Questions Regarding Free Particles - Griffith's QM (I)

[WWCY]

Homework Statement

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1) I don't quite understand what 2.94 means on its own. It was derived from 2.93, yet it doesn't show a superposition of any sort. The author then takes 2.94, and attempts to normalise it by stating

$\int \Psi_k^* \Psi_k dx = \mid A^2 \mid\int dx = \infty$

What would be the purpose of taking only 1 term from 2.93 and attempting to normalise it? Was this a simpler way of illustrating the un-normalizability of the wavefunction rather than attempting it with 2.93 as a whole?

2) The author then goes on to say (after the integral) "In the case of the free particle, then, the separable solutions do not represent physically realizable states. A free particle cannot exist in a stationary state (...) there is no such thing as a free particle with a definite energy".
a) What does a physically realizable state entail?
b) What led to the conclusion of "A free particle cannot exist in a stationary state / there is no such thing as a free particle with a definite energy"?
c) Also, what does "not having definite energy" mean? Does it mean that measurements on free particles will see variance in the energy levels observed (unlike that of stationary states)?

3) The author then formulates a general solution to the time-dependent Schrodinger equation by integrating over k:

$\Psi(x,t) = \frac{1}{\sqrt{2\pi}}\phi (k) e^{i(kx - \frac{\hbar k^2}{2m}) t}$

and stating that this was a normalizable solution.
a) Does this mean that though a free particle can't exist as a stationary state, it can exist as a wave-packet?
Also, what are the fundamental physical differences between the two?

Thanks very, very much in advance!

P.S: I am learning QM on my own for now, and the exposure I have to QM is limited to the first 2 chapters of Griffith's book. Explanations that are not too technical would be greatly appreciated!

Homework Equations

The Attempt at a Solution

 

[Nugatory]

1) 2.93 and 2.94 are the same equation, just written in two different ways. If you want $k$ to be a positive number ($0\le{k}\le\infty$), than you must use 2.93; if you will allow $k$ to range from $-\infty$ to $+\infty$ you can use 2.94. Either way, you can have superpositions when you consider sums or integrals over different values of $k$.

2) The separable states (either 2.93 or 2.94 for a single value of $k$) are not physically realizable because they describe an infinite plane wave. Look at either one for a fixed $k$ and imagine that the $y$ axis happens to pass through the Earth and Andromeda galaxy; then at any given time $\psi$ has the exact same value here and in the Andromeda galaxy (and everywhere else in $y-z$ plane throughout the entire universe). This state is pretty clearly absurd and there's no way of putting a particle into that state so it's not especially surprising that it's not normalizable. However, these solutions are very convenient because superpositions of them are mathematically easy to analyze and some superpositions of them are normalizable and have the sensible property of going to zero at infinity so can describe states that we can put a particle into.

3) The function you wrote down is a plane wave; that $\phi(k)/\sqrt{2\pi}$ term is just making the amplitude of the wave a function of $k$. It's important because the superposition that you get by integrating across all values of $k$ (that is, superposing the plane wave for every value of $k$) is normalizable and describes a particle whose position and momentum are more or less precisely known.

 

[WWCY]

Hi there, thank you for helping out. Might I ask a few follow up questions?

RE 1)
I still don't quite understand how he took $\Psi_k$ for say, positive k, failed to normalize it, and therefore concluded that $\Psi (x,t) = Ae^{ikx - \frac{-iEt}{\hbar}} + Be^{-ikx- \frac{-iEt}{\hbar}}$ was un-normalizble too. Why does he not take $\int \Psi \Psi^*$ instead?

RE 2)

Look at either one for a fixed kkk and imagine that the yyy axis happens to pass through the Earth and Andromeda galaxy; then at any given time ψψ\psi has the exact same value here and in the Andromeda galaxy (and everywhere else in y−zy−zy-z plane throughout the entire universe

Why does the value of $\psi$ here not evolve with time?

RE 3)

The author then goes on to state that a wide spread of $\phi(k)$ when $\phi(k)$ is plotted against $k$ means greater in uncertainty in k and momentum. How did he come to this conclusion? Also, how does a nice localized peak of $\phi$ a) provide more certainty b) allow us to calculate momentum?

Also, I'm not quite sure what the term definite energy means.

In the past chapter (infinite square well), definite energy seemed to refer to the fact that for a particle in state $\Psi = \psi_n$, all measurements of total energy will return $E_n$. How does the equation $\Psi = \frac{1}{\sqrt{2\pi}} \int \phi(k) e^{i(kx- \frac{\hbar k^2}{2m} t)}$ tell us that free particles do not have definite energy?

