Compilation of severe errors in famous textbooks

[vanhees71]

A solution what for? I don't quite understand the problem obviously. Of course, I always can make the ansatz with $x=A \exp(\mathrm{i} \omega t)$, but if you solve for a problem with a negative oscillator postential, $V(x)=-\alpha^2 x^2/2$, you get the EoM
$$m \ddot{x}=\alpha^2 x$$
and Plugging in your ansatz you get
$$-m \omega^2=\alpha^2 \; \Rightarrow \; \omega=\pm \mathrm{i} \alpha/\sqrt{m},$$
and again we get the general solution indicated above
$$x(t)=A_1 \exp(\alpha t/\sqrt{m}) + A_2 \exp(-\alpha t/\sqrt{m}).$$
Energy is of course conserved.

The trajectory is always unbound (the potential has no minima, and the energy is not bounded from below). So there's no surprise here at all.

 

[andresB]

he has a short discussion of the Quantum Zeno effect. His conclusion is "Thus the Quantum Zeno effect actually occurs for this system. [This refers to what is reported in https://doi.org/10.1103/PhysRevA.41.2295 ] But it is misleading to explain it as being due to a "collapse of the wave function" caused by measurement. No "collapse" actually occurs; rather the excitation of the atom is impeded by the string perturbation of the optical pulses and the coupling to the radiation field. Moreover [...], the effect occurs regardless of whether or not any measurement of the emitted photons is actually made."

Do you think there is anything wrong with these statements? If not, what other explicitly wrong statements do you have in mind?

What about interaction-free measurements?
https://www.nature.com/articles/ncomms7811

 

[Demystifier]

The problem was that in his example the solution was of the form $x_k=A_ke^{i\omega_k t}$, where the omegas are different and you get a cancellation.

No, my example is sufficiently general to include the case in which the omegas are equal (which happens when the alphas are equal), and the cancellation happens even in this case.

 

[martinbn]

No, my example is sufficiently general to include the case in which the omegas are equal (which happens when the alphas are equal), and the cancellation happens even in this case.

Not general enough to claim that their argument is wrong.

 

[Demystifier]

Then he said that with the same $\omega$ you cannot get a solution, which is also wrong.

It's not wrong, the omegas in the solution are the same as the omegas in the Hamiltonian (because the Hamiltonian is diagonalized i.e. there is no coupling between the two degrees of freedom), so if the omegas in the Hamiltonian are different then so are the omegas in the solution.

 

[Demystifier]

Not general enough to claim that their argument is wrong.

Even if that was true (which was not), you missed that I mentioned also a much more general argument (actually a theorem) that the energy is conserved whenever the Hamiltonian does not have an explicit dependence on time, which in our case is true because the omegas in the Hamiltonian are time independent.

 

[bhobba]

That sounds interesting. Where can we read more about it? By the way, I am not sure about the history, but I think that Grothendick was not involved here. You need to credit Schwartz.

Its in the beginning - you can't miss it.

Grothendick worked on Nuclear spaces that are used in Rigged Hilbert spaces eg Schwartz spaces or the space of good functions as Littlewood called it:
https://en.wikipedia.org/wiki/Nuclear_space
Thanks
Bill

 

[vanhees71]

Even if that was true (which was not), you missed that I mentioned also a much more general argument (actually a theorem) that the energy is conserved whenever the Hamiltonian does not have an explicit dependence on time, which in our case is true because the omegas in the Hamiltonian are time independent.

I'm still not understanding all these arguments. I guess one should move this discussion to the mechanics forum, but isn't LL discussing the case of small oscillations around a minimum of a potential. Then the eigenfrequencies are of course all real.

That's easy to see. Without loss of generality we can shift the coordinates such that the minimum of the potential is at $\vec{x}=0$. Then we can approximate the potential in a neibourhood of the origin by
$$V(\vec{x})=\frac{1}{2} A_{jk} x_j x_k,$$
where $A_{jk}$ is a symmetric positive definite matrix, i.e., there's a orthogonal transformation, $\vec{x}'=\hat{O} \vec{x}$ such that $A_{jk}' x_j' x_k'=m \sum_{j} \omega_j^2 x_j^{\prime 2}$ with all $\omega_j^2>0$, and then the solutions for the normal coordinates $x_k'$ are just
$$x_k'(t)=A_k \exp(\mathrm{i} \omega_k t) + B_k \exp(-\mathrm{i} \omega_k t).$$
So for this case LL seems right to me.

