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Struggeling with quantum mechanics

  1. Apr 4, 2013 #1
    Hi. I'm studying electronics at uni and like every other engineering degree, we have to learn a little physics and chemistry. The chemistry feels almost too easy for me and the thermodynamics was not a problem either. But then the lecturer began talking about quantum mechanics (a subject that isn't even 1/6 of the curriculum) and for the first time in my life I sat there not understanding a single thing. I attended every lecture and I've read through the chapter 1.5 times, but I'm still completely in the dark. I often get a feeling that I understand what I'm reading, but then I look at some questions and realize that I still don't have a single clue. I have a mandatory assignment that's due to Monday, and I'm really worried that I will fail the whole class because of this.

    We are working with particles in a one dimensional plane (for simplicity), either free particles or particles affected by a force (particles with a potential energy). I have understood that the ψ (x,t) function is pretty central as it describes the wave's velocity, wavelength etc. If I have gotten it correctly, (ψ(x,t))^2 gives the probability of finding a particle at a given place at a given time. This is something I don't understand though. The book talks about normalization of the wave function so that the integral from -∞ to +∞ would equal exactly 1 (100% chance of finding the particle as the particle has to exist somewhere). If this is not true, the wave function is said to not be normalized. I then get a question where five wave functions are represented and I am to figure out which of them are normalized and not. The functions are as follows: Ae^-x, Acos(x), Ae^x (0≤x≤L. 0 for all other x), A (-L≤x≤L. 0 for all other x), Ax(x≤L. 0 for all other x). I have no idea on how to interpret this. If you make an integral from -∞ to +∞ for any of these functions you will get ∞ in return! Also, why is it telling me that the value of x is either more or less than L when L isn't given? L can be anywhere then, thus giving the possibility for the integral to be infinite as well. Lastly, how can a plain constant like 'A' be a function? Would be stupid to try to integrate that :S

    This was the very first question I came a cross, and I am completely lost. This is the most demotivating thing I have ever encountered in any school I've been to :(
     
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  3. Apr 4, 2013 #2

    kith

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    Welcome to Physics Forums, Plecto! Cheer up, QM may be a bit more formal than what you have encountered so far, but the basics are not difficult.

    Almost. Actually, it is |ψ(x,t)|^2 because ψ is complex and probabilities have to be real. Also strictly speaking, the probability to find the particle at a single point x is zero. The correct statement would be "|ψ(x,t)|^2 dx is the probability to find it in the interval [x,x+dx]". But that's probably not so important for your case.

    The idea is that some of these functions are non-zero only in a small interval which is bounded by L, so the integral is not infinite for them. L is an arbitrary real number (if your problem doesn't state this, it is a bit sloppy here). The integral and the normalization condition will depend on L.

    Don't you know a function who yields A if you differentiate it wrt x?
     
  4. Apr 4, 2013 #3

    DrClaude

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    Hi Plecto, wecome to PF!

    Quantum mechanics is fascinating, but the learning curve can be steep! Hang in there and you will get a glimpse of a jewel of our understanding of the natural world.

    The wave function completely describes the physical system. If you know the wave function of a particle, you know all that there is to know about it.

    Be careful, the wave function is complex, so you have to take the absolute value: ##\left| \psi(x,t) \right|^2##.

    Have you actually tried to integrate ##\left| \psi(x,t) \right|^2##? And are you sure that you must say of they are normalized or is it normalizable? If you don't know the value of A, you cannot tell if it is normalized.

    You can always write it as a function of ##L##.

    Mathematically, it actually makes no sense to set ##L = \infty##. You calculate the limit as ##L \rightarrow \infty##, but that is not what is asked for here. Just take ##L## to be finite.

    Why is it not a function? What is wrong with me writing ##f(x) = A##? And integration of function is usually so hard, what not just be glad that this function is trivial to integrate. (Again, I get the feeling that you havent even tried these integrations.)

    Hope this helps.
     
