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Basic understanding

  1. Jun 25, 2008 #1
    A friend of mine who studies physics recently tried to explain some basic quantum mechanics to me. He told me quite a lot of interesting stuff but it got me confused somehow and I think I really didn't get it all. Thus I am here to recapitualte my newly gained knowledge and ask questions.

    What I have learned is this:
    1. The state of a quantum mechanical system is described as a vector in a Hilbert space, written in this weird notation |v>
    2. For each physically measurable variable (like energy, momentum or the like) there exists an operator which describes the results of measurement of that variable in this strange way:
    If the system is in the state |v>, and |n> is an eigenvector to the operator H with eigenvalue E_n the probability that E_n will be the result of measurement is |<n|v>|². By measurement the system's state is changed into the corresponding eigenvector to the result.

    What I do not yet understand is what exactly that has to do with wave functions. My friend wasn't able to make that clear to me.
    Let's say we have an electron in free space and we are only interested in its position and momentum. So we have a complex-valued wave function for the position, [itex]\phi: \mathbb{R}^{3+1} \to \mathbb{C}[/itex], and one for the momentum [itex]\varphi: \mathbb{R}^{3+1} \to \mathbb{C}[/itex], which we can put together to one function depending on one time coordinate and three spatial coordinates each for momentum and position: [itex]\psi: \mathbb{R}^{6+1} \to \mathbb{C}, (x_p, y_p, z_p, x_m, y_m, z_m, t) \mapsto \phi(x_p, y_p, z_p, t)\varphi(x_m, y_m, z_m, t)[/itex]. The state at a particular time t is then described by a wave function obtained from [itex]\psi[/itex] by setting t=t_0 to be some constant: [itex]\psi_{t_0}:=\psi_{|t=t_0}[/itex].
    I am guessing now that the Hilbert space discussed above is just the space of wave functions like the latter. Am I right? Clearly [itex]\int_{\mathbb{R}^{6n}} |\psi_{t_0}|^2 d(x_p, y_p, z_p, x_m, y_m, z_m)=1[/itex] must hold for all wave functions since their squares are meant to describe probability distributions. Thus the space of such wave functions is a Hilbert space with the inner product given by [itex]<\phi, \psi>=\int_{\mathbb{R}^{6n}} \phi(x)\overline\psi(x)dx[/itex].

    If I look at it this way all that arcane stuff looks quite straightforward to me and gives me a picture consistent with many things I have heard of but never understood.
    So for example, if I perform a measurement of, say, the position of the electron in our system, the wave function "collapses" and the state [itex]\psi_{t_0}[/itex] is suddenly transformed into a different state [itex]\psi'_{t_0}[/itex] which is an eigenvector of the position operator. But what about the time evolution of the changed wave function now? Where do we get that from? Or can't we? Just having determined the position of our electron, its momentum must be completely unknown according to Heisenberg's relation of uncertainty. So how does our new wave_function [itex]\psi'_{t_0}[/itex] look actually? It can't be a wave anymore because the probability of the particle being at the position we measured is 1 and the probability of it being anywhere else is 0. And the momentum has to be completely unknown with equal probability for every possible value. The wave function of the momentum alone would have to be a constant function then, but which constant? The integral of the momentum's absolute squared "wave function" over whole 3-space would have be infinity then, and the integral of the positions absolute squared wave function would be zero because the wave function has to be zero for every single value except the one we measured. Is it here where Dircac's "distributions" kick in, which I have heard of but don't understand?

    But let's go back to the stuff my friend told me: The measurement puts the system in an eigenstate of the measured observable's operator. This state certainly has to be an element of the Hilbert space we are talking about all the time. But obviously if I think of the elements of that Hilbert space just as wave functions like I did above there occur problems, because after measurement our wave function is not even a function anymore. I'm really getting confused here. It looked so straightforward when I thought the vectors in our Hilbert space were just the wave functions. But now I see that performing a measurement which, according to what my friend told me, should transform the former state of the system to an eigenstate of the measurement operator, in fact would transform such a wave function to something which is not even a function anymore.

    Can someone try to explain the dirty details to me, please?

  2. jcsd
  3. Jun 25, 2008 #2
    I think you should read the classic book of sakurai called modern quantum mechanics.

    But let me try to point something out that I think you didn't get explained completely by your friend.

    Lets say that you have a state, that is a vector [tex]|\Psi>[/tex] in a hilbert space.
    Know we need some postulates of quantum mechanics. First For a finite dimensional hilbert space we know that a hermitian matrix (operator) gives us an eigenbasis for the space, in quantum mechanics we postulate that this works in any dimensions. Every physical observabel are hermitian operators, so these gives us different bases.

