewhere in space is equal to 1 (that is, it will be found somewhere); this generally means solving for some constant, subject to the above constraint that the probability is equal to 1.

(" an example from wikipedia, if the above description didn't make sense.
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  • #3
One could say that normalizing a wave function to not really have any physical meaning. It just means transforming the wave function into a form in which it is easier to evaluate observables. E.g., you have
[tex]\langle \hat O \rangle = \langle\Psi|\hat O|\Psi\rangle[/tex]
instead of the more elaborate
[tex]\langle \hat O \rangle = \frac{\langle\Psi|\hat O|\Psi\rangle}{\langle\Psi|\Psi\rangle}[/tex]
you'd need to use otherwise.
  • #4
I always pictured normalization in the easiest way possible...normalizing the wavefunctions gives them a coeffecient, such that the coeffecient squared is the probability of finding the particle etc. in that eigenstate (which is represented by the wavefunction).
  • #5
Since [tex]|\psi|^2[/tex] is a probability density, integrating it over all space has to equal 1...otherwise its not a probability density.
  • #6
In order for a wavefunction to be a valid description of reality it MUST be normalizable since its square represents a probability. HOWEVER, one doesn't actually need to do the normalization (it just has to be doable in principle). Thus normalization is really just a matter of convenience since a state in quantum mechanics is only defined up to a phase constant we can add phase factors all we want until the cows come home and not change the physics (geometrically we can say that if a quantum state is a vector, its DIRECTION and not its length is all that matters). Thus, normalizing our wavefunction is often just a matter of convenience so you have 1's that disappear in your calculations instead of arbitrary phase factors that you have to cart around with you.
  • #7
To normalize a vector f is to multiply it by a complex number of magnitude [itex]1/\|f\|[/itex] (so that the result is a vector with norm 1).

Steger: The last thing you said sounds a bit weird. Since the norm is unchanged under multiplication by a phase factor (a complex number of magnitude 1), normalization won't make phase factors go away.
  • #10
Help with what? :confused:

I think the confusion is that I'm just sloppy with my language. How about "wavefunctions are only defined up to a phase factor and/or a normalization constant" (with the understanding that phase factors are subsets of normalization constants)
  • #11
Yes, I assumed that you were just a little sloppy with the details, not that you had misunderstood anything major.

I like to say that it's the one-dimensional subspaces that represent pure states, not the vectors. A one-dimensional subspace is called a ray. If R is a ray, the subset of R that consists of all members of R with norm 1 is called a unit ray. Since the set of rays can be mapped bijectively onto the set of unit rays, it's just as fine to say that the unit rays represent pure states. The only difference between two members of the same unit ray is a phase factor.

A wavefunction is a specific vector, not a ray or a unit ray. Every wavefunction is of course a member of a ray, and every normalized wavefunction is a member of a unit ray.
  • #12
The distinction between rays and vectors is crucial. Without describing states through rays, one couldn't obtain a theoretical explanation of the concept of spin.
  • #13
phase factors are not normalization values as they have modulus one
  • #14
I know what you really mean. It is fair enough to say that the integral of such and such equals one but what does it ACTUALLY mean?

To normalise something is to put the wavefunction into terms that everyone understands.
Like != factorial, A= amps etc.

To express a wavefunction as a normalisation is to generalise it so that someone from another part of the world is able to understand what you are writing.

It all boils down to probability. You know that that probability can only exist between 0 and 1. You can't get a probability of 2, it's just not right and it's the rule that the whole world uses. So if you can express a wavefunction in a normalised manner then you can say that this holds considering that the probability is one i.e. it actually happens, and people from everywhere know what you are on about.

My explaining isn't the best but I hope you get a feel for what I am trying to say!
  • #15
I think you guys are dwelling a little too much on the mathematical nature of normalization here. I think the OP wanted a more physical intuition to the problem. Since he said that he can, in fact, "normalize wave functions all day", I assume that he knows what normalization means mathematically at least.

Physically, normalization just means that there must be a probability of 1 of finding a particle in SOME state out of ALL the possible states that it could be in. For a continuous observable, like the position of a particle, then the particle must exist somewhere in space (so that you can't get a probability not equal to 1 if you integrate over all of space). For a discrete observable, the probabilities it has in each state must add up, over all possible states, to be 1. For example, if a spin 1/2 particle has a .4 probability of its z-component spin-vector pointing up, then there must be a .6 probability that its z-component points down, as these are the only 2 possibilities.
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