  5. Apr 4, 2013 #4
    Thank you for the quick responses. I took a picture of the problem mentioned, perhaps it's easier to understand:

    http://i46.tinypic.com/33vyn2f.jpg

    I didn't try to integrate them because I wanted to just plot them first to see how they looked like. Like the second one, A cos(x), plotting (A cos(x))^2 won't give me an instance where the area under the graph adds up to exactly 1. I know it depends on the value of A, but the area will still not be 1. The same goes for plotting (Ae^-x)^2, (Ae^x)^2, A^2 and (Ax)^2. Am I supposed to see if the function can be normalized if I chose L to be a certain calculated value?

    The math is also an issue here. I had integration and differentiation at school, but that's six years a go now. I also learned about complex numbers, trigonometry, differential equations and matrix calculations when trying to get another degree at uni, but this is three years a go now and I only got a D on that exam :(
     
  6. Apr 4, 2013 #5

    jtbell

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    Consider the one function in my quote. Hint:

    $$\int^{+\infty}_{-\infty} {|\psi(x)|^2 dx} = \int^{0}_{-\infty} {|\psi(x)|^2 dx}
    + \int^{L}_{0} {|\psi(x)|^2 dx} + \int^{+\infty}_{L} {|\psi(x)|^2 dx}$$

    What is ##|\psi(x)|^2## in the three different regions? What does that tell you about the first and third "sections" of the integral?
     
  7. Apr 4, 2013 #6

    DrClaude

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    Sorry to say this, but you won't be able to solve that problem if you don't perform the integrations.
     
  8. Apr 4, 2013 #7

    kith

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    No, a function is normalizeable if the integral is smaller than infinity. They can either be normalized for all values of L or for no value of L. The process of normalizing is explained here: http://en.wikipedia.org/wiki/Normalizing_constant
     
  9. Apr 4, 2013 #8

    jtbell

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    Not quite. You're supposed to assume that L is some definite (although unknown) value to start with. Then you decide whether it's possible (in principle) to choose a value of A that makes the integral equal to 1. You should be able to do this by looking at a graph of ##|\psi(x)^2|##. Make sure the graph takes into account the regions where ##\psi(x)## is defined to be zero.

    Of course, the value of A will depend on the value of L. For example, in the "infinite square well" a.k.a. "particle in a box" problem which your textbook surely discusses, you get ##A = \sqrt{2/L}##.
     
  10. Apr 4, 2013 #9
    I think I understand, but I feel that I'm still getting something wrong. If L were given to me, I'm supposed to calculate A so that the function becomes normalized regardless of the value of L, right? I think I can make that happen for every one of the functions given :S I don't understand what it means by defining for which values of x this accounts for though, as in "0≤x≤L 0, for all other x". x is a variable, why is it defining limits for it? Is it just saying that L is bigger than 0? And in "-L ≤ x ≤ L" it says that L is both negative and positive? But then again, the last function says that x is bigger than L :S

    I took me several days before making this thread as I feel really stupid doing so. I'm just asking for more patience as this is very important to me :(
     
  11. Apr 4, 2013 #10

    kith

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    This doesn't define limits for x. x runs from -∞ to +∞. The function ψ(x) is defined to be zero almost everywhere, except for the interval [0,L] where it is A²exp(2x). For example chosing A=1 and L =2, this is what your function looks like. For every other x, it is zero. Do you see that the integral is finite then?

    No, there are three constants involved: A is the amplitude of your function, L defines the interval where your function is non-zero and N is the value of the integral. So the relationship is ∫f(x,A)dx = N(A,L). You will see that N is either infinite regardless what value L and A take, or finite. If it is finite, your function is normalizable, because ∫ 1/N f(x)dx = 1. If you still have trouble, just chose values for A and L and try to do the calculation then. jitbell's first post describes how to incorporate the L.
     
  12. Apr 4, 2013 #11

    DrClaude

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    This is typical of the "particle in a box" type of problem. Imagine you have a box (in 1D) of length L. The box has infinitely high walls (in terms of energy), such that the particle has to be confined to the box. Put one end of that box at x=0, the other at x=L. The wave function for the particle will then have values for 0≤x≤L, but has to be 0 outside this range, as you cannot find the particle outside the box.

    (Aside: infinite walls are necessary, otherwise quantum mechanics allows for the particle to tunnel out of the box, so you could have a small probability of finding it outside the box.)

