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Basic QM question: Why do we describe free particles using Aexp(ikx) ? 
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#1
Sep908, 01:11 PM

P: 9

If you solve the 1D time independent schrodinger equation for a system with V(x)=0, you get a solution like:
Ψ(x) = Aexp(ikx) + Bexp(ikx) Applying certain boundary conditions would give you the solution to the Simple harmonic oscillator or infinite well etc. But I often see the wavefunction for a free particle written as Ψ(x) = Aexp(ikx) So is there some boundary condition which causes B=0 ? I have tried the periodic BC Ψ(0) = Ψ(L) and dΨ/dx (x=0) = dΨ/dx (x=L) but this only seems to give me certain values of k, it does nothing with A or B. I see that each term represents a wave in each direction, but I dont see how the B is lost if we start with the Schrodinger equation. I am under the impression that periodic boundary conditions result in the solution Ψ(x) = Aexp(ikx) but cant get rid of my B.. Thanks 


#2
Sep908, 02:25 PM

P: 242

The state [itex]\Psi (x)=Ae^{ikx}[/itex] is not only a solution to the Schrodinger Equation (the eigenvalue equation for energy), but also to the eigenvalue equation for momentum. Thus it is a state with definite energy and momentum both. It is a particle that has had its momentum measured. In particular, its momentum is k.
A and B in your solution are not determined by boundary conditions. They are arbitrary, up to the requirement that [itex]\int \Psi ^* \Psi \; dx = 1[/itex] 


#3
Sep908, 02:30 PM

P: 160

There is no reason that the free wave solution can't be written
Ψ(x) = Aexp(ikx) + Bexp(ikx). Actually the full free wave solution is (General solution to the TDSE with V=0) : [tex] \Psi(x) = Ae^{i(kx\omega t)} + Be^{i(kx+\omega t)} [/tex] The two terms just correspond to a free particle (wave) moving in opposite directions in 1D. When people write [tex] \Psi =Ae^{ikx} [/tex] k could be negative (if particle has momentum in opposite direction) This just saves writing the B term explicitley everytime. Also these are the eigenfunctions of the momentum operator, so you may have seen it in that context (again k can be positive or negative). 


#4
Sep908, 02:56 PM

P: 24

Basic QM question: Why do we describe free particles using Aexp(ikx) ?
[tex] \psi (x) \rightarrow 0[/tex] when [tex] x \rightarrow \infty[/tex]
If an exponential term, depending the area you study, is going to infinity then you have to set its amplitude = 0. In your case now: [tex] \psi = Aexp(ikx) + Bexp(ikx)[/tex] This is the wave equation for the regime [tex]( \infty , 0)[/tex]. [tex] exp(ikx) \rightarrow \infty [/tex] if [tex] x \rightarrow  \infty[/tex] So, you have to set B=0. This is because the wavefunction should not go to infinity for any x. 


#5
Sep908, 03:03 PM

P: 160

No, these are not exponential terms, they are sinusoidal: [tex] e^{ikx}=cos(kx)+isin(kx) [/tex]. So theres no requirements that any of the terms be dropped for normalization reasons. 


#6
Sep908, 04:52 PM

P: 9

ZikZak: Ok so that explains why a particle wont have both terms, because it has had its momentum measured so it must be going in one direction or the other, determined by k.
But does that mean, for the 1D SE, with V=0 and periodic boundary conditions Ψ(0) = Ψ(L) and dΨ/dx (x=0) = dΨ/dx (x=L), there is no way of getting to the Ψ(x) = Aexp(ikx) solution? Also, about BCs not determining A or B, what about in the 1D infinite well the coefficient of coskx dissappears because of BCs? h0dgey: But saying Ψ(x) = Aexp(ikx), and k can be ve or +ve is not the same as having both terms as I understand it because with both terms you can have a superposition of both, like in the 1D infinite well, where their superposition gives standing waves. So you are losing something by just having one term and saying k can be +ve or ve, I think. Xma: I think hodgey is right, complex exponentials just go round and round as x goes to infinity. 


#7
Sep908, 05:38 PM

P: 160

OK, so lets say we start with the free particle shrodinger equation then a solution for a particular value of k is [tex] \Psi(x,t) = Ae^{i(kx\omega t)}[/tex]. But in fact since the TDSE is linear the most general solution is actually a superpositon of these eigenfunctions i.e. [tex] \Psi(x, t) =\int_{\infty}^{\infty} A(k) e^{i(kx\omega t)}dk [/tex]. (since k is continuous in totally free space).
Now if you make a measurement on the momentum, that general wavefunction will collapse into an eigenfunction with definite momentum (k), and the system will be in eigenstate [tex] \Psi(x,t) = Ae^{i(kx\omega t)}[/tex]. where k can be positive or negative. But if you want you could take the limits from 0 to [tex] \infty [/tex] and express the most general state as: [tex] \Psi(x, t) =\int_{0}^{\infty} A(k) e^{i(kx\omega t)} +B(k)e^{i(kx+\omega t)}dk [/tex]. It's pretty much arbitrary. 


#8
Sep1108, 06:41 AM

P: 9

OK.
So forgetting 'free space / free particle', can we arive at Ψ(x) = Aexp(ikx) purely from the periodic BCs I wrote above? Thanks again :) 


#9
Sep1108, 07:07 AM

Sci Advisor
P: 8,788

1) What is a solution to the Schroedinger equation? 2) Why is exp(ikx) a free particle? As pointed out by others above, the reason for the second is that it is a state with definite momentum, so it just represents a particle moving with constant velocity. That any solution to the Schroedinger equation can be writtten as a superposition of momentum eigenstates is just a convenience. Just like for ordinary functions: should we use a Taylor series or a Fourier transform  it depends on the problem. 