Thank you for your patience.

 

[vanhees71]

I find this exposition of the free-particle wave function a bit confusing. The confusion comes from not properly stating the choice of the bases to define the wave function in the two cases.

I guess the case of a free non-relativistic particle moving in one space dimension is discussed. The Hamiltonian is
$$\hat{H}=\frac{1}{2m} \hat{p}^2.$$
Now there are two obvious possibilities to choose a convenient basis to solve the initial-value problem of time evolution of the wave function,
$$\mathrm{i} \partial_t \psi(t,x)=-\frac{1}{2m} \partial_x^2 \psi(t,x).$$

(a) Using momentum eigenstates
$$\hat{p} u_p(x)=-\mathrm{i} \partial_x u_p(x)=p u_p(x).$$
The solutions are
$$u_p(x)=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x), \quad p \in \mathbb{R}.$$
The normalization is chosen such that
$$\langle p|p' \rangle=\int_{\mathbb{R}} \mathrm{d} p u_p^*(x) u_{p'}(x)=\delta(p-p').$$
These are obviously also eigenstates of $\hat{H}$,
$$\hat{H} u_p(x)=\frac{p^2}{2m} u_p(x),$$
and thus the general solution of the Schrödinger equation for the free particle reads
$$\psi(t,x)=\int_{\mathbb{R}} \mathrm{d} p \exp \left [-\mathrm{i} E(p) t \right] u_p(x) \tilde{\psi}_0(p),$$
where $E(p)=p^2/(2m)$ and
$$\tilde{\psi}_0(p)=\langle p|\psi(t=0) \rangle=\int_{\mathbb{R}} \mathrm{d} x u_{p}^*(x) \psi(t=0,x).$$

(b) using energy eigenstates
$$\hat{H} u_E(x)=E u_E(x) \;\Rightarrow \; -\partial_x^2 u_E(x)=2 m E u_E(x).$$
Obviously the eigenvalues are $E \geq 0$, and for each $E$ there are 2 solutions (except for $E=0$),
$$u_E^{(+)}(x)=N_+(E) \cos(k x), \quad u_E^{(-)}(x)=N_-(E) \sin(k x),\quad k=\sqrt{2mE}.$$
We can choose the normalization factors again such that
$$\int_{\mathbb{R}} \mathrm{d} x u_E^{(\pm)*}(x) u_{E'}^{(\pm)}(x)=\delta(E-E'), \quad \int_{\mathbb{R}} \mathrm{d} x u_E^{(\pm)*}(x) u_{E'}^{(\mp)}(x)=0.$$
Then the general solution for the Schrödinger equation reads
$$\int_0^{\infty} \mathrm{d} E [\tilde{\psi}_0^{(+)}(E) u_E^{(+)}(x) + \tilde{\psi}_0^{(-)}(E) u_E^{(-)}(x)] \exp(-\mathrm{i} E t),$$
where
$$\tilde{\psi}_0^{\pm}(E)=\int_{\mathbb{R}} \mathrm{d} x u_{E}^{(\pm)}(x) \psi(t=0,x).$$
The energy eigenvalues are obviously degenerate, and I've chosen as another observable to specify the energy eigenstates the parity, i.e., I've used even and odd functions under space reflections, $x \rightarrow -x$.

 

[WWCY]

I find this exposition of the free-particle wave function a bit confusing. The confusion comes from not properly stating the choice of the bases to define the wave function in the two cases.

I guess the case of a free non-relativistic particle moving in one space dimension is discussed. The Hamiltonian is
$$\hat{H}=\frac{1}{2m} \hat{p}^2.$$
Now there are two obvious possibilities to choose a convenient basis to solve the initial-value problem of time evolution of the wave function,
$$\mathrm{i} \partial t \psi(t,x)=-\frac{1}{2m} \partial_x^2 \psi(t,x).$$

Hi @vanhees71 , I really appreciate the help! It's embarrassing to say, but I'm afraid I couldn't understand the stuff on Eigenstates as I've not gone through that type of material. Is there another way to explain this? Apologies!

 

[vanhees71]

Don't look for another way, but learn the eigenvalue/eigenstates stuff as soon as possible. Without it there's no way to understand QM!

 

[WWCY]

Don't look for another way, but learn the eigenvalue/eigenstates stuff as soon as possible. Without it there's no way to understand QM!

However all stuff "eigen" comes in the next chapter, which I am looking to get to ASAP. In the meantime, could I trouble you to help provide clarification with a couple of my queries?

Also, thanks for the advice!

