Semantics of "Wave" 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%80%93particle_duality Any ideas are welcome, Thanks, Kcodon
There a number of 'interpretations' of quantum mechanics (summarised here), all with differing idea's of what the wave function actually is. However, for the purposes of this post I will refer to the Copenhagen [non-real wave function] Interpretation since this is generally, the most widely accepted (and taught) interpretation, although the 'Many Worlds' interpretation is gaining ground. Roughly speaking, the wave function is just a complex function or a 'mathematical abstraction' or tool used to describe the state of a physical system. In other words, the wave function itself has no physical observables. So, whereas classically the wave function of a vibrating string describes the periodic variation of real physical observables (amplitude etc.) there is no such corresponding observable for a quantum mechanical wave-function. Therefore, the wave function of a particle tells us nothing of how it actually travels. 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. The de Broglie hypothesis states that all particles have a wave-like nature (wave-particle duality). The de Broglie hypothesis does not say anything about the wave function of a particle only that a particle can be describe by a classical wave of angular frequency [itex]\omega = E/\hbar[/itex]. However, the de Broglie hypothesis can be used to find the wave function of a 'free particle' (via the dispersion relation [itex]\omega =k^2\habr/2m[/itex]), that is a particle that has a non-zero constant probability to be over all space. As you say, this is related to the Heisenberg Uncertainty Principle, since to apply the dispersion relation we must fix the momentum of the particle and therefore by HUP, the uncertainty in the position of the particle approaches infinity. 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.
Thanks for the in depth reply Hootenanny! Good, good. Ok now I am somewhat confused. You say the wavefunction represents the physical state of a system, however this is not a variable...is it not comprised of many variables...position, momentum, spin etc? So when you say the shape of the wavefunction can be a certain shape...in what way have you obtained this shape? If it is drawn on the graph, then what are the two axis variables? I don't believe you can plot quantum physical state on the y, and time on the x... 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: I feel this is true in the sense that photons have wave-like nature, but it does not mean the photon wiggles along through space. So I don't think the masses wiggle through space, even though they have wave-like nature. Ok moving further with this idea then, would HUP and duality pretty much be same thing? If one knows the momentum of something, its position is undefined, thus a wave. If one knows the position, it is a particle (with undefined momentum). So in a sense could not wave-like nature simply represent uncertainties in position? With respect to de Broglie wavelength...bigger something is (more momentum) means smaller wavelength...i.e. a more defined position, as one would expect by HUP and intuition. Thanks again for your help, Kcodon
To determine the wave function for a particular system you must solve the Schrödinger equation (SE) for that particular system. When you solve the SE you will obtain a set of allowed energy states (Eigenvalues) and the wave function (Eigenfunction) for that system, it is the quantum numbers (n,l,m etc.) that determines the state, and hence the energy, of the system. In the one-dimensional, time-independent case (the state of the system has no time evolution) the wave function is a function of a single variable (position), [itex]\psi = \psi(x)[/itex]. Now, a wave function that is purely a function of position is known as a probability amplitude^{(1)}, and the values of this function are probability amplitudes. It is these values that we plot, in other words when we say that a wave function is a certain shape (sinusoidal, Gaussian etc.), we mean that when we plot the values of the wave function against position we obtain a certain shape. So the oscillations represent how the probability amplitude varies with position. For example, the wave function for a particle undergoing one-dimensional simple harmonic oscillations in the ground state is given by; [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 complex-valued). I hope that answered your first two questions. Indeed I think you've got the general idea. However, in general you should note that there will be some uncertainty in both position and momentum of a particle described by quantum mechanics. For example, as I said in my previous post we can construct a wave packet by integrating of a range of values ([itex]\Delta k[/itex]) of the wave vector k. A wake packet is characterised by a zero probability amplitude over all space except for a small region [itex]\Delta x[/itex]. According to de Broglie the spread in the wave vector k results in a spread of momentum [itex]\Delta p_x[/itex]. Hyperphysics - further reading and excellent pictorial representations. I think you're confusing the issue a little here. The actual value of the momentum of a particle says nothing about the uncertainty in either position nor momentum, a smaller wavelength does not mean less uncertainty in position, the uncertainty will remain unchanged. For example, let us take the [1D] free particle solution; [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.
