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Luboš Motl
Luboš Motl
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everything you write is perfectly OK, except for the statement that "it doesn't seem plausible". Yes, it is plausible and yes, the exponential is equal to one. That's the whole reason why the contribution for $G=k'-k+q$ to the interaction amplitude is nonzero. All the terms with different values of $G$ lead to oscillating functions in the complex plane that sum up to zero.

Yes, the integral of 1 over the whole space is infinite. And yes, you also know the way to make it finite: restrict the interval to the solid itself. It is not surprising - and it is physically correct - that the integral is infinite for an infinite solid. It's because the interaction Hamiltonian between the two plane wave states is proportional to the volume of the solid as well - the greater volume, the greater chance that the interaction will occur somewhere.

If you were calculating the things for a finite volume of the solid - e.g. a box - you would get manifestly finite numbers everywhere. However, some formulae would be unnecessary awkward and they would depend on the size and shape of the solid. That's why adult physicists learned from Paul Dirac how to use the distributions such as $\delta(G-G_0)$. For your problem, $G_0=k'-k+q$ is treated as a constant. The defining property of the $\delta$-function is that $$\int_{-\infty}^\infty dG\,f(G) \delta(G-G_0) = f(G_0)$$ for any function $f(G)$. So the integral of $f(G)$ over $G$, weighted by the delta-function, only picks the value at $G=G_0$. That's because $\delta(G-G_0)$ vanishes for all values for which $G\neq G_0$. But for $G=G_0$, it is infinite and so large that the integral of $\delta(G)$ over $G$ is equal to one, and if you insert $f(G)$, it simply picks $f(G_0)$.

The object $\delta(G-G_0)$ is not a function in the usual sense but it is extremely helpful and consistent in dealing with the integrals that appear in the Fourier transformations. The delta-function may be approximated by a function that is equal to $1/\epsilon$ for the argument being between $-\epsilon/2$ and $\epsilon/2$, and otherwise is equal to zero, in the limit where $\epsilon$ goes to zero. The delta-function can also be written as the Fourier transform of the function $1$ divided by $2\pi$: $$\delta(G-G_0) = \frac{1}{2\pi}\int_{-\infty}^\infty dR\,\exp[i(G-G_0)R] $$ You may always imagine that the delta-function exercises are translated to a calculation at a finite volume of the space or solid; then the momenta such as $G$ become discrete - number of zeros of a standing wave, for example, times $1/R_{solid}$ - and the integral over $G$ is replaced by a summation. The function $\delta(G-G_0)$ is then replaced by a simple Kronecker delta symbol $\delta_{G,G_0}$ which is only nonzero - equal to one - if $G=G_0$. But you will pay the price that most formulae will contain powers of the volume and other things, and you need to remember the spacing of the momenta etc.

All these extra things in the formulae will depend on the size and shape of the solid - or space(time) - that you picked. But at the end, you know very well that there exists a sensible physics in the infinite-volume limit that should be independent of the volume (because it was sent to infinity). For you, to learn how to deal with the Dirac distributions is important to do all these calculations effectively because the volume $V$ and all the problematic features about the spacing of the momentum evaporate from the formulae, and the fact that the formulae hold for any large piece of the solid (or space) becomes self-evident.

Best wishes Lubos

everything you write is perfectly OK, except for the statement that "it doesn't seem plausible". Yes, it is plausible and yes, the exponential is equal to one. That's the whole reason why the contribution for $G=k'-k+q$ to the interaction amplitude is nonzero. All the terms with different values of $G$ lead to oscillating functions in the complex plane that sum up to zero.

Yes, the integral of 1 over the whole space is infinite. And yes, you also know the way to make it finite: restrict the interval to the solid itself. It is not surprising - and it is physically correct - that the integral is infinite for an infinite solid. It's because the interaction Hamiltonian between the two plane wave states is proportional to the volume of the solid as well - the greater volume, the greater chance that the interaction will occur somewhere.

