TD perturbation - state initially in continuous part

  • #1
79
2

Main Question or Discussion Point

Hi everyone,

I am doing a time dependent perturbation theory, in a case when the electron is prepared in a state of the continuous part of the energy spectrum. Existence of the discrete part and the degeneracy of the continuous part is irrelevant at the moment and will not be considered.

First, the perturbed wave function is given as a superposition of the unperturbed wave functions:

[tex] \Psi(t) = \int a_E(t) \Psi_E dE [/tex]
where [itex] \Psi_E=\psi_E e^{-i\omega_Et}[/itex] and the integral is done over the whole spectrum of energies. The unperturbed wave functions are normalized to the Dirac delta with energy as an argument: [itex] \int \Psi_{E_i}^*\Psi_{E_j}dx=\delta(E_i-E_j)[/itex] and have units of [itex] \frac{1}{\sqrt{m⋅J}}[/itex], while the perturbated wave function has the units of [itex] \frac{1}{\sqrt{m}}[/itex]. The unit of [itex] a_E(t)[/itex] is [itex] \frac{1}{\sqrt{J}}[/itex], while the product [itex]|a_{E_0}(t)|^2dE[/itex] is the probability that the electron has the energy [itex] E_0[/itex] at the time [itex] t [/itex], which implies [itex] \int|a_{E_0}(t)|^2dE=1[/itex]. For more details one can look in the Landau Lifshitz course - link, introductory chapter about operators.

Now comes the issue:

If the electron is initially in the state with the energy [itex] E_0[/itex] , then it has to be that [itex]|a_{E_0}(t)|^2=\delta(E-E_0)[/itex], while it also holds that [itex] a_{E_0}=\int \Psi_{E_0}^*\Psi dx[/itex]. My question is, if one can write any expression for the value of [itex] a_{E_0}[/itex] only? The problem for me is that the [itex]\sqrt{\delta(E-E_0)}[/itex] does not seem to be well defined.

This would be very important as it would be used in the first order of the perturbation theory.

In other case, when the electron is initially in a discrete state, this problem is rather trivial and that's why I think I am missing something obvious here.

Thanks for your help!
 
Last edited:

Answers and Replies

  • #2
123
37
I think there are a few things off in your notation (either because of conceptual reasons or a typographic mistake). I will correct a few things to what I believe should be there to avoid confusion--correct me if you don't agree.


Hi everyone,
while the product [itex]|a_{E_0}(t)|^2dE[/itex] is the probability that the electron has the energy [itex] E_0[/itex] at the time [itex] t [/itex], which implies [itex] \int|a_{E_0}(t)|^2dE=1[/itex].
The product [itex]|a_{E_0}(t)|^2dE[/itex] is the probability that the electron has an energy in an infinitesimal shell around [itex] E_0[/itex] with size ##dE## at the time [itex] t [/itex], which implies [itex] \int|a_{E}(t)|^2dE=1[/itex].

If the electron is initially in the state with the energy [itex] E_0[/itex] , then it has to be that [itex]|a_{E_0}(t)|^2=\delta(E-E_0)[/itex],
I would rather say [itex]|a_{E}(0)|^2=\delta(E-E_0)[/itex]
----------------------------------------------------------------------------

I am not sure what you are really after. But if it is only the initial state you are interested in, global phase can be chosen arbitrarily.
 
  • #3
79
2
Thanks for the corrections! You are absolutely right about both of them :) I was not rigorous enough, although meant the same as you.

As you pointed, there is no problem with the phase. The problem that I have is that in the later calculations I will have to use [itex]a_E(0)[/itex] instead of [itex]|a_E(0)|^2[/itex], and [itex]a_E(0)=\sqrt{\delta(E-E_0)}[/itex] is not well defined.

E.g. How would one calculate the following:

[tex]\int a_E(0)\Psi_E dE=\int\sqrt{\delta(E-E_0)}\Psi_E dE \quad ?[/tex]

I have a feeling that having a square root should not have extremely big influence, but I do not have a knowledge of a mathematician of distributions, and therefore it is a puzzle for me..
 
  • #4
PeterDonis
Mentor
Insights Author
2019 Award
28,298
8,047
If the electron is initially in the state with the energy ##E_0## , then it has to be that ##|a_{E_0}(t)|^2 = \delta(E - E_0)##
This doesn't make sense. The squared modulus of the wave function is a number, not a delta function.

