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Semantics of Wave 
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#1
Dec2007, 10:51 PM

P: 81

Hi all,
I have two questions related to the use of the word "wave"...and I would like know whether this actually represents a physical wave nature. #1 The "wave"function. This is what little I think I know about the wavefunction...it represents certain values about the subject (i.e. electron quantum numbers) and how they evolve with time, and the amplitude squared represents the probability of these values occurring. This of course may be wrong. Now I was wondering how the wavefunction is actually related to a sinusoidal wave. I don't think it means that subject travels along a wavelike path (could someone confirm this please)  but is the shape of the wavefunction on a graph actually sinusoidal, or gaussian? As it is related to probability, I would have said it was Gaussian, and this seems confusing...because it is then not exactly a "wave". #2 De Broglie wavelength. So all I know about this is that it implies all matter has a specific wavelength, related to it's momentum. I was wondering again how to interpret this. Does it mean that the mass actually "wiggles" along, travelling a sinusoidal path through spacetime (again a yes/no here would be helpful)? Or is it again somewhat like the wavefunction above, related to probabilities? Or is it another way of putting Heisenburgs Uncertainty principle...I thought of this possibility after reading the following from http://en.wikipedia.org/wiki/Wave%E2...rticle_duality Thanks, Kcodon 


#2
Dec2107, 05:59 AM

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As for the actual shape of the wave function, this very much depends on the system that the wave function is describing and furthermore, the specific state that the system is in. For example, the wave functions of a particle confined to a box (potential well) are generally sinusoidal; however, the wave function of a harmonic oscillator (e.g. diatomic molecule) can be Gaussian. To describe a localised particle, that is a particle which is restricted to some finite region in space, we must make use of quantum wave packets, which are analogous to classical wave packets. To construct a wave packet we must integrate over all possible values of the wave vector k, which can be related back to the momentum of the particle. Since we are integrating over many values of k, we do not have a definite value of momentum to assign the particle and hence, although we have reduced our uncertainty in the position of the particle (by localising it), we have increased the uncertainty in the momentum of the particle. I hope that make sense and apologise if it's a bit verbose in parts. 


#3
Dec2107, 04:28 PM

P: 81

Thanks for the in depth reply Hootenanny!
My next question stems from the answer to the previous one, but if you have a sinusoidal wavefunction...what is oscillating? Lets take a photon for example. When treated as a wave, the sinusoidal nature represents the oscillating EM field. If your particle in a box has a sinusoidal nature, what does this imply? Changing from a particle to an antiparticle (lol I know this isn't true)? Or does it indicate the probability of being at that point (for position in this case, opposed to whole quantum state), is changing from zero to a maximum? Again, somewhat confused. And with de Broglie: Thanks again for your help, Kcodon 


#4
Dec2207, 05:29 AM

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Semantics of Wave
[tex]\psi_0(x) = A_0\exp\left(\frac{x^2}{2\sigma^2}\right)[/tex] So, if we plot [itex]\psi(x)[/itex] against [itex]x[/itex] we obtain a Gaussian curve (see this figure). To reiterate, the wave function doesn't have a physical observable and hence the oscillations of the wave function don't have any physical significance (since the wave function is complexvalued). I hope that answered your first two questions. Hyperphysics  further reading and excellent pictorial representations. [tex]\Psi(x,t) = Ae^{i\left(kx\omega t\right)}[/tex] And the us find the probability density; [tex]P = \Psi\cdot\bar{\Psi} = Ae^{i\left(kx\omega t\right)}\cdot Ae^{i\left(kx\omega t\right)}[/tex] [tex]P = A^2[/tex] Hence, the probability density is constant throughout all space and is independent of the wave vector k and hence the momentum. So to conclude it is the width of the wave packet that determines the uncertainty in position, rather than the wavelength. [tex]\hrule[/tex] ^{(1)}It should be stressed that the probability amplitude is not equivalent to the probability density. The probability amplitudes are simply the values of the wave function at some position in space and are therefore complex values and as such have no physical observables. 


