Does Multiplying Wavefunctions Affect Their Radius?

[Bob Dylan]

When we multiply psi sub x, psi sub y, psi sub z and psi sub t together to get a function of all four variables, does each separate wavefunction have a radius of one such that the radius is unchanged after the multiplication or is their radius far smaller than one? Secondarily, can this multiplication only occur in the initial state and if so can one say that the p hat, E hat and x hat eigenstates are necessarily separated?

 

[BvU]

Hi,
What exactly do you mean with the radius of a function ?
And with 'hat' ?

 

[Bob Dylan]

By radius and function I mean amplitude and wavefunction and I am talking about the x hat and p hat operators etc which scale the amplitude by the eigenvalue and thus seem to alter the way these would multiply.

 

[BvU]

each separate wavefunction have a radius of one

Of course not.

Can you please clearly distinguish (I refer to the notation here but use $\phi$ for eigenfunctions)

wave function $\ \psi\$. e.g. $\ \psi(x,y,z,t)\$

the complex value of a wave function at a given argument, e.g. $\ {\bf a} = \psi(x_1,y_1,z_1,t_1)$

to me, amplitude of a wavefunction means $\ \sqrt{\bf a^* a}\$ and 'radius' is not a term used anywhere in this context
Note: $\ \sqrt{\bf a^* a}\$ at a given argument. The $\ ^* \$ indicates complex conjugation.

In general: amplitude is $\ \sqrt{\psi^*\psi} \$ and coincides with the probability density.​

$\mathstrut$
eigenfunctions $\ \phi_n\$ of a particular operator $\ \hat O\$ for which $\ \hat O \phi = \lambda_n \phi$​

Let's try to make your question more concrete: you are worried that applying an operator on a wave function changes something that you would want to keep unchanged ? Example ?

Secondarily, can this multiplication only occur in the initial state

No. Why should it ? It depends (operator? state ?)

 

[Bob Dylan]

I'll admit my hazy terminology reflects my hazy understanding. At any rate, what I'm currently trying to do at the most general level is tie together what seem to me to be unrelated symnols of amplitude. The first symbol being that it determines the probability, the second that it can be the eigenvalue and the third that the ampliude is a measure of length (or has a length at any rate). Are all of these things on the same scale? Or do we use one scale for a value of 1 when we are talking about the length or eigenvslue and a separate scale for 1 when talking about the probability?

 

[BvU]

I'll admit

Kudos for your efforts to learn and understand. In general, early stages in the curriculum for QM lay quite a claim on one's imagination (witness he numerous threads in PF). At the same time your imagination easily leads you astray from the beaten path of well-established science (even more threads in PF, however hard the moderators try to lock 'em )

It is hard to attribute a physical meaning to the wave function $\psi$ itself ("a complex-valued probability amplitude" ), other than that $\ \psi^*\psi\$ repesents a probability density (*).

(*) Already here I have to correct post #2, where I wrote that the amplitude $\ \sqrt{\psi^*\psi }\$ repesents a probability density. The amplitude is a probability amplitude. ( Sounds logical, doesn't it. I looked it up in my Merzbacher, QM 1970 ; fortunately wiki agrees )​

Scale is determined by normalization: like $\ \int \psi^*\psi = 1$ (a resonable requirement for a probability density function) .

Since $\ \psi^*\psi\$ is a density, its dimension is 1/"whatever it is you are integrating over":$\ \int_{\rm \text whole \ space} \psi^*\psi \; d\tau = 1$

Dimensions come in with the operators. Physically relevant operators are Hermitian and (phew...) therefore have real-valued eigenvalues that are potentially observable. So expectation values of operators have the corresponding dimension.

You'll learn and after a while the lingo becomes almost natural. Usually that's where genuine theoreticians come into declare it's all nonsense and should be re-built from scratch