1D wavepacket scattering simulation, momentum distribution formula

[LucKy]

TL;DR Summary

Question regarding unusual additional term added to the gaussian momentum distribution formula (used in computer simulation of 1D wavepacket scattering).

Hello everybody at the forum

I'm from Ukraine, I have Chemistry degree, and last year I began to self studying Quantum Mechanics.

I'm reading this article:
R. Garcia, A. Zozulya, and J. Stickney, “MATLAB codes for teaching quantum physics: Part 1,” [Online]. Available: http://arxiv.org/abs/0704.1622.

And I have a question about gaussian momentum distribution formula introduced in Program 6, section VII. WAVEPACKETS AND STEP POTENTIALS of the atticle.

The formula for gaussian momentum distribution (used in next steps to generate initial wave packet by Fourier Transform) in the above article is
$\phi(k) = e^{-a(k-k_0)^ 2}e^{-i6k^2}$

Please, can anybody expalain why the term $e^{-i6k^2}$ has been included into the wavepacket's gaussian momentum distribution frormula?

I have checked Townsend and Zettili books, but wavepacket's momentum distribution fromulas introduced in these books has no such term included.

When I modify the programm by deleting this term or by setting coefficient value less than 6, a second "mirror" wavepacket appears on a plot (and it moves in towards initial wavepacket).

I have two assumptions on this matter:
1) Adding this term is just a trick to run this particular computer simulation well, and there is no general need to include such term if I plan to use different algorithm or different calculation method for scattering simulation;
2) It is some type of "phase factor" for gaussian momentum distribution, and it was ommited in Townsend and Zettili books, but it is introduced in more advanced books. In this case, please give me advise where i can read about it;

 

[vanhees71]

This is just a factor corresponding to a time evolution at some fixed positive time. Note that for a free particle in the momentum representation you simply have
$$\psi(t,p)=\exp[-\mathrm{i} \hat{H} (t-t_0)/\hbar)] \psi_0(p)=\exp[-\mathrm{i} p^2 (t-t_0)/(2m \hbar)] \psi_0(p),$$
because
$$\hat{H}=\frac{1}{2m} \hat{p^2}.$$

 

[LucKy]

This is just a factor corresponding to a time evolution at some fixed positive time. Note that for a free particle in the momentum representation you simply have
$$\psi(t,p)=\exp[-\mathrm{i} \hat{H} (t-t_0)/\hbar)] \psi_0(p)=\exp[-\mathrm{i} p^2 (t-t_0)/(2m \hbar)] \psi_0(p),$$
because
$$\hat{H}=\frac{1}{2m} \hat{p^2}.$$

Thank you, got it! understand now!

 

[Davide]

This is just a factor corresponding to a time evolution at some fixed positive time. Note that for a free particle in the momentum representation you simply have
$$\psi(t,p)=\exp[-\mathrm{i} \hat{H} (t-t_0)/\hbar)] \psi_0(p)=\exp[-\mathrm{i} p^2 (t-t_0)/(2m \hbar)] \psi_0(p),$$
because
$$\hat{H}=\frac{1}{2m} \hat{p^2}.$$

I would like to detail:

a wavepacket (a superposition of plane waves each of them with a different frequency) is a solution (one of many) of the Schrodinger equation, that it's a wave equation. Solutions of a wave equation must have this form: $f(x \pm vt)$. And that's true for this case:
$$
\psi(t,p) = exp[-i p^{2}(t-t_{0})]\cdot \psi_{0} (p)
$$
Where $\psi_{0} (p)$ is the gaussian function, that is, the form of the wavepacket. Its Fourier transform has the form $f(x -vt)$. However $\psi_{0} (p)$ will never have this form since is not function of the time.

Wonderful notes about this topic: https://scholar.harvard.edu/files/schwartz/files/lecture11-wavepackets.pdf

Also in the Griffiths -> Section 2.4.- The free particle.

 

[vanhees71]

Well, for the exact solution of the time-dependent Schrödinger equation you don't have the simple solution $f(x-vt)$ as for the 1D wave equation, because you have dispersion, i.e., $\omega$ is not proportional to $k$ but to $k^2$. This implies that the wave packet doesn't keep its original shape. In the case of a Gaussian wave packet of course it stays a Gaussian wave packet, but the width changes (that's why it's called "dispersion".