    Lets take two different the hamiltonian ( [tex]\hat{H}[/tex]) which gives the energy and the postion operator [tex]\hat{x}[/tex]. Lets assume that the energy is a discrete spectrum (that is often the case), that is the eigenvalues of [tex]\hat{H}[/tex] is countable, so we can write [tex]\hat{H} |E_n> = E_n |E_n>[/tex], that is we have all the eigenvectors and eigenvalues, we simply (this need not to ve simple :-)) solved the eigenvalue problem.

    We can do this for the postion operator too

    [tex]\hat{x} |x> = x |x>[/tex]

    notice how I denote the eigenvector by its eigenvalue. This spectrum is not discrete.

    We now use a closure reltaion, that is we can write the identity as

    [tex]\hat{I} = \sum_n |E_n><E_n|[/tex]

    this is for the discrete case, for the continous case we get

    [tex]\hat{I} = \int |x><x| dx[/tex]

    this is not obvious, but it works, it is related to the spectral theorem for operators.

    Now lets get the wavefunction to appear. First we expand our state in the energy basis

    [tex]|\Psi> = \hat{I} |\Psi> = \sum_n |E_n><E_n|\Psi> = \sum_n <E_n|\Psi> |E_n>[/tex]

    The coefficients [tex] <E_n|\Psi> [/tex] gives you the probability, as you mentioned for the state to be measured to be in the energy state [tex] |E_n> [/tex].

    In the continoues case we get

    [tex]|\Psi> = \hat{I} |\Psi> = \int |x><x|\Psi> dx = \int <x|\Psi> |x> dx [/tex]

    the countionues coeficients [tex]\Psi(x) = <x|\Psi>[/tex] is dependent on x and this is what we call the wavefunction, you could also write the wavefunction in momentum basis or any other, in principel the coefficients in the discrete energy basis is also a wavefunction, but not from the reals, this is a function from the natural numbers [tex]\Psi(n) = <E_n|\Psi>[/tex].

    As you can se the wavefunctions are the coefficients when you expand the full state in some basis, the exisetens of these basis are a little problematic (they might not exist), but this is one of the postulates of quantum mechanics.

    Anyway hope this clarifies some things, but you should really read at least the first couple of chapters of sakurai then all your questions will be answered (and maybe alot of new arise), it is very well writen and a classic.

    Can I ask what your background in physics and math is ?
  4. Jun 25, 2008 #3
    by the way, the experiment you are talking about, sounds like a free particle, one should be very carefull handeling this example. Lets say you measure the particles position to be x, then as you say the particle will go into the state [tex] |x> [/tex], lets try to expand this as before

    [tex] \int |x'><x'|x> dx' = \int |x'> \delta(x-x') dx' [/tex]

    whre I used [tex] \Psi(x) = <x'|x> = \delta(x-x') [/tex]. Comparing with the things in my first post, you see that the wavefunction is the deltafunction. If you want the wavefunction in the momentum basis you need to take the fouirer transform of the wavefunction in the position basis, this will give you a function of the form

    [tex] \Psi(p) = e^{i x p A} [/tex] where A is some constant.

    That is we can write

    [tex] |x> = \int \Psi(p) |p> dp =\int e^{i x p A} |p> dp [/tex]

    Know lets try to find the probability of measuering the particles momentum to be any value p'

    [tex] <p'|x> = \int \Psi(p) <p'|p> dp =\int e^{i x p A} \delta(p-p') dp = e^{i x p' A} [/tex]

    you see this is a wave. Now lets find the probability by taking the norm |<p'|x>|^2 this is simply one, you could argue that this gives you even probabilty for measuring any momemtum, that is as you say it is completely unknown, but there is also a problem, that the probability integrated over the whole momentum space i infinit. This is because the free particle is a very dangerous example.

    The problem is that the delta function is actually not a function. One way to get out of this problem is to say that, in any experiment you always have some container where you particles are in, such that you need not to integrate over all space but only in the space of the container. This may not be satisfactory but quantum mechanics are full of flaws and other problems, and as far as I know there is never anyone who have made it completely consistent. But as you see if you interpret the result you get, as even probality of all momentas it makes sence.