Also, an energy eigenstate is not described by a delta function like ##\delta(E - E_0)## with energy as an argument; that only works for a position eigenstate like ##\delta(x - x_0)## (in the position representation) or a momentum eigenstate like ##\delta(p - p_0)## (in the momentum representation). I suspect you are misunderstanding something in whatever source you are using; what source (textbook or paper) are you trying to learn time-dependent perturbation theory from?

while it also holds that ##a_{E_0}=\int \Psi_{E_0}^*\Psi dx##.
I don't know what you mean by this.
 
  • #5
PeterDonis
Mentor
Insights Author
2019 Award
28,298
8,047
I would rather say |aE(0)|2=δ(E−E0)
This doesn't make sense; the squared modulus is a number, not a delta function. See my response to the OP just now.
 
  • #6
A. Neumaier
Science Advisor
Insights Author
2019 Award
7,221
3,110
The electron cannot be initially in an unnormalizable state.
 
  • #7
123
37
This doesn't make sense; the squared modulus is a number, not a delta function. See my response to the OP just now.
Both on the left and the right hand side is a function of ##E##.

Also, ##\delta(E-E_0)## is sometimes used to define a microcanonical ensemble in quantum statistical mechanics. I think one usually doesn't take the continuum limit all the way though, but assumes a very dense discrete spectrum and thinks of the delta function as having a small width compared to all relevant energy scales, but still finite such that a sufficient amount of eigenstates is captured. If you explicitly won't allow for degeneracies, I don't see any problem with using the delta function for an individual energy eigenstate.
 
  • #8
79
2
For the simplicity and trying to avoid a chaos or confusion let's go one by one:

This doesn't make sense. The squared modulus of the wave function is a number, not a delta function.
Any physical quantity [itex]f[/itex], characterized with the operator [itex]\hat{f}[/itex] has eigenvalues [itex]f[/itex] and eigenfunction [itex]\Psi_f[/itex] . Let's assume it can take continuous values. Then we can represent any wave function [itex]\Psi[/itex] as [itex]\Psi=\int a_f \Psi_f df[/itex]. Now, [itex]\Psi_f[/itex] is normalized in such a way that the product [itex]|a_f|^2 df[/itex] is a probability for [itex]f[/itex] to have a value in between [itex]f[/itex] and [itex]f+df[/itex].

Here I tried to clarify the idea of what I have written beforehand. Can you now explain how can [itex]|a_f|^2[/itex] be a number, when it is not a dimensionless quantity?

Second, concerning the normalization of a wave function, it can be shown that, again for any quantity [itex]f[/itex] it holds:

[tex]a_f=\int a_{f_0}× \int \Psi_{f_0}*\Psi_f^*dx df_0[/tex]

This is true if and only if
[tex]\int \Psi_{f_0}*\Psi_f^* dx=\delta(f_0-f)[/tex]

Could you please explain me your point again? This works for any quantity, we can simply use the case [itex]\hat{f}=\hat{H}[/itex], and everything holds just for energy instead of a generic quantity [itex]f[/itex].
 
  • #9
79
2
The electron cannot be initially in an unnormalizable state.
Thank for the answer. But could you please explain this in greater detail? I could think of an electron starting in a discrete state and later being excited into one of the states of the continuous spectrum. The time after the excitation would be something I consider.
 
  • #10
PeterDonis
Mentor
Insights Author
2019 Award
28,298
8,047
Both on the left and the right hand side is a function of ##E##.
Meaning that ##a_E## is actually a function of ##E##?

Can you now explain how can ##|a_f|^2## be a number, when it is not a dimensionless quantity?
Lots of numbers in physics are not dimensionless. My weight is a number, but it's not a dimensionless quantity.
 
  • #11
79
2
This doesn't make sense. The squared modulus of the wave function is a number, not a delta function.

Also, an energy eigenstate is not described by a delta function like ##\delta(E - E_0)## with energy as an argument; that only works for a position eigenstate like ##\delta(x - x_0)## (in the position representation) or a momentum eigenstate like ##\delta(p - p_0)## (in the momentum representation). I suspect you are misunderstanding something in whatever source you are using; what source (textbook or paper) are you trying to learn time-dependent perturbation theory from?



I don't know what you mean by this.
PS are you aware of how the delta function is changed if its argument is a function of another quantity?
 
