Implication of energy measurement for wave function as a sum of stationary states


by DocZaius
Tags: energy, function, implication, measurement, states, stationary, wave
DocZaius
DocZaius is offline
#1
Oct17-12, 12:30 PM
P: 287
In Griffiths' page 36/37 he says "As we'll see in Chapter 3, what |Cn|^2 tells you is the probability that a measurement of the energy would yield the value En (a competent measurement will always return one of the "allowed" values - hence the name - and |Cn|^2 is the probability of getting the particular value En."

This statement comes at the end of the infinite square well section and it concerns any sum of stationary states.

Wouldn't measuring one of the allowed energies "En", mean collapsing the wave function into the nth stationary state (and that state only!)? Doesn't that mean that the wave function has then assumed a single stationary state, thereby guaranteeing all subsequent energy measurements of the wave function to yield that particular En?

This seems to disagree with my intuition about what happens to a wave function after measurement (that it would "delocalize" in energy space) and it disagrees with my professor's interpretation who literally takes issue with the statement that an energy measurement must yield one of the allowed values. He says that it can yield any value, and that a mix/sum of specific n states are responsible for returning a non-"allowed" En.
Phys.Org News Partner Physics news on Phys.org
Physicists design quantum switches which can be activated by single photons
'Dressed' laser aimed at clouds may be key to inducing rain, lightning
Higher-order nonlinear optical processes observed using the SACLA X-ray free-electron laser
The_Duck
The_Duck is offline
#2
Oct17-12, 12:45 PM
P: 790
Quote Quote by DocZaius View Post
Wouldn't measuring one of the allowed energies "En", mean collapsing the wave function into the nth stationary state (and that state only!)? Doesn't that mean that the wave function has then assumed a single stationary state, thereby guaranteeing all subsequent energy measurements of the wave function to yield that particular En?
Yes.

Quote Quote by DocZaius View Post
This seems to disagree with my intuition about what happens to a wave function after measurement (that it would "delocalize" in energy space)
I always liked the saying, "the great thing about physical intuition is that it can be adjusted to fit the facts."

Quote Quote by DocZaius View Post
and it disagrees with my professor's interpretation who literally takes issue with the statement that an energy measurement must yield one of the allowed values. He says that it can yield any value, and that a mix/sum of specific n states are responsible for returning a non-"allowed" En.
The *expectation value* of energy can be a non-allowed value. This is the same as the fact that the expected (mean) value of a six-sided die is 3.5. But you'll always get an integer if you roll the die.
DocZaius
DocZaius is offline
#3
Oct17-12, 12:48 PM
P: 287
So just to be clear - once I measure the energy of my infinite square well wave function, I have set it into a stationary state corresponding to that allowed energy and it will remain there subsequently? All future measurements, for all time, of the energy of that wave function will be my first energy measurement? Doesn't that violate some version of the uncertainty principle? The delta E delta t one?

This very much disagrees with my professor so I want to make sure! And I understand your point about the expectation value but we were talking about an actual single measurement.

The_Duck
The_Duck is offline
#4
Oct17-12, 01:26 PM
P: 790

Implication of energy measurement for wave function as a sum of stationary states


Quote Quote by DocZaius View Post
So just to be clear - once I measure the energy of my infinite square well wave function, I have set it into a stationary state corresponding to that allowed energy and it will remain there subsequently? All future measurements, for all time, of the energy of that wave function will be my first energy measurement?
Yes, as long as you don't disturb the system by doing something other than an energy measurement.

Quote Quote by DocZaius View Post
Doesn't that violate some version of the uncertainty principle? The delta E delta t one?
No, it's a consequence of the energy-time uncertainty principle. Griffiths has an explanation of the meaning of this uncertainty relation; have you read it?

The idea is that (delta E)(delta T) ~ hbar, where (delta E) is the spread in energy and (delta T) is some measure of the characteristic time over which the system changes. If we reduce (delta E) to zero, then (delta T) must go to infinity; that is, it takes infinitely long for the system to undergo any changes; that is, the system never changes. Of course, this is why we call energy eigenstates "stationary states"!
DocZaius
DocZaius is offline
#5
Oct17-12, 01:52 PM
P: 287
So all one needs to do to create a wave function that is exactly one stationary state is to create any wave function at all first (which is going to be a sum of stationary states), then measure its energy. Now you have a very simple single function of time and space rather than a sum of many. Can you give an example of an energy measurement in the real world which might be unperturbative enough to be a "pure" energy measurement?

What about measuring the wavelength of a photon emitted from an electron. If a specific allowed jump in energy of an electron caused this photon to be emitted, might not my measurement of that photon's wavelength be considered a "pure" energy measurement of the electron? Would that very measurement of the photon not put the electron in a stationary state? I am of course assuming that one could express the wave function of an electron as a sum of stationary states, which assumes that the time dependent Schrodinger equation of the electron (in some potential) would be separable, which is not necessarily true.
The_Duck
The_Duck is offline
#6
Oct17-12, 10:01 PM
P: 790
To prepare a photon in a highly precise energy state you can use an optical cavity: http://en.wikipedia.org/wiki/Optical_cavity

An optical cavity consists of a pair of partially transparent mirrors separated by a certain distance. If you shoot a photon at it, it will only pass through if it has a wavelength (thus energy) in a very narrow range. You can think of this as an energy measurement measurement than detects whether the photon has a certain energy or not.

A problem with thinking about stationary states of real-world electrons is that the only stationary state of an electron in an actual atom is the ground state. The excited states aren't pure energy eigenstates: they have a finite lifetime, after which they transition to the ground state + a photon, so they aren't stationary.
DrewD
DrewD is offline
#7
Oct17-12, 10:23 PM
P: 427
You should make sure that you are not misunderstanding your prof. In any real world situation, the particle will not be perfectly isolated and there is no infinite well, so over time outside influences will cause the state to evolve from the stationary state. However, in the theoretical perfect world of the infinite square well, a perfect measurement will leave the particle in a fixed state.

Whenever real world situation come in, there will be many complications to this, but don't confuse those complications with the fundamental theory.

There are more complicated situations when it comes to a measurement of the energy, but a measurement that gives you zero uncertainty for the energy of a single particle will necessarily be an eigenvalue of the state.


Register to reply

Related Discussions
Energy of a QM system, stationary states Quantum Physics 1
Definition of stationary state (for wave function quantum) Introductory Physics Homework 1
Stationary state of two body wave function Advanced Physics Homework 1
Stationary States and Spreading of Wave Function Quantum Physics 3
Wave function of Stationary State Quantum Physics 16