TD perturbation - state initially in continuous part

[Demystifier]

The problem for me is that the $\sqrt{\delta(E-E_0)}$ does not seem to be well defined.

See
https://www.physicsforums.com/threa...on-including-dirac-delta.873711/#post-5487662
for practical heuristic calculus with such objects and
https://www.physicsforums.com/threa...luding-dirac-delta.873711/page-2#post-5488516
for a hint how it can be made more rigorous.

 

[Nemanja989]

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 $a_E$, but my claim was about $|a_E|^2$. In the initial post, I said that it has to be $|a_E|^2=\delta(E-E')$ 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 $|a_E|^2=\delta(E-E')$, which was my original statement.

For me, the problem is the expression for $a_E$, as one would have to use the expression for it and not $|a_E|^2$ in the perturbative expansion.

 

[Nemanja989]

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 $\delta(E-E')$ 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 $E$ of the continuous spectrum after time $t$. Furthermore, this would mean that if one tries to calculate the integral $\int \Psi^{(1)*}_E \Psi^{(1)}_Edx$ of the continuous state after time $t$ , it would not be the delta function, but $\frac{2\pi}{\hbar}|F_{En}|^2t$.

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?

 

[Nemanja989]

See
https://www.physicsforums.com/threa...on-including-dirac-delta.873711/#post-5487662
for practical heuristic calculus with such objects and
https://www.physicsforums.com/threa...luding-dirac-delta.873711/page-2#post-5488516
for a hint how it can be made more rigorous.

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!

 

[Demystifier]

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!

 

[Demystifier]

You are the first one I hear saying anything bad of any LL book.

LL are books based more on physical intuition than on rigorous math. Not surprisingly, people who prefer mathematical rigor over physical intuition are not big fans of LL. But most physicists (including me) think that, in physics, physical intuition is more important than mathematical rigor, and such people usually don't have serious complaints on LL.

 

[Nemanja989]

LL are books based more on physical intuition than on rigorous math. Not surprisingly, people who prefer mathematical rigor over physical intuition are not big fans of LL. But most physicists (including me) think that, in physics, physical intuition is more important than mathematical rigor, and such people usually don't have serious complaints on LL.

Thanks! That sounds reasonable, although I still like rigor to a certain extent. There is a very good and serious video that at one point shows in a funny way a difference between a theoretical physicist and a mathematician - link. The part I am referring to begins at 28:04 and culminates at 31:51.

Further, what do you think of the way the normalization of the continuous spectrum of any operator was done in LL? You can see the link in one of my previous comments. This would be very nice for me to hear.

 

[Demystifier]

If physicists always waited to define math rigorously before attempting to make computations, then Dirac would not introduce his delta function because at that time there was no mathematical theory of distributions. Moreover, today particle physicists would still not use quantum field theory - the most successful theory in the history of science. So my advice is - first find a way how to use theory, and if it works, only then try make it rigorous.

 

[Demystifier]

Further, what do you think of the way the normalization of the continuous spectrum of any operator was done in LL? You can see the link in one of my previous comments. This would be very nice for me to hear.

I think it's OK in most cases of practical interest. Perhaps in some pathological cases it may not work, but such cases are rare.

Anyway, if you are interested in cases in which naive physical intuition does not work and rigorous math must be used, see https://arxiv.org/abs/quant-ph/9907069

 

[Nemanja989]

I think it's OK in most cases of practical interest. Perhaps in some pathological cases it may not work, but such cases are rare.

Anyway, if you are interested in cases in which naive physical intuition does not work and rigorous math must be used, see https://arxiv.org/abs/quant-ph/9907069

For me the problem is that PeterDonis and DrDru say that it cannot be done like in LL, I would be very happy if they could justify their statements. For me, I don't see any problem whatsoever. Maybe only as you say in some pathological cases, that are more relevant for a mathematician than for a physicist.

 

[PeterDonis]

I asked about $a_E$, but my claim was about $|a_E|^2$.

In your initial post you made claims about both, and the point you seemed to be struggling with was here:

My question is, if one can write any expression for the value of $a_{E_0}$ only? The problem for me is that the $\sqrt{\delta(E-E_0)}$ does not seem to be well defined.