If you were calculating the things for a finite volume of the solid - e.g. a box - you would get manifestly finite numbers everywhere. However, some formulae would be unnecessary awkward and they would depend on the size and shape of the solid. That's why adult physicists learned from Paul Dirac how to use the distributions such as $\delta(G-G_0)$. For your problem, $G_0=k'-k+q$ is treated as a constant. The defining property of the $\delta$-function is that $$\int_{-\infty}^\infty dG\,f(G) \delta(G-G_0) = f(G_0)$$ for any function $f(G)$. So the integral of $f(G)$ over $G$, weighted by the delta-function, only picks the value at $G=G_0$. That's because $\delta(G-G_0)$ vanishes for all values for which $G\neq G_0$. But for $G=G_0$, it is infinite and so large that the integral of $\delta(G)$ over $G$ is equal to one, and if you insert $f(G)$, it simply picks $f(G_0)$.

The object $\delta(G-G_0)$ is not a function in the usual sense but it is extremely helpful and consistent in dealing with the integrals that appear in the Fourier transformations. The delta-function may be approximated by a function that is equal to $1/\epsilon$ for the argument being between $-\epsilon/2$ and $\epsilon/2$, and otherwise is equal to zero, in the limit where $\epsilon$ goes to zero. The delta-function can also be written as $$\delta(G-G_0) = \frac{1}{2\pi}\int_{-\infty}^\infty dR\,\exp[i(G-G_0)R] $$ You may always imagine that the delta-function exercises are translated to a calculation at a finite volume of the space or solid; then the momenta such as $G$ become discrete - number of zeros of a standing wave, for example, times $1/R_{solid}$ - and the integral over $G$ is replaced by a summation. The function $\delta(G-G_0)$ is then replaced by a simple Kronecker delta symbol $\delta_{G,G_0}$ which is only nonzero - equal to one - if $G=G_0$. But you will pay the price that most formulae will contain powers of the volume and other things, and you need to remember the spacing of the momenta etc.

All these extra things in the formulae will depend on the size and shape of the solid - or space(time) - that you picked. But at the end, you know very well that there exists a sensible physics in the infinite-volume limit that should be independent of the volume (because it was sent to infinity). For you, to learn how to deal with the Dirac distributions is important to do all these calculations effectively because the volume $V$ and all the problematic features about the spacing of the momentum evaporate from the formulae, and the fact that the formulae hold for any large piece of the solid (or space) becomes self-evident.

Best wishes Lubos

everything you write is perfectly OK, except for the statement that "it doesn't seem plausible". Yes, it is plausible and yes, the exponential is equal to one. That's the whole reason why the contribution for $G=k'-k+q$ to the interaction amplitude is nonzero. All the terms with different values of $G$ lead to oscillating functions in the complex plane that sum up to zero.

Yes, the integral of 1 over the whole space is infinite. And yes, you also know the way to make it finite: restrict the interval to the solid itself. It is not surprising - and it is physically correct - that the integral is infinite for an infinite solid. It's because the interaction Hamiltonian between the two plane wave states is proportional to the volume of the solid as well - the greater volume, the greater chance that the interaction will occur somewhere.

If you were calculating the things for a finite volume of the solid - e.g. a box - you would get manifestly finite numbers everywhere. However, some formulae would be unnecessary awkward and they would depend on the size and shape of the solid. That's why adult physicists learned from Paul Dirac how to use the distributions such as $\delta(G-G_0)$. For your problem, $G_0=k'-k+q$ is treated as a constant. The defining property of the $\delta$-function is that $$\int_{-\infty}^\infty dG\,f(G) \delta(G-G_0) = f(G_0)$$ for any function $f(G)$. So the integral of $f(G)$ over $G$, weighted by the delta-function, only picks the value at $G=G_0$. That's because $\delta(G-G_0)$ vanishes for all values for which $G\neq G_0$. But for $G=G_0$, it is infinite and so large that the integral of $\delta(G)$ over $G$ is equal to one, and if you insert $f(G)$, it simply picks $f(G_0)$.

The object $\delta(G-G_0)$ is not a function in the usual sense but it is extremely helpful and consistent in dealing with the integrals that appear in the Fourier transformations. The delta-function may be approximated by a function that is equal to $1/\epsilon$ for the argument being between $-\epsilon/2$ and $\epsilon/2$, and otherwise is equal to zero, in the limit where $\epsilon$ goes to zero. The delta-function can also be written as the Fourier transform of the function $1$ divided by $2\pi$: $$\delta(G-G_0) = \frac{1}{2\pi}\int_{-\infty}^\infty dR\,\exp[i(G-G_0)R] $$ You may always imagine that the delta-function exercises are translated to a calculation at a finite volume of the space or solid; then the momenta such as $G$ become discrete - number of zeros of a standing wave, for example, times $1/R_{solid}$ - and the integral over $G$ is replaced by a summation. The function $\delta(G-G_0)$ is then replaced by a simple Kronecker delta symbol $\delta_{G,G_0}$ which is only nonzero - equal to one - if $G=G_0$. But you will pay the price that most formulae will contain powers of the volume and other things, and you need to remember the spacing of the momenta etc.