 

[Davide]

Well, for the exact solution of the time-dependent Schrödinger equation you don't have the simple solution f(x−vt) as for the 1D wave equation, because you have dispersion, i.e., ω is not proportional to k but to k2. This implies that the wave packet doesn't keep its original shape. In the case of a Gaussian wave packet of course it stays a Gaussian wave packet, but the width changes (that's why it's called "dispersion".

That's true, you have dispersive solutions such as $f(x-v(k) \cdot t)$ but it is still a wave (solution of a wave equation), isn't it?

 

[vanhees71]

$$\newcommand{\ii}{\mathrm{i}}$$
The exact solution reads
$$
\psi_t(t,x_1) = \frac{N'}{2 \pi} \sqrt{\frac{4 \pi m \alpha}{m^2+2
\ii \hbar \alpha t}} \exp \left [-\frac{m^2 \alpha}{m^2+4\hbar^2
\alpha^2 t^2} \left (x_1-\frac{\hbar k_{01} t}{m} \right)^2 \right]
\exp[\ii \Phi(t,x_1)]$$
with
$$
\Phi(t,x_1) = \frac{m \alpha}{m^2 + 4 \hbar^2 \alpha^2 t^2} \left (2
\hbar \alpha t x_1^2 + \frac{m x_1 k_{01}}{\alpha} - \frac{\hbar
k_{01}^2 t}{2 \alpha} \right), \quad N'=\left (\frac{2 \pi}{\alpha} \right)^{1/4}.$$
The initial condition is an "minimal-uncertainty" Gaussian with $\Delta x \Delta p=\hbar/2$. Of course, $\Delta p=\hbar \sqrt{\alpha}=\text{const}$ and $\Delta x$ increases with time:
$$\Delta x^2=\frac{m^2 + 4 \alpha^2 \hbar^2 t^2}{4 \alpha m^2}.$$

 

[Davide]

$$\newcommand{\ii}{\mathrm{i}}$$
The exact solution reads
$$
\psi_t(t,x_1) = \frac{N'}{2 \pi} \sqrt{\frac{4 \pi m \alpha}{m^2+2
\ii \hbar \alpha t}} \exp \left [-\frac{m^2 \alpha}{m^2+4\hbar^2
\alpha^2 t^2} \left (x_1-\frac{\hbar k_{01} t}{m} \right)^2 \right]
\exp[\ii \Phi(t,x_1)]$$
with
$$
\Phi(t,x_1) = \frac{m \alpha}{m^2 + 4 \hbar^2 \alpha^2 t^2} \left (2
\hbar \alpha t x_1^2 + \frac{m x_1 k_{01}}{\alpha} - \frac{\hbar
k_{01}^2 t}{2 \alpha} \right), \quad N'=\left (\frac{2 \pi}{\alpha} \right)^{1/4}.$$
The initial condition is an "minimal-uncertainty" Gaussian with $\Delta x \Delta p=\hbar/2$. Of course, $\Delta p=\hbar \sqrt{\alpha}=\text{const}$ and $\Delta x$ increases with time:
$$\Delta x^2=\frac{m^2 + 4 \alpha^2 \hbar^2 t^2}{4 \alpha m^2}.$$

I got lost.

Time-Dependent Schrödinger Equation (TDSE) is a wave equation, its solutions have the form of a wave (or a wavepacket, that it's a superposition of waves).

In the case of $V = 0$, we have analytical solutions: plane waves (or, again, a superposition of plane waves).

And my clarification was about the main point: where does this therm $e^{-i 6 k^{2}}$ comes from?

The path to follow is:

1.- Solutions of the TDSE in $V=0$ are plane waves (or superposition of plane waves... as you prefer)

2.- Plane waves have this form:

$$ \psi(t, x)=A e^{i(k \cdot x-\omega t)} = \psi(x-vt)$$

3.- By superposition of plane waves, you can build the wavepacket or the wave with the form that you want. For example, a Gaussian:

$$ \psi(x,t) = exp[i\cdot((p_{0}x-E(P_{0}))] \cdot F(x,t) $$

Where $F(x,t)$ is the Gaussian function. As you can see, the Gaussian is just the shape of the wavepacket. $F(x,t)$ is not a solution of the TDSE.

4.- The Gaussian function is amazing because his Fourier transform is a Gaussian too. So, the complex exponential in the function $\psi(p,t)$ in the $p$ space has to come from another term. This term is the "plane wave term", which its the first term in the $\psi(x,t)$ formula.

