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Bob Dylan

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Bob Dylan

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BvU

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Hi,

What exactly do you mean with the radius of a function ?

And with 'hat' ?

What exactly do you mean with the radius of a function ?

And with 'hat' ?

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Bob Dylan

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BvU

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Of course not.each separate wavefunction have a radius of one

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)##

eigenfunctions ##\ \phi_n\ ## of a particular operator ##\ \hat O\ ## for which ##\ \hat O \phi = \lambda_n \phi##

the complex

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.

##\mathstrut##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.

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 ?

No. Why should it ? It depends (operator? state ?)Secondarily, can this multiplication only occur in the initial state

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Bob Dylan

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BvU

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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 )I'll admit

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

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

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