  • #12
123
37
Meaning that ##a_E## is actually a function of ##E##?
Yes. up to some trivial isomorphism at least if you insist to be a purist.
 
  • #13
PeterDonis
Mentor
Insights Author
2019 Award
28,298
8,047
are you aware of how the delta function is changed if its argument is a function of another quantity?
Yes, my issue is not with the use of delta functions.
 
  • #14
79
2
Yes, my issue is not with the use of delta functions.
Ok, may I ask you for your opinion on how would you describe [itex] a_E(0)[/itex], if you know that
[tex]\int |a_E(0)|^2dE=1[/tex]

and for all [itex]E[/itex] except [itex]E=E_0[/itex], [itex] a_E(0)=0[/itex]?

PS Can you answer the second question in my post #8?
 
  • #15
PeterDonis
Mentor
Insights Author
2019 Award
28,298
8,047
may I ask you for your opinion on how would you describe ##a_E(0)##,
I wouldn't. The only thing that I would say has a well-defined physical meaning is any integral in which ##|a_E|^2## and ##dE## appear in the integrand. The normalization integral you give is an example of such an integral (but of course not the only possible one). But the integral having a well-defined meaning does not require that ##a_E## itself have a well-defined meaning.
 
  • #16
79
2
I wouldn't.
But still, having on one side a delta function with properties that

[itex] \int\delta(x-x')dx=1[/itex] and [itex] \delta(x)=0[/itex] for [itex]x\neq x'[/itex]

and on the other

[itex] \int|a_E|^2dE=1[/itex] and [itex] |a_E|^2=0[/itex] for [itex]E\neq E'[/itex]

doesn't tell you that [itex]|a_E|^2=\delta(E-E')[/itex]?

And again, can you answer the second question in my post #8, concerning the normalization of a wave function in the continuous spectrum? What I am doing there is to take any operator [itex]\hat{f}[/itex], find its eigenfunctions [itex]\Psi_f[/itex] and normalize these eigenfunction as [itex]\int\Psi_{f'}^*\Psi_{f}dx=\delta(f-f')[/itex]. I think that this is perfectly valid and good for any operator and so as for [itex]\hat{H}[/itex].
 
  • #17
DrDu
Science Advisor
6,023
755
And again, can you answer the second question in my post #8, concerning the normalization of a wave function in the continuous spectrum? What I am doing there is to take any operator [itex]\hat{f}[/itex], find its eigenfunctions [itex]\Psi_f[/itex] and normalize these eigenfunction as [itex]\int\Psi_{f'}^*\Psi_{f}dx=\delta(f-f')[/itex]. I think that this is perfectly valid and good for any operator and so as for [itex]\hat{H}[/itex].
No, it is not. Operators are defined on the Hilbert space and if and operators do not have eigenstates corresponding to values in their continuous spectrum in this space. You can extend the analysis to rigged Hilbert spaces, where these eigenvectors can be defined, but still, this normalisation condition will not arise.
 
  • #18
79
2
No, it is not. Operators are defined on the Hilbert space and if and operators do not have eigenstates corresponding to values in their continuous spectrum in this space. You can extend the analysis to rigged Hilbert spaces, where these eigenvectors can be defined, but still, this normalisation condition will not arise.
Thanks. For my own clarifications may I ask you the following:

On the link, equation (5.4) states what I have written in my previous post. It is either that I misunderstood something there or that the statement in that link is wrong?

(PS I have no problem when pointed of being wrong, to me it is a way of progressing and understanding the subject better and better.)
 
  • #19
A. Neumaier
Science Advisor
Insights Author
2019 Award
7,221
3,110
Thank for the answer. But could you please explain this in greater detail? I could think of an electron starting in a discrete state and later being excited into one of the states of the continuous spectrum. The time after the excitation would be something I consider.
An electron starting in a discrete eigenstate would evolve into a weighted integral over continuum eigenstates. Then the problems with the delta function disappear.
 
  • #20
79
2
Thanks. For my own clarifications may I ask you the following:

On the link, equation (5.4) states what I have written in my previous post. It is either that I misunderstood something there or that the statement in that link is wrong?

(PS I have no problem when pointed of being wrong, to me it is a way of progressing and understanding the subject better and better.)
Also, on link Landau and Lifshitz discuss in detail on how to change the normalization condition from momentum to energy. I really think there is not much to be misunderstood there, but still it's either that or that the problem is in communication between us or in the end that the book is not right.
 