#5
Dec2207, 04:29 PM

P: 81

Thanks, Kcodon 


#6
Dec2307, 05:29 AM

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To simplify things, lets step away from the harmonic oscillator and stick to a particle in a box. Now classically, if you put a single particle in a sealed box away from any other influences you know whats going to happen. The particle will travel with a uniform velocity until it collides with a the wall of the box, in which case it will bounce off the wall (accelerate) and then proceed with uniform motion once again. Therefore, there would be equal probability to find the particle anywhere in the box, it is equally probably to find the particle at any point in the box. However, if we use quantum mechanics to describe the 'particle in a box' system we find that the system doesn't behave as classically predicted. For a particle in a bound state (localised), the probability density will oscillate as a function of position and there will be points where the probability amplitude vanishes (nodes) and points where it is maximal (antinodes). How do we know this? We know this because when we solve the Schrödinger equation for a bound state, the solution we obtain does indeed oscillate. There is not classical explanation for this nor any intuitive explanation of why this is case, it is a purely quantum mechanical effect. There are some 'analogies' I've seen (and indeed, was taught at undergraduate level) but they tend to only confuse the matter further. Let us now take a concrete example so that you can see how we determine the probability density from a wave function. Let us take the example of the onedimensional case of a particle in a infinite potential well (particle in a box) of width a. In this case the Schrödinger equation has solutions of the form; [tex]\psi_n(x) = \sqrt{\frac{2}{a}}\sin\left\{\frac{n\pi}{a}\left(x+\frac{a}{2}\right)\r ight\}[/tex] Now we find the probability density; [tex]P_n(x) = \psi_n(x)\cdot\overline{\psi_n}(x) = \psi_n^2(x)[/tex] [tex]P_n(x) = \frac{2}{a}\sin^2\left\{\frac{n\pi}{a}\left(x+\frac{a}{2}\right)\right\ }[/tex] So, we have essentially squared the wave function to obtain the probability density, which effectively squares the amplitudes and reflects the portion of the curve below the xaxis in the xaxis. Hence, the probability density still oscillates, but is positive and has a greater amplitude than the wave function. Although you may not be satisfied with the explanation, hopefully now you can understand why (mathematically at least) the probability density of a localised particle oscillates. I should also mention that the solution for a free particle (post #4) is not a true wave function, since a requirement for a valid wave function is that it should be square integrable^{(2)}. Hence, the solution given in post #4 is unphysical and the reason for this relates to my comment earlier; [tex]\hrule[/tex] (2)If a function is square integrable over some interval, then the integral of the square of it's absolute value over that interval must be finite. In the case of a free particle the interval in question would be [itex](\infty,\infty)[/itex]. 


#7
Dec2307, 03:07 PM

P: 81

However, I have one last question to do with this probability oscillation. With your particle in a box example, you get the oscillating probability density. I must assume the placement of nodes and antinodes is independent of your point of reference for position? I.e. the wavefunction will remain in the same place relative to the walls of the box, even if you take the position values from varying reference points  outside the box? Otherwise the wavefunction would move relative to the box, due to moving reference points; which, from the particles point of view, should have nothing to do with it or its placement of the wavefunction in the box. Just looking over the equation you gave previously, I just saw the value a, as the width of the well. Therefore I assume position is measured relative to the box, so I assume my prior question is somewhat irrelevant, as if reference is the box, then there should be no problem. Thanks again, Kcodon 