    The good thing about quantum mechanics is that it gives alot of good results, but you have to be carefull, when interpreting it.
  5. Jun 25, 2008 #4
    Um, actually, if you're trying to read more into quantum mechanics, I wouldn't recommend Sakurai... that's more of a graduate level text. I recommend Liboff's Introductory Quantum Mechanics, which I am currently using. It's a really good text, but doesn't have solutions to the problems (that's why in the Quantum Physics section of these forums in homework, like 10 of the created posts are mine :smile:), heheh.
  6. Jun 25, 2008 #5
    Actually i think it is possible to learn quantum mechanics directly from sakurai, but you are right it is a bit hard, the good thing about sakurai is that he starts with the hilbert space and then introduce wave function in different bases, if you start with fx. griffiths introduction to quantum mechanincs you start with wavefunctions in position space and I think that can be a bit confusing later on. Don't know the book you recommend though.

    But maybe you are right that sakurai is to hard to begin with, but wouldn't say that introduction books gonna answer his questions very good.
  7. Jun 25, 2008 #6


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    Sakurai defines the wave function corresponding to state [itex]|\alpha\rangle[/itex] as [itex]\psi_\alpha(\vec x)=\langle \vec x|\alpha\rangle[/itex]. If we include the time dependence, it's [itex]\psi_\alpha(\vec x,t)=\langle \vec x|e^{iHt}|\alpha\rangle[/itex]. It satisfies the Schrödinger equation because the time evolution operator does.

    Note that

    [tex]\langle\beta|\alpha\rangle=\int d^3x\langle\beta|\vec x\rangle\langle\vec x|\alpha\rangle=\int d^3x \psi_\beta(\vec x)^*\psi_\alpha(\vec x)[/tex]

    This matches the definition of the scalar product of two wave functions.

    If you want to explicitly construct the Hilbert space, you can take it to be the set of solutions of the Schrödinger equation, if you're dealing with a single spin-0 particle in non-relativistic QM. More complicated quantum systems require more complicated Hilbert spaces. That's why you will see things like [itex]\psi(x)\chi[/itex] in textbooks, where the [itex]\chi[/itex] represents the spin state.
    Last edited: Jun 25, 2008
  8. Jun 25, 2008 #7


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    Position measurements in the real world don't quite manage to do that. They will squeeze the wave function into a state that's zero outside of some region, but they won't turn the wave function into a delta function (which as you said yourself, isn't even a function).

    We can't. This is the "measurement problem" of quantum mechanics. I think some would say that decoherence answers the question, but I don't think it really does.
  9. Jun 25, 2008 #8
    Hi, mrandersdk! Thanks for explaining.
    To answer your question: I'm a math student in 6th semester, but a relatively lazy and ignorant one, I guess, so you should perhaps think of me as a 4th semester or so... :p Unfortunately I haven't learnt anything about operator algebra until now because I didn't know what it is good for. And I suck at solving differential equations. That's why I don't have physics as a secondary. My phyiscal background is only school stuff basically.

    So, let me recapitulate again:
    The Hilbert spaces that describe quantum mechanical systems do not exactly consist only of wave functions but of a more abstract kind of vectors that can be seen sometimes as wave functions but in other cases they can't be represented as wave functions (when the wave function has just collapsed due to measurement for example). The vectors that can be written as wave functions are obviously a linear subspace of the whole Hilbert space, and so are the vectors that can be written as discrete Energy state probability distributions because one only uses the normalized vectors, operating only in projective space effectively, so the whole probability of everything stays 1 after addition of two states.
    I actually learnt something. Yay!
    But is it always possible to give a probability distribution of the possible Energy states? I mean is the Energy state space the whole space? Or are there states from which one can't extract information about the energy, as there are cases in which one can't turn a state vector into a wave function? Can the discrete energy "wave function" collapse in the same way? I read on Wikipedia that analogous to the impulse/position uncertainty there is a time/energy uncertainty relation. But a time measuring operator doesn't seem to make any sense since in our state space time is set to be a constant.
    [itex]\hat{I} = \sum_n |E_n><E_n| = \int |x><x|dx[/itex]
    This equation is really interesting. So that means that the function [itex]\varpsi(x)[/itex] holds the same information as [itex]\varpsi(n)[/itex] in some way, doesn't it? But there's still something I don't understand.
    As you can see the wavefunctions are the coefficients when you expand the full state in some basis, the exisetens of these basis are a little problematic (they might not exist), but this is one of the postulates of quantum mechanics.[/itex]
    The axiom of choice (or Zorn's Lemma) tells me that every vector space has a basis. And the states which can be represented as wave functions obviously form a linear subspace of the state space. So there must be a basis of this subspace. And some other theorem tells me that this basis can be expanded to a basis of the whole state space. So why does quantum mechanics have to postulate the existence of such a basis? Is the axiom of choice rejected by physicists or is there some other subtlety I did not see here? Of course it's difficult to see for me that there should be a countable basis for the wave function subspace. And as the Energy state base is assumed countable and if we assume the Energy probability function cannot "collapse", then the wave function space must be a subspace of the energy state space and has to be countable, too. Is that the issue?