Ok thats kind of what I expected about the position and time etc, after reading a bit on probability and the like but I'm still confused with the sinusoidal wave nature of the wavefunction. When we square the wavefunction, for the case you are referring too, this then gives the probability for finding the particle at that certain position. But how come the probability oscillates from zero to a maximum and back again? For example in the link you gave previously... http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/imgqua/hoscom2.gif I would expect the shape of the bottom graph, but not the graphs above it. How come these sinusoidal ones implie that there is zero chance of finding a particle at some positions and a maximum at others? I would have thought the probability of finding a particle would be shaped like the bottom graph...? Sorry I can't really put my problem into words...I hope you know what I'm asking. Wow ok thats really neat...I totally like thought of that idea myself and you (and the hyperphysics page) just confirmed it! That is really cool!! Ok now after reading the hyperphysics page, and your explanation I am somewhat confused. In the Hyperphysics page, it shows that summing some waves together (is this to simulate the uncertainties in momentum?) will give an interference pattern - the "wavepacket", and its width represents the uncertainty of position. However these waves were infinite weren't they, as they represented exact momentums? Therefore should not the intereference pattern repeat itself, and it too be infinite...thus giving infinitely many wavepackets, and thus giving the particle an infinitely undefined position? Thanks, Kcodon
From what I can gather, you believe that the probability density of a localised particle (i.e. Harmonic Oscillator, particle in a box etc.) should be Gaussian and you can't see why it would be otherwise? Furthermore, you can't understand how when we find the probability density from a wave function, the probability density oscillates. If I'm wrong, please correct me. 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 (anti-nodes). 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 one-dimensional 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)\right\}[/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 x-axis in the x-axis. 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 apologies for confusing you, I will attempt to qualitatively clarify the idea of wave packets here. Firstly, the reason we construct a wave packet is not to simulate the uncertainty in momentum, we do so to localise the wave function (and hence the particle) into some small region Δx. The resultant uncertainty in momentum is a necessary 'bi-product' if you like of this process. In addition, we can't just 'choose' any old waves to superimpose, the wave packet must be constructed from the eigenfunctions of the system, that is the waves must be solutions of the Schrödinger equation. As for your final point, yes, when the waves 'interfere' they will generate more than one 'wave packet'. However, the probability amplitudes of these 'secondary wave packets' will be small compared to the probability amplitude of the 'primary wave packet', analogous to the amplitudes of the diffraction pattern observed for the double slit experiment. Furthermore, the probability amplitudes rapidly decrease away from the center of the wave packet. Take note that wave packets are difficult to accurately explain without going into the mathematics, and the above is a 'rough and ready guide'; but I hope I managed to answer your questions. If you would like to study wave packets more rigorously I can recommend some texts, but beware that the mathematics required is not trivial. 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; In other words, a free particle cannot have an exactly defined momentum, this implies that there must be sum uncertainty in the momentum of the particle and hence sum spread in the wave vector k. Therefore, physically acceptable solutions take the form of a wave packet. [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].
That is correct Yup, you figured out my problem. I was thinking like your classical example...there should be an equal probability of finding the particle anywhere in the box. I'm glad you pointed out then that there is no intuitive explanation for the probability osciallation, because thats what I was looking for. And unfortunately it seems that QM can't explain it, but I'm betting that in experiments QM will predict observations correctly. Thanks for this...normally I just pass over the maths because its way over my head, but this I can follow. QM is a mathematical model, so I suppose looking at the mathematics occasionally will probably help me understand it . 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 anti-nodes 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. I see what you are saying about localizing the position by adding the waves, but this must only apply if the waves are finite...i.e. already have a somewhat localised position. Ah, this must be where this comes in: Previously I was thinking these waves that you add together were infinite. Adding simple sine waves like in the Hyperphysics example, would not - I don't believe - give wavepackets that diminished, similiar to diffraction patterns. There would be some point where they would all be in phase again and the process would repeat itself all over again. However if these waves are not infinite (as have some certainty in position), then I can see how this could happen. However, then again, looking over the Schrodinger solution you showed me above, it seems to indicate that the wave is infinite...a normal sine wave, so maybe my train of thought is completely wrong. Apart from this little point on adding the waves, the answer to the following would be yes...thanks a bunch. I suspect that in the real world, one does not actually add the waves together...there is some other mathematical process that gives the wave packets (that are also diminishing). However as you said, the maths would be difficult, in which case I will probably avoid those books you refer to until my maths can keep up. Thanks again, Kcodon