You're right, it isn't, and the way to fix this "problem" is to not even try to take the square root to begin with. And the way to understand that is to stop thinking of $|a_E|^2$ as having any meaning in its own right, but only when it appears in an integral with $dE$.

After your last post, I have impression that you agree with $|a_E|^2=\delta(E-E')$, which was my original statement.

Formally, yes, you can write $|a_E|^2=\delta(E-E')$; but the issue is not writing down a formal equation but understanding what it does and does not tell you. This formal equation is not telling you that $|a_E|^2$ has any meaning by itself. It is telling you that it only has meaning when it appears in an integral with $dE$. (Just as any expression containing a delta function only has meaning in an appropriate integral.) That is what the people who are saying that $|a_E|^2$ is only a probability density, not a probability, are trying to tell you.

For me the problem is that PeterDonis and DrDru say that it cannot be done like in LL

As far as I can tell, LL do not assign any meaning to $|a_E|^2$ by itself, only when it appears in an integral with $dE$. Which is what I've been telling you as well.

LL is an advanced textbook, and as @A. Neumaier has pointed out, you have to know not to interpret what they say literally. They are not giving you a cookbook recipe for cranking out quantum field theories. They are trying to give you some clues which will help you to build your own understanding.

 

[PeterDonis]

For me, the problem is the expression for $a_E$, as one would have to use the expression for it and not $|a_E|^2$ in the perturbative expansion.

Why?

 

[Nemanja989]

Formally, yes, you can write $|a_E|^2=\delta(E-E')$; but the issue is not writing down a formal equation but understanding what it does and does not tell you. This formal equation is not telling you that $|a_E|^2$ has any meaning by itself. It is telling you that it only has meaning when it appears in an integral with $dE$. (Just as any expression containing a delta function only has meaning in an appropriate integral.) That is what the people who are saying that $|a_E|^2$ is only a probability density, not a probability, are trying to tell you.

Exactly, the integral has the meaning. But I have nowhere said that $|a_E|^2$ is a probability. Please check my first post! From it is clear that I regard $|a_E|^2$ as a probability density and $|a_E|^2 dE$ as probability for it to be in the energy interval $E+dE$ (I originally said only $E$, but @thephystudent had corrected me). This is one of the reasons why I was giving units for each of the terms I define.

I sincerely ask you to quote me where have I said that $|a_E|^2$ is a probability. This is something that causes a lot of unnecessary discussion in this topic.

As far as I can tell, LL do not assign any meaning to $|a_E|^2$ by itself, only when it appears in an integral with $dE$. Which is what I've been telling you as well.

They are not assigning any meaning to it, they are simply calculating it in a formal way. They only say that $a_{E}$ is the expansion coefficient, that is dependent on time. Nothing more, nothing less.

LL is an advanced textbook, and as @A. Neumaier has pointed out, you have to know not to interpret what they say literally. They are not giving you a cookbook recipe for cranking out quantum field theories. They are trying to give you some clues which will help you to build your own understanding.

This would be a huge underestimate of their books, which I am not doing and I ask you not to imply it. Every paragraph has a lot of useful information and honestly, I am amazed how they cover "every single corner" of the topic they are discussing. Also the simplicity of the arguments they are using to make the foundations on which the more complex theory is built is again, just amazing.

 

[Nemanja989]

Why?

Because I got curious of the problem they didn't mention in the book. Namely, if you have the electron starting from the continuous spectrum, but not from discrete, and then if I do perturbative expansion and try to solve the equations the only coefficient in the zeroth order that is non zero is $a^{(0)}_{E_0}$, which then appears in the integral of the following form:

$$\int a^{(0)}_E f(E)dE$$

Not as a modulus square, but like that. I was thinking of a possibility to multiply it somewhere along with the corresponding complex conjugate, but it doesn't seem to be an option. I still did't try it with @Demystifier's idea.

PS Do you think that it is physically valid to assume that the electron starts in continuous spectrum with a definite energy?

 

[Nemanja989]

@A. Neumaier

Have you seen from my first post that the wave function I started with is normalized to $1$ and not to delta function?