 

[bhobba]

In the meantime, could I trouble you to help provide clarification with a couple of my queries?

We are always happy to help people having trouble with QM, whether its the beginner, student at the level of Griffiths like you - indeed any level.

Thanks
Bill

 

[WWCY]

We are always happy to help people having trouble with QM, whether its the beginner, student at the level of Griffiths like you - indeed any level.

Thanks
Bill

Cheers Bill, really appreciate the help!

Of all other concepts in this thread, the 3 that I'm really struggling to grasp are these:

1) The author took $\Psi_k = Ae^{i(kx-\frac{\hbar k^2}{2m})}$ for say, positive k, failed to normalize it, and therefore concluded that $\Psi (x,t) = Ae^{ikx - \frac{-iEt}{\hbar}} + Be^{-ikx- \frac{-iEt}{\hbar}}$ was un-normalizble too. Why does he not take $\int \Psi \Psi^*$ instead?

2) The author later states that a wide spread of $k$ values when $\phi(k)$ is plotted against $k$ means greater in uncertainty in k and therefore momentum. How did he come to this conclusion?

3) I'm not quite sure what the term definite energy means.

In the past chapter (infinite square well), definite energy seemed to refer to the fact that for a particle in state $\Psi = \psi_n$, all measurements of total energy (for an ensemble) are sure to return $E_n$. Does the meaning of "free particles cannot exist in a state of definite energy" mean that, measurements carried out on a wave-packet always yield varying energy levels that are dependent somehow, on the $\phi (k)$ term?

Thank you!

 

[bhobba]

Hi

You have run into a deep mathematical difficulty.

Try to take the integral - what happens? That's right you can't do it. The answer lies in what's called Rigged Hilbert Spaces - but first before delving into that you need to understand what's called Distribution Theory, which all applied mathematicians should know:
https://www.amazon.com/dp/0521558905/?tag=pfamazon01-20

For now simply think of such pathological states as for all practical purposes some function that is the same except at some very large positive and negative number where it goes to zero.

The second one is the famous Heisenberg Uncertainty Principle - to understand it you really need to see a proper statement of it. Suppose you have a large number of similarly prepared systems ie all are in the same quantum state. Divide them into two equal lots. In the first lot measure position to a high degree of accuracy. QM places no limit on that accuracy - its a misunderstanding of the uncertainty principle thinking it does. The result you get will have a statistical spread. In the second lot measure momentum to a high degree of accuracy - again QM places no limit on that. It will also have a statistical spread. The variances of those spreads will be as per the Heisenberg Uncertainty principle.

For definite energy think eigenvalue of energy operator.

Don't worry if you don't fully get Griffiths - its at intermediate level and leaves some issues up in the air so to speak. Good students will spot it and that cause issues. The answer is after Griffiths read a slightly more advanced text - I like Sakurai - Modern QM:
https://www.amazon.com/dp/0805382917/?tag=pfamazon01-20

Then you are prepared for Ballentine which leaves nothing open - it is an axiomatic development where everything is explained. But its too advanced to start with - you need to build up to it gradually:
https://www.amazon.com/dp/9810241054/?tag=pfamazon01-20

Thanks
Bill

 

[vanhees71]

Cheers Bill, really appreciate the help!

Of all other concepts in this thread, the 3 that I'm really struggling to grasp are these:

1) The author took $\Psi_k = Ae^{i(kx-\frac{\hbar k^2}{2m})}$ for say, positive k, failed to normalize it, and therefore concluded that $\Psi (x,t) = Ae^{ikx - \frac{-iEt}{\hbar}} + Be^{-ikx- \frac{-iEt}{\hbar}}$ was un-normalizble too. Why does he not take $\int \Psi \Psi^*$ instead?

2) The author later states that a wide spread of $k$ values when $\phi(k)$ is plotted against $k$ means greater in uncertainty in k and therefore momentum. How did he come to this conclusion?

3) I'm not quite sure what the term definite energy means.

In the past chapter (infinite square well), definite energy seemed to refer to the fact that for a particle in state $\Psi = \psi_n$, all measurements of total energy (for an ensemble) are sure to return $E_n$. Does the meaning of "free particles cannot exist in a state of definite energy" mean that, measurements carried out on a wave-packet always yield varying energy levels that are dependent somehow, on the $\phi (k)$ term?

Thank you!

Well, as any unbound states (the free particle of course hasn't any bound states, because there's no binding potential) they are not Hilbert-space vectors but live in a larger space of distributions and thus are not normalizable in the usual sense. You have to normalize them to a Dirac $\delta$ distribution.