High School Is energy really probabilistic in quantum mechanics?

[cube137]

in classical physics.. energy forms a continuum.. for example.. water gently gets warm and hot and boils then cools down gradually.. but in quantum mechanics.. it's like water suddenly gets ice cold and then boils..

I mean, can someone shows experimental proof that energy is really probabilistic in quantum mechanics? I need an actual setup or example..

by the way. if energy is probabilistic in QM, why is mass not probabilistic since mass is related to energy?

thank you..

 

[PeterDonis]

in quantum mechanics.. it's like water suddenly gets ice cold and then boils

Where are you getting that from?

can someone shows experimental proof that energy is really probabilistic in quantum mechanics?

Sure, just measure the energy of any system that is not in an energy eigenstate.

if energy is probabilistic in QM, why is mass not probabilistic since mass is related to energy?

Mass in the sense of energy divided by $c^2$ is probabilistic in QM, because it's just energy in different units.

What is not probabilistic is "mass" in the sense of an inherent property of particular particles, such as the "mass" of the electron. But you need quantum field theory to really understand what this "mass" is and why it's not the same as energy.

 

[cube137]

Where are you getting that from?
Sure, just measure the energy of any system that is not in an energy eigenstate.
Mass in the sense of energy divided by $c^2$ is probabilistic in QM, because it's just energy in different units.

What is not probabilistic is "mass" in the sense of an inherent property of particular particles, such as the "mass" of the electron. But you need quantum field theory to really understand what this "mass" is and why it's not the same as energy.

How do you measure the energy of any system? I'd like to know if when you measure them consecutively, the energy really is probabilistic.. do you have actual example that is already done?

 

[PeterDonis]

How do you measure the energy of any system?

Strictly speaking, we measure differences in energy. For example, we measure the frequency of light emitted by an atom when it changes state; that tells us the difference in energy between the states of the atom involved.

I'd like to know if when you measure them consecutively, the energy really is probabilistic

It will be if the system is not in an eigenstate of energy. For example, if you take a large sample of identically prepared radioactive atoms, some will decay and some will not. The energy of the atom after decay is different from before, so the atoms when they are prepared cannot be in an eigenstate of energy (if they were, they would never decay). So this is a probabilistic measurement of energy.

 

[cube137]

Strictly speaking, we measure differences in energy. For example, we measure the frequency of light emitted by an atom when it changes state; that tells us the difference in energy between the states of the atom involved.
It will be if the system is not in an eigenstate of energy. For example, if you take a large sample of identically prepared radioactive atoms, some will decay and some will not. The energy of the atom after decay is different from before, so the atoms when they are prepared cannot be in an eigenstate of energy (if they were, they would never decay). So this is a probabilistic measurement of energy.

You stated that we don't measure the energy of a branch because it's meaningless.. Right now.. we are in one such branch.. and you can indeed measure a radioactive atom and you can get certain values. So in what sense do you mean it's meaningless. How can you tell if the radioactive atom is in one branch or it is the whole quantum state itself where you apply the Hamiltonian operator?

 

[PeterDonis]

Right now.. we are in one such branch.. and you can indeed measure a radioactive atom and you can get certain values.

Yes, but you aren't observing what all the other branches are doing. So you can't say that the measurement is only taking place in one branch.

Also, the measurement itself (if the system is not in an energy eigenstate) creates branches, in each of which a different energy value is observed.

in what sense do you mean it's meaningless

I mean that it's meaningless to ask how the energy of the system is "split up" between the different branches. That's not how it works.

How can you tell if the radioactive atom is in one branch or it is the whole quantum state itself where you apply the Hamiltonian operator?

This question doesn't make sense.

We are to the point where a "B" level thread simply can't address the issue you want to discuss. As I said before, you need to spend some time working through a QM textbook. When you have enough background to participate in an "I" or "A" level thread, you can start one--but by that time I think the questions you think you need to ask now will have evaporated as your knowledge increases.

Thread closed.