Observables of position and momentum have a continuous spectrum

asdf1
Messages
734
Reaction score
0
could someone explain this paragraph taken from "concepts of modern physics" by arthur beiser pg175? I'm having trouble understanding it...

"A dynamical variable G may not be quantized. In this case, measurements of G made on a number of identical systems will not yield a unique result but instead a spread of values whose average is the expectation value
<G>=(integrate) G(psi^2)dx"

and why if the electron's position in the hydrogen atom isn't quantized, we have to think of the electron in the vicinity of the nuvleus with a ceratian probability?
 
Physics news on Phys.org
Some physical observables can take on discrete (quantized) values, like the energy of a particle in an infinite potential well, harmonic oscillator, or the energy of the electron in an hydrogen atom. In this case the eigenvalue-spectrum of the corresponding observable is discrete. This is not always the case though. The observables of position and momentum have a continuous spectrum (ie, not quantized).
The expectation value is calculated the same way as with any observable:
&lt;G&gt;=&lt;\psi|G|\psi&gt;

So the position of the electron in a hydrogen atom in not quantized (it's after all described by a continuous wavefunction) and thus given by a probability density |psi|^2
 
thank you very much for explaining!:)
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top