TD perturbation - state initially in continuous part

  • #1
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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!
 
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Answers and Replies

  • #2
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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
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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
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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.
 
  • #6
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The electron cannot be initially in an unnormalizable state.
 
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  • #7
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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.
 
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  • #20
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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
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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
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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
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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
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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
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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.
 
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  • #27
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Because of your statement earlier. You said something that I found contradicting some other fact. Further, you said that very thing will solve all of my problems.

I am not playing any game, but honestly trying to understand something. I get your last comment as a bit sarcastic, but I still want to ask you if you agree with my previous post or not, and why if possible?
 
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  • #28
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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.

And I would really like to hear a critical opinion of the others on the method that Landau and Lifshitz used for normalizing a continuous state, as this is something as far as I understood they were strongly opposing.

(In the end for me, this forum serves as a place for a constructive discussion, where if possible the problems get solved and confusion one may have goes away)
 
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  • #29
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doesn't tell you that ##|a_E|^2=\delta(E-E')##?

You didn't ask about##|a_E|^2##, you asked about ##a_E##. The whole issue you have been struggling with is that you don't know how to take the square root of a delta function. The correct answer to that problem is "don't do that". You don't need to.
 
  • #30
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if you agree with my previous post or not
Not at all. You simply dismiss my explanations, making me feel having wasted my time.

Any S-matrix is unitary hence preserves norms. Thus it cannot turn a discrete eigenstate (of norm 1) into a continuous one (of norm infinity). Thus there are no such transitions. The language in L/L is to be understood cum grano salis.

Nemanja989 said:
they did not make a mistake in labeling it as a probability density.
This is not a mistake but a widely used convention. Probabilities cannot have delta-functions as values, hence whenever you see such a ''probability'' you should conclude that a probability density is meant.

Take your time to see how both what L/L write and what I wrote makes sense.
 
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  • #32
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You didn't ask about##|a_E|^2##, you asked about ##a_E##. The whole issue you have been struggling with is that you don't know how to take the square root of a delta function. The correct answer to that problem is "don't do that". You don't need to.

Yes, I asked about [itex] a_E[/itex], but my claim was about [itex] |a_E|^2[/itex]. In the initial post, I said that it has to be [itex] |a_E|^2=\delta(E-E')[/itex] and to this you said in the comment #4 that it makes no sense and that I am misunderstanding the source I am using. After your last post, I have impression that you agree with [itex] |a_E|^2=\delta(E-E')[/itex], which was my original statement.

For me, the problem is the expression for [itex] a_E[/itex], as one would have to use the expression for it and not [itex] |a_E|^2[/itex] in the perturbative expansion.
 
  • #33
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Not at all. You simply dismiss my explanations, making me feel having wasted my time.

Any S-matrix is unitary hence preserves norms. Thus it cannot turn a discrete eigenstate (of norm 1) into a continuous one (of norm infinity). Thus there are no such transitions. The language in L/L is to be understood cum grano salis.


This is not a mistake but a widely used convention. Probabilities cannot have delta-functions as values, hence whenever you see such a ''probability'' you should conclude that a probability density is meant.

Take your time to see how both what L/L write and what I wrote makes sense.

I am sorry if you felt like wasting your time, however I highly appreciate yours and anyone else's time spent on the topic. Also, it's hard to communicate through messages like here and one unfortunately may get a different impression of what was meant (there is no intonation or gesticulation as when communicating in person).

Now, to come back to the topic.

You are the first one I hear saying anything bad of any LL book. That is quite a statement and I am happy to hear an opinion like that, do you have any other experience like that in the other LL books? My impression is that classical stuff and the application of QM was done well. What is your opinion on the way the normalization of the continuous spectrum of a operator was done in the LL? See the comment #28.

About the probabilities in my comment #25, it is the probability density which is proportional to [itex]\delta(E-E')[/itex] and once we take the energy integral and obtain the probability, it has no delta function in it and I would interpret that number as a chance of the electron being in that state with energy [itex]E[/itex] of the continuous spectrum after time [itex]t[/itex]. Furthermore, this would mean that if one tries to calculate the integral [itex] \int \Psi^{(1)*}_E \Psi^{(1)}_Edx [/itex] of the continuous state after time [itex]t[/itex] , it would not be the delta function, but [itex]\frac{2\pi}{\hbar}|F_{En}|^2t[/itex].

May I ask you for a bit of help, if the final conclusion is not correct, it has to be that a mistake has already been made somewhere beforehand. Having the impression that you have read LL, could say where the problem is emerging actually?
 
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  • #34
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Thank you very much for your comment. I read your post and like the idea of what you have done and how you defined it. However, as I do not have a very good knowledge of distributions, I am not fully confident of it, specially because later in the same topic member samalkhaiat said that it was mathematically proven that it cannot be done. Could you please give me your opinion on that? Thanks!
 
  • #35
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later in the same topic member samalkhaiat said that it was mathematically proven that it cannot be done. Could you please give me your opinion on that?
Well, samalkhaiat did not say what exactly goes wrong, except in a special case which is of no relevance here. You can always work with finite but small ##\epsilon## as in my second link. Then at the end of calculation I would expect that you can put ##\epsilon\rightarrow 0##, even if sometimes you cannot do that at all intermediate steps. The problem is that the meaning of the expression "the end of calculation" depends on the exact problem one has in hand and I don't know how to define the general prescription. Here I use physical intuition rather than rigorous math. So I would suggest you to work with finite but small ##\epsilon## and see what happens if at the end of calculation you put ##\epsilon\rightarrow 0##. Try it!
 

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