In conclusion: the $e^{-i 6 k^{2}}$ it's due to wavepackets solutions of the TDSE (with $V= 0$) are a superposition of plane waves, that is to say: complex exponentials.

 

[hutchphd]

Why is it written as that particular function? What if it were $e^{i45k^2}$ Would the result be different? Why $k^2$ in particular?

 

[vanhees71]

What I've posted above is the exact solution of non-relativistic time-dependent Schrödinger equation for a free particle moving in one dimension. It is not a wave equation in the proper sense, because that reads
$$\frac{1}{c^2} \partial_t^2 \phi(t,x)-\partial_z^2 \phi(t,x),$$
and this has the general solution
$$\phi(t,x)=u(c t-x)+v(ct+x).$$
The Schrödinger equation is of different form (1st order in the time derivative and 2nd order in the spatial derivative). That's why its solutions do not keep their shape, i.e., there is dispersion already in the free-field case.

 

[hutchphd]

(My question was intended for the OP.) I'm not sure what the OP is trying to say but I don't think the relation of the dispersion relation to the fixed time offset was part of his explanation That was his original query in part.

 

[Davide]

What I've posted above is the exact solution of non-relativistic time-dependent Schrödinger equation for a free particle moving in one dimension. It is not a wave equation in the proper sense, because that reads
$$\frac{1}{c^2} \partial_t^2 \phi(t,x)-\partial_z^2 \phi(t,x),$$
and this has the general solution
$$\phi(t,x)=u(c t-x)+v(ct+x).$$
The Schrödinger equation is of different form (1st order in the time derivative and 2nd order in the spatial derivative). That's why its solutions do not keep their shape, i.e., there is dispersion already in the free-field case.

$\bullet$ About terminology. Yes, there is a problem with the terminology:

One thing is THE wave equation, as you posted:

$$\frac{\partial^{2} u}{\partial t^{2}}=c^{2} \frac{\partial^{2} u}{\partial x^{2}}$$

And another thing is A wave equation, that is, an equation (PDE) whose solutions are waves: $f(x-vt)$. TDSE is A wave equation.

$\bullet$ About the solutions of the TDSE. I will repeat what I said before again:

1.- Solutions of the TDSE in V=0 are plane waves (or superposition of plane waves... as you prefer)

2.- Plane waves have this form:

ψ(t,x)=Aei(k⋅x−ωt)=ψ(x−vt)

You can check that plane waves are the solutions of the TDSE for a free particle in any quantum physics introductory book. For example:

In the S. Gasiorowicz - Quantum Physics you can check all this information in sections: 2-2.- Plane waves and wave packets and 2-4.- The Schödinger equation.

 

[hutchphd]

You can check that plane waves are the solutions of the TDSE for a free particle in any quantum physics introductory book. For example:

`I commend your choice of Gasiorowicz. I once taught a pretty good Quantum course from it.
But your statement

2.- Plane waves have this form:

ψ(t,x)=Aei(k⋅x−ωt)=ψ(x−vt)

is not correct. As pointed out by @vanhees71 $Ae^{ikx-\omega t}$ is a solution to the free space S. Eqn. but that does not imply an arbitrary form $\Psi (x-vt)$ will be a solution. This is true only if $\omega /k$ is a constant (independent of k).
For free space Schrodinger solution one must have $$\omega =\hbar k^2/2m$$ and so the unfettered Fourier synthesis is not allowed.

 

[Davide]

`I commend your choice of Gasiorowicz. I once taught a pretty good Quantum course from it.
But your statement

2.- Plane waves have this form:

ψ(t,x)=Aei(k⋅x−ωt)=ψ(x−vt)

is not correct. As pointed out by @vanhees71 $Ae^{ikx-\omega t}$ is a solution to the free space S. Eqn. but that does not imply an arbitrary form $\Psi (x-vt)$ will be a solution. This is true only if $\omega /k$ is a constant (independent of k).
For free space Schrodinger solution one must have $$\omega =\hbar k^2/2m$$ and so the unfettered Fourier synthesis is not allowed.