    And to everyone: Thanks for your reading tips. Maybe I will get some of those books once upon a time. But my laziness prohibits me from reading too much :) At the moment I'm reading Feynman's lectures on physics. I've not yet come to the part where quantum mechanics is covered. I think that's in volume 3 and now I'm reading volume 1. I hope I will some day arrive at volume 3. :)
    Last edited: Jun 25, 2008
  10. Jun 25, 2008 #9
    Think the postulate is to make sure that the eigenvectors to any hermitian operator has a basis that spans the whole space. For finite vactor spaces, we know from simple linear algebra that this is true, but for unbounded hermitian operators on infinite vector spaces i'm not sure this is true.

    Just to make something more precise. You have a state, this is described in some abstract hilbert space. Wave functions (descrete og non-descrete) are the coefficients you get when you write that vector in some basis. This could be the position, momemtum or energy basis.

    When you have a basis an work in that, fx. the position basis (the one most people learn quantum mechanincs in), you see that knowing the wavefunction give you all information about the system, because you know the basis vectors, so given you the expansion coefficients give you the complete state. Determening the time evolution of the coefficients then give you how the state evolves in time.

    The schrödinger equation as writen here:


    is the schrödinger equation writen in the position basis, you could write this in any complete basis. If you want to write it in the momemtum basis you simply need to take the fourier transform with respect to the postion coordinate.

    But remember knowing the wavefunction in posistion space, give me the state, and I can as in normal linear algebra change to another basis, as follows

    Asume I know [tex] \Psi(x)[/tex], then the state is given by

    [tex] |\Psi> = \int \Psi(x) |x> dx[/tex]

    Now I want this in the energy basis, so i simply insert the identity

    [tex] |\Psi> = \int \Psi(x) \hat{I} |x> dx = |\Psi> = \int \Psi(x) \sum_n |E_n><E_n|x> dx = \sum_n \int \Psi(x) |E_n><E_n|x> dx [/tex]
    [tex]= \sum_n \int \Psi(x) <E_n|x> dx |E_n> = \sum_n \int \Psi(x) \phi_n(x) dx |E_n> = \sum_n c_n |E_n> [/tex]

    where [tex] \phi_n(x) = <E_n|x> [/tex], of cause you have to be calculated, and may not be easy. And [tex]c_n = \int \Psi(x) \phi_n(x) dx [/tex]. You see in principle i have changed to the energy basis, changes like these can be made in the same way to all kind of bases. An element like [tex] <E_n|x> [/tex] can be hard to find, if it even is possible, but as a theoretical tool it can be very powerfull. In most practical cases, it is clear what basis to choose depending on the problem, but that is another story.
  11. Jun 25, 2008 #10
    A wave function of, say, the position X, is the inner product of the state |[itex]\Psi[/itex]> and the position eigenvector |x>, which is an eigenvector of the observable X. I.e.

    [itex]\Psi[/itex](x) = <x|[itex]\Psi[/itex]>

    The wave function of the momentum P is the inner product of the state |[itex]\Psi[/itex]> and the momentum eigenvector |p>, which is an eigenvector of the observable P. I.e.

    [itex]\Psi[/itex](p) = <p|[itex]\Psi[/itex]>

    If you measure the position of a particle then, ideally, the wave function <x|[itex]\Psi[/itex]> collapses to an eigenfunction of position. However this physically can't happen. A postulate of quantum mechanics requires that all wavefunctions be continuous and square integrable. An eigenfunction of position does not fit this requirement. There must therefore be a finite value of uncertainty in the measured position. However this uncertainty can be as small as you'd like so long as it is not zero. The corresponding <p|[itex]\Psi[/itex]> then becomes a function with a very wide peak.
    Sorry but this doesn't make any sense to me.
    If the system is conserved (i.e. the potential is not a function of time) then multiply the state vector by the time evolution operator. This gives a wave function which is the initial wave function multiplied by e-iEt/h.
    Therefore you can't determine the position with a zero uncertainty.
    [itex]\Psi[/itex](x,t) = e-iEt/h[itex]\Psi[/itex](x,t0)
    Physically that is true. Mathematically, in the ideal sense, no. The basis of a Hilbert space need not be elements in the Hilbet space. Stange as it seems! :bugeye:
    When studying quantum mechanics that is often considered to be a good sign. :tongue:

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