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Left hand side of the Schrödinger equation

  1. Mar 21, 2014 #1
    The time dependent Schrödinger equation is:

    [itex]i\hbar\partial_{t}\Psi=\hat{H}\Psi[/itex]

    Does it mean that the operator [itex]i\hbar\partial_{t}[/itex] has the same eigenstates and eigenvalues as any Hamiltonian?
     
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  3. Mar 21, 2014 #2

    ChrisVer

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    the operator, you're asking about, is the Hamiltonian...
    the generator of time translations...
     
  4. Mar 21, 2014 #3
    The problem is that I'm having difficulties finding the correct commutator between the position and the Hamiltonian. Usually it is written that these 2 operators do not commute. See for example here on the second page, expression (306) and (307).

    But when I calculate the commutator between position [itex]\hat{x}[/itex] and the [itex]i\hbar\partial_{t}[/itex] I get zero:

    [itex][\hat{x},i\hbar\partial_{t}]=xi\hbar\partial_{t}ψ-i\hbar\partial_{t}(xψ)=xi\hbar\partial_{t}ψ-(i\hbarψ\partial_{t}x+xi\hbar\partial_{t}ψ)=xi\hbar\partial_{t}ψ-xi\hbar\partial_{t}ψ=0[/itex]

    So I'm wondering whether [itex]i\hbar\partial_{t}[/itex] and [itex]\hat{H}[/itex] are identical operators or what kind of error am I doing.
     
  5. Mar 21, 2014 #4

    ChrisVer

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    Why did you drop away the partial of x?
    In general if you don't drop it away, you are getting Heisenberg's equation for the time evolution of a quantity/operator.

    [itex] \dot{A}\propto[A,H] [/itex]

    at the same time, if you know the form of hamiltonian (you know H(p,x)) you can replace the hamiltonian in the commutator to find out what you are looking for:
    [itex] [x,H(p,x)]= [x,f(p)]+[x,g(x)]=[x,f(p)] [/itex]
    (if you can make this separation like in [itex]H= \frac{p^{2}}{2m}+V(x)[/itex] )


    They are identical though...
     
    Last edited: Mar 21, 2014
  6. Mar 22, 2014 #5
    I was thinking that [itex]\partial_{t}x=0[/itex], because [itex]\hat{x}=x[/itex] does not seem to depend on time. The operator [itex]\hat{x}[/itex] is just multiplication by the coordinate [itex]x[/itex] and it does not change in time, so [itex]\partial_{t}x[/itex] should be zero. It would be non-zero only if [itex]\hat{x}=x(t)[/itex]. But I'm probably missing something.
     
  7. Mar 22, 2014 #6
    The OP is asking whether iℏ∂t is an operator。
    I think I also asked the same quesetion a while back, is iℏ∂t an operator? It certainly looks like one from the schroedinger equation and from the fact that it fits every criteria to be one.
    I haven't seen this question addressed anywhere
     
    Last edited: Mar 22, 2014
  8. Mar 22, 2014 #7

    ChrisVer

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    If you take the solution of the shr's equation you gave above you find:
    [itex] \psi(t)= e^{-\frac{iHt}{\hbar}} \psi(0)[/itex]
    from where you can see indeed that the Hamiltonian is the generator of time translations. Eg you can expand the exponential up to the first order:
    [itex] \psi(t)= (1-\frac{iHt}{\hbar}) \psi(0)=\psi(0) - \frac{iHt}{\hbar} \psi(0) [/itex]
    which tells you how [itex]\psi(0)[/itex] transforms after time [itex]t[/itex], and you can see this happens by the action of the Hamiltonian on your initial state.
    Working on Heisenberg's picture
    In general, an operator is time-independent only if it commutes with the Hamiltonian (since the commutation leads to symmetry under time translations- so if you know the operator at time t you also know the operator at time t+a...). In many cases this is not true. For example the free particle having a Hamiltonian of the form [itex]H=p^{2}/2m[/itex] the position will be time dependent...
    [itex]\hat{\dot{x}}(t) \propto [x,H] \propto \frac{\hat{p}(t)}{m}[/itex]
    while the momentum will be time independent...
    [itex]\hat{\dot{p}}(t) \propto [p,H] =0[/itex]
    so you have that [itex]\hat{p}(t)=\hat{p}(0)[/itex]
    and thus inserting it in the x we can solve the differential:
    [itex]\hat{\dot{x}}(t)= A \frac{\hat{p}(0)}{m}[/itex]
    [itex]\hat{x}(t)= \hat{x}(0) + A \frac{\hat{p}(0)t}{m} [/itex]