According to Gasiorowicz 2-1 equation:

$$ \psi_{k}(x,t) = Ae^{i(kx-wt)}+Be^{-1(kx-wt)} $$

$\psi_{k}(x,t)$ it is a plane wave. So, why is my statement not correct?About the waveform of the solutions: I'm saying that knowing that the initial condition of the paper is a solution of the TDSE for $t=0$, $e^{-i 6 k^{2}}$ comes from the plane-wave factor (equation 2-7 and the lines below of the Gasiorowicz) and without that factor there is no Gaussian wavepacket but only a Gaussian function that is not a solution of the TDSE.

 

[hutchphd]

An arbitrary function of (x-vt) is NOT a solution to TDSE.
The factor $e^{-ik^2 6 }$ provides an offset in time (as does any factor of the form $e^{iw_kt_0}$) in the Fourier Transform. This works only because $k^2 \sim \omega _k$. The function is a solution without the factor but may not match the t=0 boundary condition.
Sorry you must have a newer edition of Gasiorowicz so I cannot follow explicit references. Please work through @vanhees71 solution: it is, as usual, quite complete

 

[vanhees71]

$\bullet$ About terminology. Yes, there is a problem with the terminology:

One thing is THE wave equation, as you posted:

$$\frac{\partial^{2} u}{\partial t^{2}}=c^{2} \frac{\partial^{2} u}{\partial x^{2}}$$

And another thing is A wave equation, that is, an equation (PDE) whose solutions are waves: $f(x-vt)$. TDSE is A wave equation.

$\bullet$ About the solutions of the TDSE. I will repeat what I said before again:
You can check that plane waves are the solutions of the TDSE for a free particle in any quantum physics introductory book. For example:

In the S. Gasiorowicz - Quantum Physics you can check all this information in sections: 2-2.- Plane waves and wave packets and 2-4.- The Schödinger equation.

Of course plane waves are solutions of the Schrödinger equation, but not all solutions of the Schrödinger equation are plane waves as the example with the Gaussian wave packet shows: it's clearly NOT of the form $f(ct-x)$.

The reason is that the plane-wave solutions are the energy eigenstates. For a free particle these are the momentum eigenstates,
$$u_p(t,x)=\frac{1}{\sqrt{2 \pi \hbar}} \exp(-\mathrm{i} \omega_{p} t+\mathrm{i} k x)$$
with
$$k=p/\hbar, \quad \omega = \hbar^2 k^2/(2m).$$
This quadratic (i.e., non-linear) dependence between $\omega$ and $k$ is the reason why, in contradistinction to solutions of the 1D wave equation, the general solutions of the 1D Schrödinger equation are not (superpositions of) plane waves (see my previous posting).

It's a long time ago that I learned some QT from Gasiorowicz's book, and if I remember right, it's a pretty good textbook. I don't think that this author claims that the general solutions of the free 1D Schrödinger equation are (superpositions of) plane waves.

 

[Davide]

The solutions of the 1D TDSE (with $V=0$) are plane waves (Or combinations of plane waves):
$$\psi_{k}(x,t) = Ae^{i(kx-wt)}+Be^{-i(kx-wt)}$$

This statement is undeniable. You can check this anywhere:

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/Schr2.html

http://physics.gmu.edu/~pnikolic/PHYS308/lectures/schrodinger.pdf

https://www.electrical4u.com/schrodinger-wave-equation/#:~:text=The%20Schr%C3%B6dinger%20equation%20(also%20known,by%20solving%20the%20Schr%C3%B6dinger%20equation.

http://www.astro.uvic.ca/~jwillis/teaching/phys215/phys215_lecture4.pdf

You can find a solution that is not a plane wave, but, since you really understand what is the Fourier Analysis you realize that any "reasonably well-behaved" function is a superposition of plane waves (check https://scholar.harvard.edu/files/david-morin/files/waves_fourier.pdf).

About the main point of the thread, that is the meaning of $e^{-ik^2 6 }$:

The factor e−ik26 provides an offset in time (as does any factor of the form eiwkt0) in the Fourier Transform.

That's right. Thank's, I didn't realize it.

 

[Davide]

Of course plane waves are solutions of the Schrödinger equation, but not all solutions of the Schrödinger equation are plane waves as the example with the Gaussian wave packet shows: it's clearly NOT of the form $f(ct-x)$.