You can of course solve everything in the original coordinates. The EoMs read
$$m \ddot{x}_j=-\partial_{j} V=A_{jk} x_k \; \Rightarrow \; m\ddot{\vec{x}}=-\hat{A} \vec{x}.$$
This you can solve by the ansatz
$$\vec{x}(t)=\vec{a} \exp(\mathrm{i} \omega t).$$
This gives
$$-m \omega^2 \vec{a} = -\hat{A} \vec{a},$$
which means that $\omega$ must be chosen such that $m \omega^2$ is an eigenvalue of $\hat{A}$ and $\vec{a}$ the corresponding eigenvector.

Since $\hat{A}=\hat{A}^{\text{T}}$ is positive definite there are $3$ positive eigenvalues (which may all be different from each other or not) you have $\omega \in \mathbb{R}$, and one can choose the eigenvectors $\vec{a}_k$ real and orthonormal. Then the general solution reads
$$\vec{x}(t) = \sum_{j=1}^3 \vec{a}_k [A_k \exp(\mathrm{i} \omega_k t)+B_k \exp(-\mathrm{i} \omega k t)].$$
This can be written in real form as
$$\vec{x}(t)=\sum_{j=1}^3 \vec{a}_k [C_k \cos(\omega_k t) + D_k \sin(\omega_k) t],$$
where $C_k,D_k \in \mathbb{R}$.

If $\hat{A}$ is not positive definite, the approximation doesn't make sense, because then there are unbound solutions as discussed above.

 

[Demystifier]

I'm still not understanding all these arguments.

What you say completely agrees with my statements and I don't understand the arguments by @martinbn either.

 

[vis_insita]

If we agree that (according to quantum Zeno) frequent measurements slow down the decay, then I see nothing absurd in the idea that quantum Zeno in the continuous measurement limit (that is, the watched pot) can, in principle, stop the decay completely.

We don't agree. The effect is caused by perturbing the system into an unstable state which almost instantly decays back into the ground state via spontaneous emission. The repeated excitation and the coupling to the EM-field cause the delayed decay. Since the metastable state isn't affected, the result of the disturbance is still a superposition and not a collapsed state. All this has nothing to do with frequent measurement or an induced collapse at all. In particular the effect is independent of whether any emitted photons are measured or not.

As far as I understand it the decay could only be halted in the unphysical limit of infinite coupling. Thus the collapse is at best an imperfect model for the real effect, and cannot be true exactly. Another way of putting it is that the collapse implies a strictly exponential decay law. But for short times the time evolution necessarily deviates from the exponential law. Thus the collapse should be expected to give wrong results in the limit of continuous observation.

 

[vis_insita]

What about interaction-free measurements?
https://www.nature.com/articles/ncomms7811

I don't know. What about them? They say they employ the Quantum Zeno effect. I'm not sure if they claim to have shown the existence of a collapse, athough this is probably their interpretation of the cause of that effect. Do you think something in their setup conflicts with the analysis of Ballentine?

 

[Keith_McClary]

what the Casimir effect is not

Is there anyone still arguing that vacuum energy is needed to explain the Casimir effect (or that the Casimir effect proves the existence of vacuum energy)?

 

[TSny]

Regarding the LL discussion, I don't see any errors in the LL argument. As some others have pointed out, it is important to keep in mind that their argument is based on the assumption that $T$ and $U$ are always non-negative. Here is how I would summarize their argument:

1. $T$ is a positive definite quadratic form $T = \frac{1}{2}\sum m_{ik}\dot x_i \dot x_k$.

2. Given: $U$ is a positive definite quadratic form $U = \frac{1}{2}\sum k_{ik} x_i x_k$.

3. The total energy $E = T+V$ is conserved. This was discussed back in chapter II of LL.

Claim: If a solution of the equations of motion has the form $x_k = A_ke^{i \omega t}$, then $\omega$ must be real. (Note that $\omega$ is assumed to be the same for all $x_k$.) We don't need to consider the trivial solution where all of the $A_k$'s are zero.