All these extra things in the formulae will depend on the size and shape of the solid - or space(time) - that you picked. But at the end, you know very well that there exists a sensible physics in the infinite-volume limit that should be independent of the volume (because it was sent to infinity). For you, to learn how to deal with the Dirac distributions is important to do all these calculations effectively because the volume $V$ and all the problematic features about the spacing of the momentum evaporate from the formulae, and the fact that the formulae hold for any large piece of the solid (or space) becomes self-evident.

Best wishes Lubos

Source Link
Luboš Motl
Luboš Motl
  • 181.3k
  • 15
  • 413
  • 643

everything you write is perfectly OK, except for the statement that "it doesn't seem plausible". Yes, it is plausible and yes, the exponential is equal to one. That's the whole reason why the contribution for $G=k'-k+q$ to the interaction amplitude is nonzero. All the terms with different values of $G$ lead to oscillating functions in the complex plane that sum up to zero.

Yes, the integral of 1 over the whole space is infinite. And yes, you also know the way to make it finite: restrict the interval to the solid itself. It is not surprising - and it is physically correct - that the integral is infinite for an infinite solid. It's because the interaction Hamiltonian between the two plane wave states is proportional to the volume of the solid as well - the greater volume, the greater chance that the interaction will occur somewhere.

If you were calculating the things for a finite volume of the solid - e.g. a box - you would get manifestly finite numbers everywhere. However, some formulae would be unnecessary awkward and they would depend on the size and shape of the solid. That's why adult physicists learned from Paul Dirac how to use the distributions such as $\delta(G-G_0)$. For your problem, $G_0=k'-k+q$ is treated as a constant. The defining property of the $\delta$-function is that $$\int_{-\infty}^\infty dG\,f(G) \delta(G-G_0) = f(G_0)$$ for any function $f(G)$. So the integral of $f(G)$ over $G$, weighted by the delta-function, only picks the value at $G=G_0$. That's because $\delta(G-G_0)$ vanishes for all values for which $G\neq G_0$. But for $G=G_0$, it is infinite and so large that the integral of $\delta(G)$ over $G$ is equal to one, and if you insert $f(G)$, it simply picks $f(G_0)$.

The object $\delta(G-G_0)$ is not a function in the usual sense but it is extremely helpful and consistent in dealing with the integrals that appear in the Fourier transformations. The delta-function may be approximated by a function that is equal to $1/\epsilon$ for the argument being between $-\epsilon/2$ and $\epsilon/2$, and otherwise is equal to zero, in the limit where $\epsilon$ goes to zero. The delta-function can also be written as $$\delta(G-G_0) = \frac{1}{2\pi}\int_{-\infty}^\infty dR\,\exp[i(G-G_0)R] $$ You may always imagine that the delta-function exercises are translated to a calculation at a finite volume of the space or solid; then the momenta such as $G$ become discrete - number of zeros of a standing wave, for example, times $1/R_{solid}$ - and the integral over $G$ is replaced by a summation. The function $\delta(G-G_0)$ is then replaced by a simple Kronecker delta symbol $\delta_{G,G_0}$ which is only nonzero - equal to one - if $G=G_0$. But you will pay the price that most formulae will contain powers of the volume and other things, and you need to remember the spacing of the momenta etc.

All these extra things in the formulae will depend on the size and shape of the solid - or space(time) - that you picked. But at the end, you know very well that there exists a sensible physics in the infinite-volume limit that should be independent of the volume (because it was sent to infinity). For you, to learn how to deal with the Dirac distributions is important to do all these calculations effectively because the volume $V$ and all the problematic features about the spacing of the momentum evaporate from the formulae, and the fact that the formulae hold for any large piece of the solid (or space) becomes self-evident.

Best wishes Lubos