The reason is that the plane-wave solutions are the energy eigenstates. For a free particle these are the momentum eigenstates,
$$u_p(t,x)=\frac{1}{\sqrt{2 \pi \hbar}} \exp(-\mathrm{i} \omega_{p} t+\mathrm{i} k x)$$
with
$$k=p/\hbar, \quad \omega = \hbar^2 k^2/(2m).$$
This quadratic (i.e., non-linear) dependence between $\omega$ and $k$ is the reason why, in contradistinction to solutions of the 1D wave equation, the general solutions of the 1D Schrödinger equation are not (superpositions of) plane waves (see my previous posting).

It's a long time ago that I learned some QT from Gasiorowicz's book, and if I remember right, it's a pretty good textbook. I don't think that this author claims that the general solutions of the free 1D Schrödinger equation are (superpositions of) plane waves.

Please, could you give a reference to the general solutions of the free 1D Schrödinger Equation that you are talking about?
Thanks.

 

[vanhees71]

Your plane waves are not even true states, because they are not square integrable! For the free field from the momentum eigenstates I derived above you find the most general solution of the TDSE as
$$\psi(t,x) = \int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi \hbar}} \tilde{\psi}_0(p) \exp \left (\mathrm{i} \frac{p x-\omega_p t}{\hbar} \right) \quad \text{with} \quad \omega_p=\frac{p^2}{2m},$$
where $\tilde{\psi}_0 \in \mathrm{L}^2(\mathbb{R})$ is the initial wave function in momentum representation, i.e., it's given by the initial wave function in position space by
$$\tilde{\psi}_0(p)=\int_{\mathbb{R}} \mathrm{d} x \frac{1}{\sqrt{2 \pi \hbar}} \psi_0(x) \exp(-\mathrm{i} p x/\hbar).$$
These proper square integrable solutions ("wave pacekts") are no a plane waves! As an example see the Gaussian-wave-packet solution quoted above!

 

[vanhees71]

Please, could you give a reference to the general solutions of the free 1D Schrödinger Equation that you are talking about?
Thanks.

I guess you find them in nearly any textbook on quantum mechanics (e.g., Merzbacher, Quantum Mechanics), but for the Gaussian wave packet, I've given above, it's also easy to check that the solution satisfies the Schrödinger equation. It's only a bit work with the derivatives ;-)).

 

[Davide]

Your plane waves are not even true states, because they are not square integrable! For the free field from the momentum eigenstates I derived above you find the most general solution of the TDSE as
$$\psi(t,x) = \int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi \hbar}} \tilde{\psi}_0(p) \exp \left (\mathrm{i} \frac{p x-\omega_p t}{\hbar} \right) \quad \text{with} \quad \omega_p=\frac{p^2}{2m},$$
where $\tilde{\psi}_0 \in \mathrm{L}^2(\mathbb{R})$ is the initial wave function in momentum representation, i.e., it's given by the initial wave function in position space by
$$\tilde{\psi}_0(p)=\int_{\mathbb{R}} \mathrm{d} x \frac{1}{\sqrt{2 \pi \hbar}} \psi_0(x) \exp(-\mathrm{i} p x/\hbar).$$
These proper square integrable solutions ("wave pacekts") are no a plane waves! As an example see the Gaussian-wave-packet solution quoted above!

This:

I derived above you find the most general solution of the TDSE as
ψ(t,x)=∫Rdp12πℏψ~0(p)exp⁡(ipx−ωptℏ)withωp=p22m,

is the general form of a wavepacket constructed by the superposition of plane waves.

Some references (I don't know if this is legal but I'm getting crazy with this thread :D):

- From the Gasiorowicz, Quantum Physics :

- From the Merzbacher, Quantum Mechanics :

About the normalization of the plane waves:

https://physics.stackexchange.com/questions/165373/normalizing-the-solution-to-free-particle-schrödinger-equation

 

[vanhees71]

What you quote from Gasiorowicz is precisely what I wrote. What's missing is just the dispersion relation, $\omega=p^2/(2m \hbar)$, but I'm sure that it's writting somewhere in the book.