LL's "physical argument”:

If $\omega$ has negative imaginary part, then the magnitudes of all the $x_k$’s and all the $\dot x_k$’s for which $A_k \neq 0$ will grow exponentially with time. Since $T$ and $U$ are positive definite forms, $E$ must necessarily grow exponentially with time. If $\omega$ has positive imaginary part, then all the $x_k$’s and all the $\dot x_k$’s for which $A_k \neq 0$ will decay exponentially with time. $E$ must then decay exponentially with time. But we know that $E$ must be a constant. So, $\omega$ cannot have an imaginary part.

 

[vanhees71]

Is there anyone still arguing that vacuum energy is needed to explain the Casimir effect (or that the Casimir effect proves the existence of vacuum energy)?

Nobody is arguing that the usual introductory-textbook treatment (using simply a boundary condition for two plates) is the infinite-charge-value limit. It's an effective theory to describe the manybody system, but the Casimir effect is not some vacuum fluctuation (since the vacuum is the one state which does not fluctuate at all) but due to the fluctuations of the charges in the plates and the em. field (the quantum pendant of van der Waals forces). See

R. L. Jaffe, The Casimir effect and the quantum vacuum,
Phys. Rev. D 72 (2005) 021301.
https://dx.doi.org/10.1103/PhysRevD.72.021301
https://arxiv.org/abs/hep-th/0503158v1

Also @Demystifier has written a PLB (open access):

https://doi.org/10.1016/j.physletb.2016.08.036
and a very nice pedagogical paper:

https://doi.org/10.1016/j.aop.2017.05.013https://arxiv.org/abs/1702.03291v2

 

[atyy]

Well, one severe conceptual mistake is in some textbooks (if I remember right even in the Feynman Lectures vol. II and in Berkeley Physics Course vol. II, which shows that also Nobel Laureates make mistakes ;-)) when treating the magnetostatics of a wire relativistically. The mistake lies in the assumption that the wire is uncharged in the rest frame of the wire. In fact it's uncharged in the rest frame of the electrons. The correct treatment has to take into account the "self-induced" Hall effect though it's academic for house-hold currents, where the drift velocities are of the order of 1mm/s, but if you want to treat it fully relativistically you must take into account the correct relativistic version of Ohm's Law, $\vec{j}=\sigma \gamma (\vec{E} + \vec{v} \times \vec{B}/c)$ with $\sigma$ the usual electric conductivity (a scalar as any transport coefficient), $\gamma=(1-v^2/c^2)^{-1/2}$ (using Heaviside-Lorentz units).

Oh no, is Schroeder (of Peskin and Schroeder) wrong too in this presentation of Purcell?
http://physics.weber.edu/schroeder/mrr/MRRtalk.html

I never realized there was a relativistic correction to Ohm's law ...

 

[Demystifier]

Is there anyone still arguing that vacuum energy is needed to explain the Casimir effect (or that the Casimir effect proves the existence of vacuum energy)?

Most still compute Casimir effect in terms of vacuum energy, without thinking whether it is really necessary to do it that way.

 

[fluidistic]

So, can we conclude that TSny's post (#83) finally settles once and for all that Landafshitz's textbook is fine on that part? I am constantly swinging my lighter beneath my physical copy of that textbook, I need to know whether I set it on fire or not!

 

[TSny]

I am constantly swinging my lighter beneath my physical copy of that textbook, I need to know whether I set it on fire or not!

NO NO NO

 

[vanhees71]

Oh no, is Schroeder (of Peskin and Schroeder) wrong too in this presentation of Purcell?
http://physics.weber.edu/schroeder/mrr/MRRtalk.html

I never realized there was a relativistic correction to Ohm's law ...

Yes it's wrong too, because he assumes the wire having $\rho=0$ in the lab frame. Of course it's the conduction electrons that are moving in the lab frame while the positive ions are at rest, and then there's the Hall effect. Taking into account the Hall effect leads to the correct relativistic Ohm's Law and to the correct conclusion that $\rho'=0$ in the rest frame of the conduction electrons. You find this treatment here:

https://www.physicsforums.com/insights/relativistic-treatment-of-the-dc-conducting-straight-wire/https://itp.uni-frankfurt.de/~hees/pf-faq/relativistic-dc.pdf
That there's a relativistic correction to Ohm's Law goes back to the seminal work by Minkowski and can be found in many textbooks on relativistic classical electrodynamics.