I want to know properties of the wave function $\Psi$, for which I know $i\hbar\frac{\partial\Psi}{\partial t}=\hat{H}\Psi$, where $\hat{H}=\hat{H_0}+\hat{H'}$ and I know that $\hat{H_0}\Psi_E(t)=i\hbar\frac{\partial\Psi_E(t)}{\partial t}$ and $\hat{H_0}\psi_E=E\psi_E$, where $\Psi_E(t)=\psi_Ee^{i\omega_{E}t}$. Again the usual normalization conditions holds $\int\Psi_{E'}^*\Psi_E dx=\delta(E-E')$ (this exact condition is given in LL, without a room for poetry). Just to be clear, I am not saying, nor did I say that in any of my previous posts, that the previous integral gives me the probability, it's not even in right units.

Now, I simply make the assumption that the $\hat{H'}$ is strong, but not enough to change the eigenvectors of $\hat{H_0}$, hence

$$\Psi=\int a_E \Psi_E dE$$

with $|a_E|^2=\delta(E-E')$. It is now easy to show that for the initial wave function $\Psi$ holds

$$\int |\Psi|^2dx=1$$

even though the electron was prepared in the specific state $\Psi_{E'}$ of the continuous spectrum. The new wave function, which is the eigenfunction of the $\hat{H}=\hat{H_0}+\hat{H'}$ has no normalization problems.

 

[Demystifier]

With proper distributions, such as $\delta(x)$, you cannot do all the operations that you can with functions. But some operations you can do, and the theory of distributions tells you which operations you can and which operations you can't do.

I expect that something similar can be said about objects such as $\sqrt{\delta(x)}$, namely that you can do some but not all operations with them that you can do with proper distributions. As far as I know, a general theory of such objects is not yet developed, but I believe it could be an interesting research program for pure mathematicians. The statement that "$\sqrt{\delta(x)}$ does not exist" really means that it does not exist as a proper distribution. But I have no doubts that it exists as some more general object, even if the theory of such objects is not yet developed.

Let me also remind skeptics that there were times when mathematicians thought that $\delta(x)$ does not exist, that $df(x)/dx$ does not exist, that $\sqrt{-1}$ does not exist, that $\sqrt{2}$ does not exist, that $-1$ does not exist, or even that $0$ does not exist. It turned out that all those objects exist by certain generalizations of the concepts that mathematicians of that time was used to.

 

[Nemanja989]

I expect that something similar can be said about objects such as $\sqrt{\delta(x)}$, namely that you can do some but not all operations with them that you can do with proper distributions.

Really the center! I like this very much. As you say, one should consider $\sqrt{\delta(x)}$ as something one layer more general than distributions and in order to make that generalization, we have to give up on something. As you say, not all the operations can be done as with distributions. Like with complex numbers, where with introduction of $\sqrt{-1}$, we had to give up on ordering them as it is the case for real numbers.

 

[Demystifier]

Like with complex numbers, where with introduction of $\sqrt{-1}$, we had to give up on ordering them as it is in case for real numbers.

Just a little nitpicking. While complex numbers cannot be ordered in the same way as real numbers can, the axiom of choice implies that complex numbers can be ordered somehow, even if nobody can give an explicit prescription how to do it.

 

[Nemanja989]

Just a little nitpicking. While complex numbers cannot be ordered in the same way as real numbers can, the axiom of choice implies that complex numbers can be ordered somehow, even if nobody can give an explicit prescription how to do it.

 

[Nemanja989]

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.

The following could come handy:

Let's consider the momentum operator $\hat{p}$, whose eigenfunctions are $\Psi_p$, which are further normalized as $\int\Psi^*_{p}\Psi_{p\ '} dx=\delta(p-p')$, where $p$ is the corresponding eigenvalue.

We know that:

$$\delta(f(x')-f(x))=\frac{1}{|\frac{df(x)}{dx}|}\delta(x'-x)$$

knowing further the relation between energy and momentum we obtain

$$\delta(p'-p)=v\delta(E'-E)$$

and one can define the relation between the momentum eigenfunction and the energy eigenfunction as $\Psi_E=\frac{1}{\sqrt{v}}\Psi_p$. This is the way to come to it if you know only the momentum representation. One did not have to go like this, in the special way, from the momentum representation. I can without a problem do it from the energy eigenfunctions and come to momentum or any other way!

Simply put, obtaining the eigenfunctions of $\hat{H}$, we can depending on application choose how to normalize them (e.g. for neutron scattering we would need the cross sections and we would choose with respect to what to normalize the wave functions so that the Fermi's golden rule gives us the cross sections).