#8
Dec2407, 04:44 AM

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[tex] \begin{picture}(200,200) \put(100,15){\vector(1,0){100}} \put(100,15){\vector(1,0){100}} \put(100,12){\line(0,1){3}} \put(170,12){\line(0,1){3}} \put(30,12){\line(0,1){3}} \put(99,0){0} \put(155,0){+a/2} \put(15,0){a/2} \color{blue} \put(30,15){\vector(0,1){100}} \put(170,15){\vector(0,1){100}} \put(160,120){$\infty$} \put(20,120){$\infty$} \end{picture} [/tex] Of course you can define your coordinate system however you like, but a symmetric system usually makes life simpler. Writing the solutions to the Schrödinger equation in that form isn't entirely correct. In this case, our potential function (V) is defined piecewise thus; [tex]V(x) = \left\{ \begin{array}{cr} \infty & \leftx\right > a/2 \\ 0 & \leftx\right \leq a/2 \end{array}\right.[/tex] Which in words means, the potential energy of the system goes to infinity if the distance from the origin is greater than a/2, otherwise, the potential is equal to zero. This restricts our particle to exist in a 'box' of width a. Equally we must define our wave function (ψ_{n}) in a similar fashion since we have two distinct cases; the case where we have zero potential, and the case where our potential tends to infinity. Hence, we write the wave function for a particle of mass m in our onedimensional potential well thus; [tex]\psi_n(x) = \left\{ \begin{array}{cr} 0 & \leftx\right > a/2 \\ \sqrt{\frac{2}{a}}\sin\left\{\frac{n\pi}{a}\left(x +\frac{a}{2}\right)\right\} & \leftx\right \leq a/2 \end{array}\right.[/tex] So you see in actual fact, the wave function of a particle in a box is finite, it only exists inside the box. Hopefully that will put your mind at rest in terms of the construction of wave packets and finite wave functions. Incidentally, something you may find interesting is if we consider a finite potential well, that is similar to the case above but where the potential energy function terminates at some finite value. We find that a particle with a kinetic energy that is less than the potential energy of the well has some nonzero probability to be found inside the walls of the potential well! This is forbidden classically and is the basis for Quantum Tunneling. I hope I managed to clear everything up for you, if I didn't I'm sure you'll be back. In the meantime have a very merry Christmas. Regards, Hootenanny 


#9
Dec2407, 03:53 PM

P: 81

Ok thanks for going through the simple maths of it, I can now see how the wavepacket is finite, of width a, and although the waves forming it (differing values of n) are infinite sine waves, the wave packet itself is only defined for within the potential well. So thats good! However I still do have some questions still... = ) #1 Well firstly I think that this means the wavepacket has width a, so this represents the uncertainty in position (as a side question, I'm guessing the "main" packet is not a wide, so the width of the "main" packet does not indicate uncertainty in position, but actually the width of the whole wave packet). Anyway this would make classical sense; as the potential well and a gets bigger, so does the uncertainty in position. So thats whats I am assuming, however my question is: what happens as a tends to infinity  mathematically? I'm guessing, that by HUP, a and thus the uncertainty in position, approaching infinity would result in an exactly determined momentum. In Hyperphysics pages this means a perfect sine wave, so I'm guessing as the value of a approaches infinity, the n value in the equations become negligible, so that all the waves produced as solutions of Schrodingers equation are the same sine wave (or conversely ones with same period) so as to produce a final wave packet that is a perfect sine wave, i.e. defined momentum. I wonder if the actual mathematical solution for when a approaches infinity, agrees with what I've stated before?? #2 I just noticed the n in the equation. I assume the differing integer values of n give the differing sine waves that are added together to make the resultant wave packet (as referred to in Hyperphysics)? However I was wondering what the value for n actually represents? I don't believe it is just to represent a general solution to a trigonometric equation...? Anyways thanks, and you have a great Christmas too, Kcodon 