 

[vanhees71]

Regarding the LL discussion, I don't see any errors in the LL argument. As some others have pointed out, it is important to keep in mind that their argument is based on the assumption that $T$ and $U$ are always non-negative. Here is how I would summarize their argument:

1. $T$ is a positive definite quadratic form $T = \frac{1}{2}\sum m_{ik}\dot x_i \dot x_k$.

2. Given: $U$ is a positive definite quadratic form $U = \frac{1}{2}\sum k_{ik} x_i x_k$.

3. The total energy $E = T+V$ is conserved. This was discussed back in chapter II of LL.

Claim: If a solution of the equations of motion has the form $x_k = A_ke^{i \omega t}$, then $\omega$ must be real. (Note that $\omega$ is assumed to be the same for all $x_k$.) We don't need to consider the trivial solution where all of the $A_k$'s are zero.

LL's "physical argument”:

If $\omega$ has negative imaginary part, then the magnitudes of all the $x_k$’s and all the $\dot x_k$’s for which $A_k \neq 0$ will grow exponentially with time. Since $T$ and $U$ are positive definite forms, $E$ must necessarily grow exponentially with time. If $\omega$ has positive imaginary part, then all the $x_k$’s and all the $\dot x_k$’s for which $A_k \neq 0$ will decay exponentially with time. $E$ must then decay exponentially with time. But we know that $E$ must be a constant. So, $\omega$ cannot have an imaginary part.

That doesn't make sense indeed. No matter which signature the matrix $k_{ik}$ has, energy is always conserved. Noether's theorem applied to time-translation invariance tells you that energy is conserved as long as the Hamiltonian (Lagrangian) is not explicitly time dependent, and that's obviously the case here.

Of course, if $k_{ik}$ is not positive definite, you have unbound (exponentially growing) solutions, but that doesn't mean that energy conservation could be violated.

In LL 1 is a much more severe mistake. Somewhere they claim all Hamiltonian systems were integrable, which is for sure wrong. There are more non-integrable Hamiltonian systems than integrable ones in fact.

Nevertheless, the LL series is among the best textbooks written, as are the Feynman Lectures though there are indeed some severe errors and some typos in there. I wish there'd be some recipe to ensure error-free scientific texts. If you have one, please tell us ;-)).

 

[TSny]

No matter which signature the matrix $k_{ik}$ has, energy is always conserved. Noether's theorem applied to time-translation invariance tells you that energy is conserved as long as the Hamiltonian (Lagrangian) is not explicitly time dependent, and that's obviously the case here.

Agreed. Any solution of the equations of motion will conserve energy whether or not $U$ is positive definite . LL are simply arguing that if $k_{ik}$ is positive definite, then there can't be solutions of the equations of motion which are exponentially growing or decaying because energy would not conserved.

Of course, if $k_{ik}$ is not positive definite, you have unbound (exponentially growing) solutions, but that doesn't mean that energy conservation could be violated.

I know. But LL are assuming that $k_{ik}$ is positive definite. This seems to be the important feature of LL's argument that some people here are overlooking.

 

[atyy]

Yes it's wrong too, because he assumes the wire having $\rho=0$ in the lab frame. Of course it's the conduction electrons that are moving in the lab frame while the positive ions are at rest, and then there's the Hall effect. Taking into account the Hall effect leads to the correct relativistic Ohm's Law and to the correct conclusion that $\rho'=0$ in the rest frame of the conduction electrons. You find this treatment here:

https://www.physicsforums.com/insights/relativistic-treatment-of-the-dc-conducting-straight-wire/https://itp.uni-frankfurt.de/~hees/pf-faq/relativistic-dc.pdf
That there's a relativistic correction to Ohm's Law goes back to the seminal work by Minkowski and can be found in many textbooks on relativistic classical electrodynamics.

So in the "standard" Purcell-Feynman-Schroeder presentation, the correct transformation is obtained because all the errors cancel out?

 

[vanhees71]

I've to check, whether at the end they get it right, but then the derivation would be self-contradictive, because in the beginning they assume the wire's charge density vanishes in the restframe in the wire to deduce that in fact that this is not the case (because of the Hall effect). I think the correct treatment is simple enough, at least not more complicated than the wrong one, to present it at the end of the introductory E&M theory lecture (usualy the 3rd semester of the theory course in German universities).

 

[Demystifier]

their argument is based on the assumption that T and U are always non-negative.