#10
Sep1208, 06:50 AM

P: 9

My issue is the 1st of those two atyy. I guess I thought I could answer that by asking the 2nd.
I understand exp(ikx) is one with definite momentum and so definite k. Im trying to understand electrons in solids, wherebye with periodic boundary conditions as stated above the derivation seems to end in exp(ikx), yet with no explanation of where the other term in the solution from the schrodinger equation goes. 


#11
Sep1208, 09:45 AM

Sci Advisor
P: 8,788




#12
Sep1208, 10:56 AM

P: 60

Hi guys,
I think this thread is confusing me. If i get it right Dizzle is asking about the solutions of the 1D time independent schrodinger equation for a system with V(x)=0. That would be the spectrum (eigenvalues + eigenvectors). Now, on a 1D domain of length L the plane wave [tex]\psi(x) = \frac{1}{\sqrt{L}} e^{i k x}, \qquad k = \frac{2 \pi}{L} n,\qquad n \in {\mathbb Z}[/tex] is an eigenvector corresponding to the eigenvalue [tex]\epsilon_k = \hbar^2 k^2/ (2 m)[/tex]. The conditions on the [tex]k[/tex] arises from the p.b.c. Now it is easy to see that [tex]\epsilon_k=\epsilon_{k}[/tex]. This means that any combination (including Dizzle's) of the two plane waves relevant to k and k is still an eigenvector for the same eigenvalue. That is, the eigenspace is doubly degenerate. This is typical of homogeneous systems with p.b.c (actually I think it has ultimately to do with the fact that the system is so symmetric you can't tell clockwise and counterclockwise apart). Now, as already pointed out, the Hamiltonian commutes with the momentum operator, whose eigenvectors are nothing but the above plane waves. This means that you can label your eigenvectors with the "direction" of the momentum. In summary I think that ZikZak's first reply said basically everything. Actually the normalization condition can be an issue when [tex]L \to \infty[/tex], because strictly speaking you could not consider single plane waves, but only superpositions. 


#13
Sep1208, 12:05 PM

P: 60

What I'm trying to say is that for a given eigenvalue you have 2 independent solutions. You can very well decide that one solution is [tex]\psi[/tex], i.e. the one you write in your first post. But you have to recall that at the same energy you have another independent solution of the same form, [tex]\psi'[/tex] whose coefficients A' and B' are fixed by the requirement of orthogonality.
So even in this case, why [tex]\psi[/tex] instead of [tex]\psi'[/tex]? You can also choose B=0 and A' = 0, which ultimately means that you're labeling your spectrum using the momentum operator. Or, you're considering a plane wave with a definite momentum, instead of a solution of the TISE with the same energy but "indefinite" momentum (in the sense that it is not an eigenvalue of the momentum operator). I hope this helps. 


#14
Sep1608, 12:53 AM

P: 91

2. If You want "boundary" condition for interacting particle and "B=0", then i can suggest one of them: In momentum representaition: U(potential)=infinity (p<0) U(potential)=constant (p>=0) 


#15
Sep1608, 06:51 AM

P: 60

I think Dizzle's question was much more basic, especially in relation to your point 2.
It seems to me Dizzle is concerned with the solutions to the time independent Schroedinger equation in the absence of a potential, i.e. U(x)=0 [and U(p)=0]. A plane wave with a given momentum is a solution of such an equation, and in this respect there is no contradiction whatsoever in seeing [tex] \psi_+^{k}{(x)=\frac{1}{\sqrt{L}}e^{i k x},\quad \psi_^{k}(x)=\frac{1}{\sqrt{L}}e^{i k x}\quad \leftrightarrow \quad \epsilon_k=\frac{\hbar k^2}{2 m} [/tex] where the list is enumerated according to k. An equivalent list is [tex] \psi_1^{k}(x)= A_1 \psi_+^{k}(x) + B_1 \psi_^{k}(x), \qquad \psi_2^{k}(x)= A_2 \psi_+^{k}(x) + B_2 \psi_+^{k}(x) \quad \leftrightarrow \quad \epsilon_k=\frac{\hbar k^2}{2 m} [/tex] with [tex] A_j^2 + B_j^2 =1, \qquad A_1 A_2 + B_1 B_2 = 0; [/tex] My feeling is that the first listing style is usually picked up because it is undeniably cleaner. Furthermore, as previously pointed out, the items of the first list have the nice feature of being eigenvectors of the momentum operator. All of this as far as one is concerned in listing the solutions of the TISE. If these solutions are used to solve the TDSE one has to employ the formulas mentioned above, choosing the coefficients according to the initial state or some similar condition. 


#16
Sep1608, 07:11 PM

Sci Advisor
P: 1,082

Dizzle  Your question is answered in almost any book dealing with waves, E&M, quantum theory, and in many books on differential equations and mathematical physics. Second order linear differential equations have two independent solutions, and any linear combination of those solutions is also a solution. That's it  the rest is determining what combinations are appropriate for the problem at hand, including boundary conditions. This matter is so basic that it is actually discussed in some freshman physics texts.
Regards, Reilly Atkinson 


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