    Finally the A I used was just the proportion constant, it has just hbars and is probably, by calculating the commutators and plugging in all the rest constants in Heisenberg's equation you can find it...

    As you can see, the time (in)dependency is subject to the commutator of the quantity/operator with the Hamiltonian. In that sense, Hamiltonian is the time translations generator (ih dt). Also the Hamiltonian is connected to the energy of the system, which you can write as a kinetic term and an interaction term, so you can have H=Hkinet+Hinter and via this you can compute the commutators... for example for the free particle you had H=p^2/2m, for the Harmonic Oscillator you have H=p^2/2m + a x^2 (since V(x)=ax^2 for H.O.) and so on...
     
  9. Mar 22, 2014 #8
    Thank you for the answer. So it was no surprise that my commutator turned up to be zero, because I put x time independent, and that ensured it commutes with the Hamiltonian. But in general the position operator does not have to be time independent (in Heisenberg picture) and does not need to commute with the Hamiltonian.

    There is however one more related question: in hydrogen atom (with one electron) the electron orbital 1s is a stationary state (time independent) and it is an eigenstate of the Hamiltonian (solution of the time-independent Schrödinger equation). In this time-independent case the position operator should commute with the Hamiltonian. So the position operator should have identical eigenstates as the Hamiltonian (commuting operators have the same eigenstates).

    But the 1s orbital (kind of a spatial "sphere") looks quite differently than a position eigenstate (ψ non-zero only at one single position, zero everywhere else, like Dirac delta). Where is this difference coming from?
     
  10. Mar 22, 2014 #9

    ChrisVer

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    why do you say that position commutes with the Hamiltonian in the Hydrogen atom? I didn't know of such a thing currently happening.
    Write the hamiltonian of a Hydrogen atom, and take the commutor [r,H]... is it zero?
     
  11. Mar 22, 2014 #10
    To answer the original question, just because$$i\hbar\frac{\partial}{\partial t} \mid \psi \rangle = H \mid \psi \rangle $$
    does not mean $$H = i\hbar\frac{\partial}{\partial t} $$
    One may not take an arbitrary expression, put H in front of it, and obtain the time derivative of that expression.
     
  12. Mar 22, 2014 #11

    ChrisVer

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    if the thing that H acts on is an eigenstate of the hamiltonian, then you'll of course get the time derivative of it....
    That's because you can do (passing from classical to quantum mechanics) the operator assignment:
    [itex] E \rightarrow i \hbar \frac{d}{dt}[/itex]
     
  13. Mar 22, 2014 #12

    Fredrik

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    It takes functions with domain ##\mathbb R^4## as input, but the Hilbert space whose elements are called "state vectors" is a set of functions with domain ##\mathbb R^3##. So ##i\hbar\partial/\partial t## can't be an operator on that space.

    The right-hand side is the one that needs interpretation, at least if we want H to denote the Hamiltonian, defined as an operator on the Hilbert space. For each t, define ##\psi_t:\mathbb R^3\to\mathbb C## by ##\psi_t(x)=\psi(x,t)## for all x. The right-hand side of the Schrödinger equation means ##(H\psi_t)(x)##.

    This looks a bit awkward, so it's probably best to take the point of view that the H that appears in the Schrödinger equation isn't the actual Hamiltonian, but the map that takes the function ##\psi## to the function ##(x,t)\mapsto(H\psi_t)(x)##. In other word, the new H is defined by ##(H\psi)(x,t)=(H\psi_t)(x)##, where the H on the right is the actual Hamiltonian.

