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Ket derivatives

  1. Feb 4, 2010 #1
    What is de real difference between parcial and total time derivatives of kets?
  2. jcsd
  3. Feb 4, 2010 #2


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    The "ket" is something abstract; it lives in a generic rigged Hilbert space. One needs to specify which representation of this abstract RHS is used. This in order to account for the <spatial variables>. Since in ordinary QM time and space are separate variables, one should constantly use partial derivatives, bu not of kets, but of wavefunctions (in case the RHS is made up of function spaces, such as L^2(R, dx).
  4. Feb 4, 2010 #3


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    The answer is the same as for functions in to [itex]\mathbb R[/itex]. None whatsoever.

    [tex]\frac{\partial}{\partial t}|\psi;t,s\rangle=\frac{d}{d t}|\psi;t,s\rangle=\lim_{h\rightarrow 0}\frac{|\psi;t+h,s\rangle-|\psi;t,s\rangle}{h}[/tex]

    There is only a small technical difference between the two operators [tex]\frac{\partial}{\partial t}[/tex] and [tex]\frac{d}{d t}[/tex]. The former acts on the function [tex](t,s)\mapsto|\psi;t,s\rangle[/tex] and the latter on the function [tex]t\mapsto|\psi;t,s\rangle[/tex].

    I wouldn't define kets that way. (This is the way I do it). Rigged Hilbert spaces are used to ensure that every self-adjoint operator has eigenvectors. This is an issue that goes beyond notation.
    Last edited: Feb 4, 2010
  5. Feb 4, 2010 #4

    George Jones

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    I'm not sure if this is what you are thinking about, but the following often trips up students.

    Let [itex]f = f \left( x, y, z, t \right)[/itex] be a function of position and time, i.e.,

    f : \mathbb{R}^4 &\rightarrow \mathbb{R}\\
    \left( x,y,z,t\right) &\mapsto f \left( x,y,z,t\right).

    Now, suppose that the position is itself a function of time, and use this to define

    [tex]\tilde{f} \left(t\right) = f \left( x\left(t\right), y\left(t\right), z\left(t\right), t \right).[/tex]


    [tex]\frac{ d \tilde{f}}{dt} = \frac{ \partial f}{\partial x} \frac{dx}{dt} + \frac{ \partial f}{\partial y} \frac{dy}{dt} + \frac{ \partial f}{\partial z} \frac{dz}{dt} + \frac{ \partial f}{\partial t}.[/tex]

    In general,

    [tex]\frac{ d \tilde{f}}{dt} \ne \frac{ \partial f}{\partial t}.[/tex]

    The function

    [tex]\tilde{f} : \mathbb{R} &\rightarrow \mathbb{R}[/tex]

    has a different domain than [itex]f[/itex], and thus is a different function. The two functions are so closely related, however, that the tilde [itex]\tilde{}[/itex] is omitted often (particularly by physicists), resulting in the somewhat nonsensical

    [tex]\frac{ d f}{dt} \ne \frac{ \partial f}{\partial t}.[/tex]
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