Energy measurement on superposition of states

  • #1
138
2

Main Question or Discussion Point

We have a 1 dimensional infinite well (from x=0 to x=L) and the time dependent solution to the wavefunction is the product of the energy eigenstate multiplied by the complex exponential:

[tex] \Psi_n(x, t) = \sqrt{\frac{2}{L}} \sin(\frac{n\pi x}{L}) e^{-\frac{iE_n}{\hbar}} [/tex]

Now, I want to create a superposition of two states, n = 1 and n = 2. I can write this as:

[tex] \Psi(x, t) = \frac{1}{\sqrt{2}}\sqrt{\frac{2}{L}} \sin(\frac{\pi x}{L}) e^{-\frac{iE_1}{\hbar}} + \frac{1}{\sqrt{2}}\sqrt{\frac{2}{L}} \sin(\frac{2\pi x}{L}) e^{-\frac{iE_2}{\hbar}} [/tex]

Before measurement, the energy is ##E_1 + E_2##.

I make a measurement of the energy and we find that the energy is ##E_1##.

Here's my question: What happened to the rest of the energy that was in the system (ie, the ##E_2## energy)? Where does it go?
 

Answers and Replies

  • #2
963
213
Before measurement, the energy is E1+E2E1+E2E_1 + E_2.
on what basis /assumption can you say that energy is E1 +E2 ?

suppose a particle is in the well and can exist in N number of states having each energy eigen functions and one superposed all the states theoretically , then practically particle can jump out of the well with all N energies added together.? something is amiss.
 
  • #3
138
2
Hey @drvrm, thanks for the response. My assumption was that I have created a system in which the particle is only in a superposition of the two states, n = 1 and n = 2.

Is that a physically impossible scenario?
 
  • #4
963
213
Is that a physically impossible scenario?
i am afraid it is so.

see the following details-

The non-classical nature of the superposition process is brought out clearly if we consider the superposition of two states, A and B, such that there exists an observation which, when made on the system in state A, is certain to lead to one particular result, a say, and when made on the system in state B is certain to lead to some different result, b say.

What will be the result of the observation when made on the system in the superposed state?

The answer is that the result will be sometimes a and sometimes b, according to a probability law depending on the relative weights of A and B in the superposition process.

It will never be different from both a and b [i.e, either a or b]. The intermediate character of the state formed by superposition thus expresses itself through the probability of a particular result for an observation being intermediate between the corresponding probabilities for the original states, not through the result itself being intermediate between the corresponding results for the original states.[1]

ref.https://en.wikipedia.org/wiki/Quantum_superposition
 
  • #5
Nugatory
Mentor
12,619
5,171
Before measurement, the energy is ##E_1 + E_2##
It is not. That state means there is a 50% chance that an energy measurement will yield ##E_1## and a 50% chance that it will yield ##E_2##. The expectation value of an energy measurement, which is the closest thing to a "energy value before measurement" we have (and it's not very close) is ##(E_1+E_2)/2##
It is worth noting that this value is guaranteed not to be the result that we actually get.
Here's my question: What happened to the rest of the energy that was in the system (ie, the ##E_2## energy)? Where does it go?
The interaction between the measuring device and the system under measurement involves an exchange of energy. Thus, the energy of the system can come out to be either ##E_1## or ##E_2##, but either way the energy of the combined system consisting of the measuring device and the system under measurement will be conserved. However, there's no way of verifying this without setting up yet another device to measure the energy of the combined system - and then the problem just reoccurs because I have to account for the energy exchanged with that device, and so on ad infinitum.

This is in stark contrast with classical physics, where I can make the energy exchange between device and system under measurement arbitrarily small, and therefore insist that there must be a premeasurement energy that was arbitrarily close to the value I actually measured.
 
  • #6
hilbert2
Science Advisor
Insights Author
Gold Member
1,343
415
A minor note: remember to include the time variable in the exponential phase factor: ##e^{-iEt/\hbar}##.
 
  • #7
It is not. That state means there is a 50% chance that an energy measurement will yield ##E_1## and a 50% chance that it will yield ##E_2##. The expectation value of an energy measurement, which is the closest thing to a "energy value before measurement" we have (and it's not very close) is ##(E_1+E_2)/2##
Why it's not very close? Could you guide me through the details please? Some book reference or article could be very great (I have Sakurai, Cohen-Tannoudgi, Griffiths and Shankar for Quantum Mechanics)
 
  • #8
Nugatory
Mentor
12,619
5,171
Why it's not very close? Could you guide me through the details please? Some book reference or article could be very great (I have Sakurai, Cohen-Tannoudgi, Griffiths and Shankar for Quantum Mechanics)
Not close in the sense that it doesn't correspond very closely to the classical notion that this system has some preexisting tangible property with that numerical value.
 
  • #9
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,395
5,160
Why it's not very close? Could you guide me through the details please? Some book reference or article could be very great (I have Sakurai, Cohen-Tannoudgi, Griffiths and Shankar for Quantum Mechanics)
The situation is simpler in the case of half-integer spin. Let's say a particle has spin up in the z-direction. We measure the spin in the x-direction, then again in the z-direction. Now, the particle may have spin down in the z-direction. It's clear from this that conservation of spin angular momentum for the particle alone does not survive the measurement process.

See the opening chapter of Sakurai.
 
  • #10
PeterDonis
Mentor
Insights Author
2019 Award
28,333
8,084
Before measurement, the energy is ##E_1 + E_2##.
No. Before the measurement, the system doesn't even have a definite energy, because it's in a superposition of states of different energy.
 
  • #11
117
0
The energy of the superposition of states is not E = E1 + E2. It is E = E1 and E2. It's not the sum of the two energies.
 
Last edited:

Related Threads on Energy measurement on superposition of states

Replies
4
Views
571
Replies
2
Views
826
Replies
5
Views
607
Replies
9
Views
3K
Replies
1
Views
3K
  • Last Post
Replies
2
Views
855
  • Last Post
Replies
13
Views
4K
Replies
30
Views
8K
Top