  • #21
79
2
An electron starting in a discrete eigenstate would evolve into a weighted integral over continuum eigenstates. Then the problems with the delta function disappear.
I can be very well wrong, but from a purely theoretical point of view, why cannot the transition happen to a single state in the continuous spectrum? In that case the weight within the integral is still a density of states in respect to energy, just given as a delta function. This is used again in the Landau Lifshitz book here, when going from eq. (42.5) to (42.6).
 
  • #22
A. Neumaier
Science Advisor
Insights Author
2019 Award
7,221
3,110
I can be very well wrong, but from a purely theoretical point of view, why cannot the transition happen to a single state in the continuous spectrum? In that case the weight within the integral is still a density of states in respect to energy, just given as a delta function. This is used again in the Landau Lifshitz book here, when going from eq. (42.5) to (42.6).
Because single continuum states are not normalizable. Ignoring this is the cause of all your problems
 
  • #23
79
2
Because single continuum states are not normalizable. Ignoring this is the cause of all your problems
Sorry for looking as if I am stubborn, but I simply want to clarify things for myself.

The fact that it is not possible to induce a transition to the single state of the continuous spectrum is in the direct contradiction with, I quote from the LL book:

"If the energy levels of the continuous spectrum are not degenerate, so that [itex] \nu [/itex] can be taken as the value of the energy alone, then the whole "interval" of states [itex] dE [/itex] reduces to a single state with energy [itex] E=E^{(0)}_N+\hbar\omega [/itex] and the probability of a transition to this state is ... "

To me, they clearly say that the transition from a discrete state to a single state of a continuous spectrum is very possible. Could you please explain me the cause of the contradiction? Thanks.
 
  • #24
A. Neumaier
Science Advisor
Insights Author
2019 Award
7,221
3,110
Sorry for looking as if I am stubborn, but I simply want to clarify things for myself.

The fact that it is not possible to induce a transition to the single state of the continuous spectrum is in the direct contradiction with, I quote from the LL book:

"If the energy levels of the continuous spectrum are not degenerate, so that [itex] \nu [/itex] can be taken as the value of the energy alone, then the whole "interval" of states [itex] dE [/itex] reduces to a single state with energy [itex] E=E^{(0)}_N+\hbar\omega [/itex] and the probability of a transition to this state is ... "

To me, they clearly say that the transition from a discrete state to a single state of a continuous spectrum is very possible. Could you please explain me the cause of the contradiction? Thanks.
They call the probability density probability,. But there is zeeo probability of making such a transition. The continuum behaves quite different from the discrete part. To get a positive probability one needs to integrate over a whole interval of states.
 
  • #25
79
2
They call the probability density probability,. But there is zeeo probability of making such a transition. The continuum behaves quite different from the discrete part. To get a positive probability one needs to integrate over a whole interval of states.
No, they do not call the probability density as probability. What they did change is calling the rate of probability as probability.

Here it goes:

[tex] |a_{fi}|^2=(2\pi/\hbar)|F_{fi}|^2\delta(E_f-E^{0}_i-\hbar\omega)t [/tex]

and then they say [itex] |a_{fi}|^2 d\nu_f[/itex] is the probability of a transition from the original state to one in the interval [itex] d\nu_f[/itex], where in case of no degeneracy [itex] d\nu_f=dE[/itex]. [itex] |a_{fi}|^2 d\nu_f[/itex] is in absolute consistency with the interpretation they gave in the earlier chapters, being the probability for something to happen, not probability density or anything similar. Now one can get the rate of the probability ("the probability of the transition per unit time") as following:

[tex] dw_{fi}=(2\pi/\hbar)|F_{fi}|^2\delta(E_f-E^{0}_i-\hbar\omega)dE[/tex]

and they integrate with respect to energy and obtain the rate of probability for a transition from a single discrete state to a single state of the continuous spectrum:

[tex] w_{Ei}=(2\pi/\hbar)|F_{Ei}|^2[/tex]

So I think they did not make a mistake in labeling it as a probability density.
 
Last edited:

Related Threads on TD perturbation - state initially in continuous part

  • Last Post
Replies
2
Views
529
Replies
1
Views
921
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
1
Views
596
  • Last Post
Replies
16
Views
3K
  • Last Post
Replies
1
Views
2K
Replies
4
Views
1K
Top