But the assumption that U is non-negative is violated by negative $\omega^2$. So the true reason why $\omega^2$ needs to be positive is positivity of U, not conservation of energy.

 

[Demystifier]

I wish there'd be some recipe to ensure error-free scientific texts. If you have one, please tell us

There is a recipe, write only about trivial stuff and don't try to be original.

 

[Demystifier]

So, can we conclude that TSny's post (#83) finally settles once and for all that Landafshitz's textbook is fine on that part?

No we can't.

 

[TSny]

But the assumption that U is non-negative is violated by negative $\omega^2$.

I don’t see how the non-negativity of U can be violated by having a negative value of $\omega^2$.

The assumption that $U = \frac{1}{2} \sum k_{ik} x_i x_k$ is a positive definite quadratic form means that U cannot be negative for any choice of real values of the $x_k$’s. So, no matter whether $\omega^2$ is positive or negative, U cannot be negative.

 

[Demystifier]

I don’t see how the non-negativity of U can be violated by having a negative value of $\omega^2$.

The assumption that $U = \frac{1}{2} \sum k_{ik} x_i x_k$ is a positive definite quadratic form means that U cannot be negative for any choice of real values of the $x_k$’s. So, no matter whether $\omega^2$ is positive or negative, U cannot be negative.

You are missing the point. $\omega^2$ are the eigenvalues of the matrix $k_{ik}/m$ (for simplicity I take all masses to be equal) so positivity of $k_{ik}$ is equivalent to positivity of $\omega^2$. So if $k_{ik}$ is positive, then $\omega^2$ cannot be negative. It is a total nonsense to consider negative $\omega^2$ if one has already decided that $k_{ik}$ is positive.

So the correct chain of reasoning is the following:
- Why is $\omega^2$ positive?
- Because $k_{ik}$ is positive.
- Fine, but why is $k_{ik}$ positive?
- Because we want potential energy to be positive.
- Are those positivity requirements related to conservation of energy?
- No, the energy is conserved for any sign of $k_{ik}$, or equivalently, for any sign of $\omega^2$.

 

[vanhees71]

It has nothing to do with kinetic energy but with the question whether the Hamiltonian is bounded from below or not. Only if the matrix $k_{ik}$ is positive (semi-)definite that's the case.

Now make the ansatz
$$\vec{x}(t)=\vec{x}_0 \exp(-\mathrm{i} \omega t).$$
The EoM reads
$$\ddot{\vec{x}}=-\hat{K} \vec{x},$$
where $\hat{K}=(k_{ij}/m)$. Plugging in the ansatz leads to
$$\hat{K} \vec{x}_0 = \omega^2 \vec{x}_0,$$
i.e., $\vec{x}_0$ must be an eigenvector of $\hat{K}$ with eigenvalue $\omega^2$. Any symmetric matrix, which we have here since we can always choose $k_{ik}=k_{ki}$ (or $\hat{K}^{\text{T}} =\hat{K}$), can be diagonalized with an orthogonal transformation. If we choose the appropriate basis we thus have
$$\hat{K}'=\hat{O} \hat{K} \hat{O}^{\text{T}}=\mathrm{diag}(\omega_1^2,\ldots,\omega_d^2),$$
where $d$ is the dimension of the system.

Now it's clear that for any vector
$$\vec{x}^{\text{T}} \hat{K} \vec{x}^{\text{T}}=\vec{x}^{\text{T}} \hat{O}^{\text{T}} \hat{K}' \hat{O} \vec{x} \geq 0$$
if and only if $\omega_j^2 \geq 0$. If all $\omega_j^2 >0$, all motions are bounded oscillations. If one or more eigenvalues are 0 you have directions, given by the eigenvectors, where the particle is unbound and can move as a free particle though the Hamiltonian is still bounded from below.

If one or more eigenvalues are negative, the motion in these directions can be unbound and the particle is accelerated exponentially with time.

In any case the total energy is conserved since the Hamiltonian is not explicitly time-dependent.

If you consider the harmonic (or pseudoharmonic if there are negative eigenvalues) potential as approximation of some other more complicated potential the approximation is only good for the bound oscillatory motion, for which the particle always stays near the equilibrium value, and that's only the case if the potential has a true minimum, and that's where the Hesse matrix of the potential $\hat{K}$ is positive definite.