Indeed, since Physics is based on Mathematics I firmly believe that one can never truly understand a phenomenon until one can follow the Mathematics. One may be able to get the general idea from a qualitative analysis, but there will often be observations that at first seem counter-intuitive, it is only when one follows through the mathematics that it becomes clear. As for QM not being able to explain oscillating probability densities, it can mathematically, it follows directly from the postulates of quantum mechanics. I shall try and answer all of the above in one fell swoop. You are entirely correct, we define our system relative to the box. So in the case of our one-dimensional infinite potential well x=0 is in the middle and at the bottom of the well and the two walls are located at x = +/- (a/2) as shown here; [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 & \left|x\right| > a/2 \\ 0 & \left|x\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 one-dimensional potential well thus; [tex]\psi_n(x) = \left\{ \begin{array}{cr} 0 & \left|x\right| > a/2 \\ \sqrt{\frac{2}{a}}\sin\left\{\frac{n\pi}{a}\left(x +\frac{a}{2}\right)\right\} & \left|x\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 non-zero probability to be found inside the walls of the potential well! This is forbidden classically and is the basis for Quantum Tunneling. The mathematical technique used to construct wave packets are called Fourier Transformations, which are part of the more general area of Fourier Analysis. If your looking to study Quantum Mechanics at any sort to depth, I would recommend taking courses in Linear Algebra and Analysis, specifically Functional Analysis. 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
I suppose I best keep taking mathematics then, in an attempt to further understand QM! Firstly I'm not sure if you're a teacher or not already, but you should consider the profession! Very helpful replies! 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...? Yes I'm considering whether to try and get into Cambridge, and take the natural sciences course. However for first year, you can do a course that is something like the mathematics involved in physics instead...and then progress to the natural sciences in 2nd year. I'm betting this would be helpful!! Anyways thanks, and you have a great Christmas too, Kcodon
Thank you for the kind words Okay, I'm going to address your second question and hopefully this will make sense of your first question. First and foremost, the solution that I gave in post #6 is not a wave packet, it is a pure sinusoidal wave. The solution to the infinite potential well does not require the use of wave packets and can be obtained trivially by directly solving the Schrödinger equation. It is a pure sinusoidal wave, with no interference or summations. The n in the solution does not refer to any summations or integrations, it simply defines the energy eigenstates of the system. The energy eigenstates for a particular system are the set of eigenvalues (energies) and eigenfunctions (wave functions) that satisfy the [time independent] Schrödinger equation. Perhaps it would have been prudent to mention it soon, but in general, there isn't just one single solution that satisfies the Schrödinger equation for each system, there can be infinitely many solutions. This set of solutions are the energy eigenstates for that particular system. 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]\left|x\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 non-zero, this phenomenon is known as zero-point 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. A good friend of mine took natural sciences at Cambridge and he had very good things to say about it. From what he said the course sounded interesting and apparently, after your first year, you have virtually free choice over which modules you take. It's good to be interested in Quantum, but I wouldn't worry about understanding it all yet, any Quantum Mechanics you take in your first two years as an undergraduate will in all probability be fairly superficial. That said, there's no harm in getting ahead of the game, especially if your interested in the subject! 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.
Ok these two confused me somewhat...in the first quote I read it as meaning the wave packet was constructed from the various solutions to the Schrodinger equation, which I then thought meant due to the differing values of n. I thought adding these waves together gave the wavepacket. However the second quote seems to go against this, and suggest the solutions to the Schrodinger equation is the wavepacket (which in this case is just sinusoidal)? Unless a wavepacket is only defined for each energy eigenstate... i.e. each n value gives a different wavepacket. In this case the solution to the Schrodinger equation is just the sinusoidal wavepacket? Actually thinking about it logically, you can't have a wavepacket for all n values, otherwise when added together it would be a mess, so to speak...as n goes to infinity. So then I'm guessing you can have a wave function to represent one variable (in this case position), however it only applies to certain eigenstates/eigenvalues...i.e. one n value? 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? Thanks for this...I now know the role of the n value much more clearly! The relation to energy eigenstates is also neat...I assume that again one must know the energy eigenstate the system is in, for the wave function to be of any use?? Am I correct to assume that HUP still applies to energy, as momentum and energy are directly proportional? However this would then imply one could not know the exact energy eigenstate the system is in, so I am thus making contradictory assumptions!! 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: It's somewhat reminiscent of aether isn't it? Anyways thanks again for your replies, Kcodon