    The main thing that prevents you from writing ##i\hbar\partial/\partial t=H## is that we haven't specified exactly what the domains of these maps are. Recall that two maps f and g are equal if and only if their domains are equal and f(x)=g(x) for all x in that domain. When we have found the vector space of all separable solutions, let's call it W, we can choose to say that ##i\hbar\partial/\partial t## and ##H## both denote maps defined on W. This makes the equality ##i\hbar\partial/\partial t=H## come true.
     
  14. Mar 23, 2014 #13
    The hydrogen eigenstates have specific angular momenta, so they must be eigenstates of the angular momentum operator. Does that operator commute with the position operator? Also check ChrisVer's point, does position really commute with the Hamiltonian in this case?
     
  15. Mar 23, 2014 #14
    Thanks to everybody for your comments, it helped me a lot. I finally understand where my confusion was coming from; I got stuck in an incorrect idea. Now it is clear. Position really does not commute with the Hamiltonian in hydrogen atom. And also I understand better the difference between [itex]\hat{H}[/itex] and [itex]i\hbar\partial_{t}[/itex].
     
  16. Mar 23, 2014 #15

    ChrisVer

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    Could you please describe how you figure out these domains?
     
  17. Mar 23, 2014 #16

    Fredrik

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    All I really did for the part you're quoting was to note that the notation ##\frac{\partial\psi(x,t)}{\partial t}## indicates that ##\psi## is defined on some subset of ##\mathbb R^4##. I also know that t can be any real number, so the domain should be ##\mathbb R^4##.

    It's part of the definition of the theory that the state vectors are equivalence classes of square integrable functions from ##\mathbb R^3## into ##\mathbb C##. (Two functions f and g are considered equivalent if the subset of ##\mathbb R^3## where f and g have different values has Lebesgue measure 0. The equivalence class that a function f belongs to is often denoted by [f]). The Schrödinger equation is supposed to tell us how a state vector changes with time. So given a state vector [f], it's supposed to find us a unique curve through [f]. We can introduce functions ft such that this curve can be written as ##t\mapsto [f_t]##. The map ##t\mapsto f_t## is then a curve in the space of square-integrable complex-valued functions on ##\mathbb R^3##. The map ##(x,t)\mapsto f_t(x)## is the one that satisfies the Schrödinger equation. This is the function we denote by ##\psi##, and since x is really three variables, this is a function from ##\mathbb R^4## to ##\mathbb C##.
     
  18. Mar 23, 2014 #17

    stevendaryl

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    Just to expand on what Fredrik said:

    Often the way that the Schrodinger equation is solved is by starting with a complete basis of functions. For example, for particles confined to the "box"

    [itex]0 \leq x \leq L[/itex]

    one example of a complete basis would be the functions [itex]\phi_n(x) = sin(\frac{n \pi x}{L})[/itex]

    Note, that the "complete basis" consists of functions of [itex]x[/itex], not [itex]x[/itex] and [itex]t[/itex]. A general solution of the Schrodinger equation for that box would be a function of [itex]x[/itex] and [itex]t[/itex] that can be written in this way:

    [itex]\psi(x,t) = \sum_n C_n(t) \phi_n(x)[/itex]

    In this way of going about solving the Schrodinger equation (it's not the only way, but it's a way), the time dependence is only in the coefficient [itex]C_n(t)[/itex]. In that case, Schrodinger's equation gives:

    [itex]H \psi = i \dfrac{\partial}{\partial t} \psi
    \Longrightarrow \sum_n C_n(t) H \phi_n(x) = \sum_n i \dfrac{d C_n}{d t} \phi_n(x)[/itex]

    The [itex]H[/itex] acts only on the [itex]\phi_n[/itex], and the \dfrac{d}{dt} acts only on the coefficients [itex]C_n(t)[/itex]
     
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