The plane waves are generalized functions (distributions) and not square integrable functions. They are "normalizable" to a $\delta$ distribution, indeed with
$$\langle x| p \rangle=u_p(x)=\frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} p x/\hbar)$$
you get
$$\langle p|p \rangle=\int_{\mathbb{R}} \mathrm{d} p u_p^*(x) u_{p'}(x) = \delta(p-p').$$
With them you can express the position-space wave function in terms of the momentum-space wave function (and vice versa):
$$\psi(x)=\int_{\mathbb{R}} \mathrm{d} p \langle x|p \rangle \langle p|\psi \rangle=\int_{\mathbb{R}} u_p(x) \tilde{\psi}(p).$$
The time evolution for the free particle is given by ##\hat{H}=\hat{p}^2/(2m), i.e., in the Schrödinger picture
$$|\psi(t) \rangle=\exp(-\mathrm{i} \hat{H} t)|\psi(0) \rangle,$$
and from this
$$\psi(t,x)=\langle x|\psi(t) \rangle = \int_{\mathbb{R}} \mathrm{d} x u_p(x) \langle p|\psi(t) \rangle =\int_{\mathbb{R}} \mathrm{d} x u_p(x) \langle p|\exp(-\mathrm{i} \hat{H} t/\hbar)|\psi(0) \rangle$$
$$=\int_{\mathbb{R}} \mathrm{d} x u_p(x) \langle \exp(+\mathrm{i} \hat{H} t/\hbar) p|\psi(0) \rangle=\int_{\mathbb{R}} \mathrm{d} x u_p(x) \exp[-\mathrm{i} p^2 t/(2m \hbar)] \tilde{\psi}_0(p).$$

 

[Davide]

What you quote from Gasiorowicz is precisely what I wrote. What's missing is just the dispersion relation, $\omega=p^2/(2m \hbar)$, but I'm sure that it's writting somewhere in the book.

The plane waves are generalized functions (distributions) and not square integrable functions. They are "normalizable" to a $\delta$ distribution, indeed with
$$\langle x| p \rangle=u_p(x)=\frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} p x/\hbar)$$
you get
$$\langle p|p \rangle=\int_{\mathbb{R}} \mathrm{d} p u_p^*(x) u_{p'}(x) = \delta(p-p').$$
With them you can express the position-space wave function in terms of the momentum-space wave function (and vice versa):
$$\psi(x)=\int_{\mathbb{R}} \mathrm{d} p \langle x|p \rangle \langle p|\psi \rangle=\int_{\mathbb{R}} u_p(x) \tilde{\psi}(p).$$
The time evolution for the free particle is given by ##\hat{H}=\hat{p}^2/(2m), i.e., in the Schrödinger picture
$$|\psi(t) \rangle=\exp(-\mathrm{i} \hat{H} t)|\psi(0) \rangle,$$
and from this
$$\psi(t,x)=\langle x|\psi(t) \rangle = \int_{\mathbb{R}} \mathrm{d} x u_p(x) \langle p|\psi(t) \rangle =\int_{\mathbb{R}} \mathrm{d} x u_p(x) \langle p|\exp(-\mathrm{i} \hat{H} t/\hbar)|\psi(0) \rangle$$
$$=\int_{\mathbb{R}} \mathrm{d} x u_p(x) \langle \exp(+\mathrm{i} \hat{H} t/\hbar) p|\psi(0) \rangle=\int_{\mathbb{R}} \mathrm{d} x u_p(x) \exp[-\mathrm{i} p^2 t/(2m \hbar)] \tilde{\psi}_0(p).$$

That's right. Thanks!

 

[LucKy]

Thanks everybody for valuable ansvers,
However I have new question, this time regarding expectation energy of the particle represented by a Gaussian wavepaket.
Please can anybody expain:
If we know that wavepacket in momentum space (or in wavenumber k-spase) peaks at particular value $k_0$,
why we need to calculate expectation value of energy as
$\langle E_p \rangle = \int dk \phi^*(k - k_0) \hat {\mathbf H} \phi(k - k_0)$, where $\hat {\mathbf H} \to \frac {\hbar^2k^2} {2m}$
but not just as
$\langle E_p \rangle = \frac {\hbar^2k_0^2} {2m}$ (because we already know that our Gaussian wavepacket peaks at $k_0$, and it's uncertainty for wavenumber k is proportional to 1/a and it is symmetrical about $k_0$ ),
also why $\hat {\mathbf H} \to \frac {\hbar^2k^2} {2m}$, but not $\hat {\mathbf H} \to \frac {\hbar^2(k-k_0)^2} {2m}$ for localized at $k_0$ wavepackets?

 

[vanhees71]

There is the Hamilton operator, which is $\hat{p}^2/(2m)$ and there is the wave function at the initial time (in momentum representation) which is $\propto \exp[-(p-p_0)^2/(4 \Delta p^2)]$. There cannot be any property of the state occurring in the operators describing observables.