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Heisenberg picture describes emission, Schroedinger picture does not

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nortonian
#1
Jul17-11, 11:14 AM
P: 59
Am I right in thinking that the Heisenberg matrix interpretation describes emission, while the Schroedinger interpretation does not?
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dextercioby
#2
Jul17-11, 03:57 PM
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Well, essentially they're one and the same thing: either wave mechanics or matrix mechanics, it doens't matter. Either the Schroedinger picture (evolution of the states) or Heisenberg picture (evolution of the observables), or interaction picture (evolution of both categories), it doesn't really matter.

The only reason we choose though one of them over the other is the easiness we seek for making the calculations.
Naty1
#3
Jul17-11, 06:37 PM
P: 5,632
Post #2 is precisely on target.

Because wave mechanics and and matrix mechanics have somewhat different starting points, some phenomena might be a bit clearer in one than the other.

For example, the Debroglie theory of matter waves leads naturally to wave mechanics and once you set boundary conditions to the Schrodinger wave equation, as in a resonant cavity or a violin string, which limit vibrational patterns, you are "smacked upside the head" with quantized resonant energies related to simple trig functions (sin and cos).

nortonian
#4
Jul18-11, 10:34 PM
P: 59
Heisenberg picture describes emission, Schroedinger picture does not

"For example, the Debroglie theory of matter waves leads naturally to wave mechanics and once you set boundary conditions to the Schrodinger wave equation, as in a resonant cavity or a violin string, which limit vibrational patterns, you are "smacked upside the head" with quantized resonant energies related to simple trig functions (sin and cos)."

I realize that the two pictures are mathematically equivalent, but in trying to visualize them I am thinking of a standing wave in the case of the Schroedinger picture (no emission) and in the Heisenberg picture there is a matrix where each off diagonal element is a transition (emission). Even though both give the same answer they make different physical assumptions. Time is discrete in the H picture and continuous in the S picture. Does this make sense?
nortonian
#5
Jul18-11, 10:36 PM
P: 59
I replied to your message
nortonian
#6
Jul20-11, 09:54 AM
P: 59
Quote Quote by dextercioby View Post
Well, essentially they're one and the same thing: either wave mechanics or matrix mechanics, it doens't matter. Either the Schroedinger picture (evolution of the states) or Heisenberg picture (evolution of the observables), or interaction picture (evolution of both categories), it doesn't really matter.

The only reason we choose though one of them over the other is the easiness we seek for making the calculations.
As I said in my previous message the two pictures may be mathematically equivalent, but the H picture uses discrete time and the S picture uses continuous time. Matrix mechanics consists of compilations of discrete transitions (photons) in arrays while the S wave equation describes a standing wave, not an emitted photon. So even if you get the same answers it seems to me that you are reading into it a physical interpretation where there is none.
dextercioby
#7
Jul20-11, 10:46 AM
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Quote Quote by nortonian View Post
As I said in my previous message the two pictures may be mathematically equivalent, but the H picture uses discrete time and the S picture uses continuous time. Matrix mechanics consists of compilations of discrete transitions (photons) in arrays while the S wave equation describes a standing wave, not an emitted photon. [...]
.

With green I've marked the right part of the quoted section of your post.
nortonian
#8
Jul20-11, 12:15 PM
P: 59
Quote Quote by dextercioby View Post
.

With green I've marked the right part of the quoted section of your post.
Sorry, I thought physics was based upon the word "physical". I must have the wrong forum.
nnnm4
#9
Jul20-11, 04:08 PM
P: 113
Well nortonian you seem to have your ideas about the Heisenberg picture wrong. Time is still a continuous parameter there. And the two pictures are mathematically equivalent in such a simple way that the the physical interpretations are not so different for them. Converting between the two pictures just makes use of the associative property built into the Dirac notation. In S, the states are time dependent, so you have your vibrating string picture, in H the operators are time dependent.
nortonian
#10
Jul21-11, 04:51 AM
P: 59
Quote Quote by nnnm4 View Post
Well nortonian you seem to have your ideas about the Heisenberg picture wrong.
It seems I did have the wrong idea about the H picture. It turns out neither picture describes emission.

Both pictures describe the "development of a physical system in time. The quantities that are not altered are the matrix elements of dynamical variables calculated for pairs of states. These provide the essential content of the theory. Any unitary transformation, whether or not it depends on time, may be applied to both states and physical variables without affecting the values of the matrix elements. Thus an arbitrarily large number of different pictures can be found." p. 171 L. Schiff Quantum Mechanics

So pictures refer to the development, continuously in time, of a physical system in Hilbert space and emission occurs discretely in time in real space. Does that sound better? I realize that this is not an important distinction to mathematicians and it does not change any calculations, but it is important to me. I need to keep a clear separation between what is real and what is not.
nnnm4
#11
Jul22-11, 10:47 AM
P: 113
Nice, I'm glad you found a good reference.

Anyway, the way I see it there is no distinction between whether H or S is more real. Neither picture, as you pointed out, makes any difference when one considers only observable quantities, i.e. the matrix elements of the operator under question. In the Schrodinger picture, we can get this abstract idea of the state of the system performing motion in the Hilbert space. That is the time evolution operator causes the states in a linear superposition of eigenstates to change their relative phase depending on the energies of those states. More energetic states change their phase more quickly. So when one calculates the matrix element of this particular state it is time dependent because you get "interference" from the changing relative phases.

Now in the H picture, we DEFINE the observable operator to be time-dependent. So the actual operator acting on states (which are no longer time dependent) changes from one moment to the next. The distinction, however, is purely interpretational, since in the S picture the matrix elements are something like

[<psi|exp(iHt)] * A * [exp(-iHt)|psi>] =defined= <psi(t)|A|psi(t)>

but in the H picture it's

<psi| [exp(iHt)*A*exp(-iHt)] |psi> =defined= <psi|A(t)|psi>.

For some problems it may easier to think of one picture or the other, but the distinction is purely formal and is just equivalent to the associative property of quantum mechanical operators and the inner product.

Also, the H picture let's us use the Heisenberg Equation of Motion

dA/dt = i[A, H] (i may have screwed up some signs here and h-bar=1)

This equation can formally allow us to connect quantum mechanics with the Hamilton equations of classical mechanics.
Dickfore
#12
Jul22-11, 10:52 AM
P: 3,014
Quote Quote by nortonian View Post
Am I right in thinking that the Heisenberg matrix interpretation describes emission, while the Schroedinger interpretation does not?
emission of what?
nortonian
#13
Jul24-11, 02:55 PM
P: 59
Quote Quote by nnnm4 View Post
Anyway, the way I see it there is no distinction between whether H or S is more real. Neither picture, as you pointed out, makes any difference when one considers only observable quantities, i.e. the matrix elements of the operator under question.
When we use classical terminology such as energy to describe quantum concepts it is easy to forget that we are describing reality as predicted by a mathematical model something that human languages weren't designed to do.


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