 

[Demystifier]

It has nothing to do with kinetic energy but with the question whether the Hamiltonian is bounded from below or not.

Of course, but it was @TSny who first framed it in terms of positivity of kinetic and potential energy, so I was replying to him.

 

[vanhees71]

It's amazing that we have such a long debate about such a simple issue. Maybe we should carefully study LL 1 again. It's information density is very high, i.e., though the text looks pretty short, it takes quite a time to study each chapter carefully and get a full understanding of it.

 

[TSny]

You are missing the point. $\omega^2$ are the eigenvalues of the matrix $k_{ik}/m$ (for simplicity I take all masses to be equal) so positivity of $k_{ik}$ is equivalent to positivity of $\omega^2$. So if $k_{ik}$ is positive, then $\omega^2$ cannot be negative. It is a total nonsense to consider negative $\omega^2$ if one has already decided that $k_{ik}$ is positive.

LL give two separate arguments for why $\omega^2$ must be real and positive (see page 67 of their text). The first argument is based on what they call “physical arguments” while the second, independent proof is “mathematical”. Your argument above appears to be similar to their second argument. They provide the mathematical steps to show how the positivity of $k_{ik}$ leads to $\omega^2$ being real and positive. I take it we agree that LL’s reasoning here is good.

LL’s “physical argument” for why $\omega^2$ must be real and positive is a separate argument which I also think is OK. It’s a simple proof by contradiction. If $\omega^2$ is not real and positive, then $\omega$ must have an imaginary part, $i \alpha$, where $\alpha$ is real. Then each $x_k = A_k e^{i \omega t}$ would have the same exponential factor $e^{-\alpha t}$. Thus, each $\dot x_k$ would also have the factor $e^{- \alpha t}$. By inspection of the forms of $T$ and $U$ (and keeping in mind that they are positive definite) you can see that this would imply that the total energy $E$ cannot be constant in time. But $E$ must be conserved for this system as shown earlier in the text in section 6, pages 13-15. So, the assumption that $\omega^2$ is not real and positive leads to a contradiction. Therefore, $\omega^2$ must be real and positive.

I don’t see anything wrong with this line of reasoning.

 

[Demystifier]

If $\omega^2$ is not real and positive, then $\omega$ must have an imaginary part, $i \alpha$, where $\alpha$ is real. Then each $x_k = A_k e^{i \omega t}$ would have the same exponential factor $e^{-\alpha t}$.

I don't think it's a valid argument because it is not clear why each $x_k$ needs to have the same exponential factor. It seems much more natural to consider $\omega_k$ with the imaginary part $i \alpha_k$, so that each $x_k = A_k e^{i \omega_k t}$ has a different exponential factor $e^{-\alpha_k t}$.

 

[TSny]

I don't think it's a valid argument because it is not clear why each $x_k$ needs to have the same exponential factor. It seems much more natural to consider $\omega_k$ with the imaginary part $i \alpha_k$, so that each $x_k = A_k e^{i \omega_k t}$ has a different exponential factor $e^{-\alpha_k t}$.

Seeking particular solutions where each $x_k$ has the same exponential factor $e^{i\omega t}$ is standard fare. (For example, see Goldstein’s text.) It can then be shown that any solution of the equations of motion can be expressed as a linear combination of these particular solutions.

 

[Demystifier]

Seeking particular solutions where each $x_k$ has the same exponential factor $e^{i\omega t}$ is standard fare.

Not really. Consider, for instance, two degrees of freedom ($k=1,2$) without a coupling between $x_1$ and $x_2$. In practice one usually starts with the ansatz $x_k(t)=A_ke^{i\omega t}$. But then from two differential equations for $x_1$ and $x_2$ one obtains a quadratic equation for $\omega^2$ with two solutions $\omega^2_1$ and $\omega^2_2$. Then, when one turns $\omega^2_1$ back into the differential equations, one finds that it is a solution only if $A_2=0$. Similarly, $\omega^2_2$ gives a solution only if $A_1=0$. Hence the actual solution is
$$x_1(t)=A_1e^{i\omega_1 t}, \;\;\; x_2(t)=A_2e^{i\omega_2 t}$$
which means that the two oscillators never oscillate with the same frequency. There is no solution of the form $x_1(t)=e^{i\omega t}$, $x_2(t)=e^{i\omega t}$ with the same $\omega$.