#10
Dec2407, 07:16 PM

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If we now go back to our particle in a box and examine our general solution together with the energy eigenvalues (just considering the case where [itex]\leftx\right \leq a/2[/itex]); [tex]\psi_n(x) = \sqrt{\frac{2}{a}}\sin\left\{\frac{n\pi}{a}\left(x +\frac{a}{2}\right)\right\} \hspace{5cm} E_n = \frac{h^2}{8ma^2}n^2[/tex] And in this case, [itex]n\in\mathbb{Z}^+[/itex], that is n must be a positive integer. Hence, we can start writing our our energy eigenstates; n=1 [tex]\psi_1(x) = \sqrt{\frac{2}{a}}\sin\left\{\frac{\pi}{a}\left(x +\frac{a}{2}\right)\right\} \hspace{5cm} E_1 = \frac{h^2}{8ma^2}[/tex] n=2 [tex]\psi_2(x) = \sqrt{\frac{2}{a}}\sin\left\{\frac{2\pi}{a}\left(x +\frac{a}{2}\right)\right\} \hspace{5cm} E_2 = \frac{h^2}{2ma^2}[/tex] And so on. You can see a visual representation of the solutions here. So rather than n representing combinations of several waveforms into a single solution (wave packet), it represents individual discrete solutions. I hope that makes more sense to you. While we're here we may as well discuss some consequences of the above solutions. Firstly, you should observe that we have quantised energy eigenvalues, the particle is only 'allowed' to have certain energies. For example, the particle can have an energy equivalent to E_{1} or E_{2} (or E_{3,4,5,...}), but can't have anything in between. We say the particle has a discrete energy spectrum, which is in stark contrast to classical physics, were the energy spectrum is continuous. Secondly, note that our lowest permitted energy state (i.e. the energy eigenstate corresponding to n=1) is nonzero, this phenomenon is known as zeropoint energy and I'm sure you've at least heard of it before. We can understand this phenomenon qualitatively in terms of the HUP, which states that the product in the uncertainty in two measurements must be of the order of [itex]\hbar[/itex]. However, if a particle has zero energy it will be at rest, and therefore it will have a uniquely define momentum (zero) and position, thus violating HUP. We can take this concept further and say that the particle in the infinite potential well of width a is restricted to [itex]x\leq a/2[/itex], and hence has an associated uncertainty in position of [itex]\Delta x \approx a[/itex] (we know that the particle must be somewhere in the well, but we don't know where). Hence, we can write; [tex]\Delta x \cdot \Delta p \approx \hbar \Rightarrow \Delta p \approx \frac{\hbar}{a}\hspace{5cm}(1)[/tex] Furthermore, we know that kinetic energy is related to momentum thus, [itex]E = p^2/2m[/itex], hence we can write; [tex]\Delta E \approx \frac{\hbar^2}{2ma^2} = \frac{h^2}{8ma^2}\pi^2[/tex] Which is 'qualitatively' in agreement with our first energy eigenvalue E_{1}. Note that although this is a very 'rough and ready' analysis, a more formal treatment can show that the associated uncertainty in the energy is in exact agreement with the energy eigenvalues. Furthermore if we examine equation (1), we find that the uncertainty in momentum is inversely proportional to the width (a) of the well, which intuitively makes sense. If we reduce the the width of the well (as a approaches zero), we are increasing the spatial localisation of the particle and hence, decreasing the uncertainty in position. Therefore, by HUP we would expect the uncertainty in momentum to increase. Conversely, if we increase the size of the well (a approaches infinity), the particle becomes less localised and behaves more like a free particle, hence the uncertainty in momentum approaches zero. I think that, partially at least, answers your first question. Well at the outset I intended that to be quite a concise post, but it seems that it ran away with me a little. I apologise if it seemed a little hard going. 


#11
Dec2407, 11:19 PM

P: 81

I am assuming then that one must know the energy eigenstate the particle is in, otherwise one would not know which solution to the Schrodinger equation to apply...? Lets clear this up too...is the wave function the same as the "wave packet" which is the same as a particular solution to the Schrodinger equation? Hmmm Fourier analysis is representing a function as the sum of sinusoidal terms, right? So in the case of the wave packet  wave function  in the Hyperphysics page, how it referred to adding waves together to get the wave packet, do these waves actually have a physical meaning (like the differing n values I implied, although as you have said, this is wrong), or are they simply the basis function in the fourier analysis? Anyways quantised energy...thats an interesting proposition. I've heard it before, but thanks for showing me the maths to prove it...well to some degree anyway. Also yes, that answers my first point. You've agreed with me on that one, so that problems settled! The zero point energy is interesting...I have heard of it briefly before. I think applying this with quantum field theory results in the so called "Vacuum energy". Not sure my classical side really likes the idea. As a completely off topic bit of info, while reading this page from Wikipedia http://en.wikipedia.org/wiki/Vacuum_energy I found the following: Anyways thanks again for your replies, Kcodon 


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