    L is what it is, and from the statement of the problem, it is positive. Here, the box is 2L in size, and centered on 0, so x goes from -L to L. (Replace L by a number if you are still not sure of the meaning of -L ≤ x ≤ L.)

    Hope this helps.
     
  13. Apr 4, 2013 #12

    jtbell

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    It's basically using different formulas to define ψ(x), in different regions of x. A more explicit way of writing this example would be:

    $$\psi \left( x \right) = \left\{ {\begin{array}{*{20}{c}}
    {0,}&{{\rm{for }}x < 0}\\
    {A{e^x},}&{{\rm{for }}0 \le x \le L}\\
    {0,}&{{\rm{for }}x > L}
    \end{array}} \right.$$
     
  14. Apr 5, 2013 #13
    Thank you so much. I would then believe that function 1, 3 and 4 are normalized. Function 2 is a cosinus wave so if L isn't set, the area will be infinite. With function 5, x≥L so it doesn't matter what value L has, the area will be infinite.

    Another question that I didn't quite understand is this one:

    "If an atom in a crystal is acted upon by a restoring force that is directly proportional to the distance of the atom from its equilibrium position in the crystal, then it is impossible for the atom to have zero kinetic energy". Not sure what is meant by a 'restoring force', but I would say that the statement is true. Isn't it impossible for a particle to have zero kinetic energy as that would mean that the temperature is 0K which again is impossible?
     
  15. Apr 5, 2013 #14

    DrClaude

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    A restoring force is just a force that acts to bring back the system to equilibrium. If you have a "restoring force that is directly proportional to the distance of the atom from its equilibrium position", you basically get a harmonic oscillator. Showing that "it is impossible for the atom to have zero kinetic energy" corresponds to demonstrating that the ground state of a quantum mechanical harmonic oscillator has a non-zero expectation value for the kinetic energy (related to the zero-point energy). It has nothing to do with the 3rd law of thermodynamics, about attainability of absolute zero.
     
  16. Apr 5, 2013 #15
    I see :)

    I'm really happy about getting a better understanding though, I've now solved enough problems to pass, but I will solve a couple of more just to be safe. One problem asked what the value of A must be if -L≤x≤L in ψ(x)=A. I figured that to be 1/2L which seems to match my calculation.

    The next one is this one:

    http://i45.tinypic.com/o05d8p.jpg

    What throws me off is that it's asking for the maximum probability per unit length. This would depend on how big this unit of length is, no? Or does it mean an infidecimal point along the x-axis? In that case the probability would be infinitely small, wouldn't it? If I were to see what the probability for a particle being at for instance exactly x=1, there would be no chance of finding it exactly at that point.
     
  17. Apr 5, 2013 #16

    DrClaude

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    Be careful: you are calculating the integral of ##| \psi(x) |^2## (don't forget there is a square).


    It just means to consider ##| \psi(x) |^2## directly. The probability of finding the particle between ##x## and ##x + \Delta x## is
    $$
    \int_x^{x+ \Delta x} | \psi(x) |^2 dx
    $$
    which is a pure number (probability). Since ##dx## has units of length, this means that ##| \psi(x) |^2## has units of inverse length, and is therefore the probability per unit length.

    Exactly. That's why ##| \psi(x) |^2## is not the probability of finding the particle at point ##x##, because that would be zero. But it can still tell you where it is most probable to find the particle.
     
  18. Apr 7, 2013 #17
    I should have read the last question I posted more carefully, was quite simple to be honest :)

    This question was a little different though:
    http://i46.tinypic.com/35ksuvq.jpg

    I found the formula E=(n^2*π^2*h^2)/(2mL^2). I can't really use this formula as both E and L are unknown. Is there a direct correlation between the energy of a photon and it's wavelength? I can't seem to find a formula that states that :(
     
  19. Apr 7, 2013 #18

    DrClaude

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    Take care that it is ##\hbar## there, not ##h##.

    There is indeed! First, ##E = h \nu##, where ##\nu## is the frequency of the photon. And as it travels at the speed of light, you get the wavelength as ##\lambda = c/\nu##.
     
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