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Replacing total derivative with partial derivative in Griffiths' book

  1. Feb 15, 2007 #1
    I'm using Griffiths' book to self-study QM and I'm having a slight problem following one of his equations. In page 11 of his "Intro to Quantum Mechanics (2nd ed.)", he gives the reader the following 2 equations:

    [tex] \frac {d} {dt} \int_{-\infty}^{\infty}|\Psi(x,t)|^2 dx = \int_{-\infty}^{\infty} \frac{\partial}{\partial t} |\Psi(x,t)|^2 dx [/tex]

    In the next line, he gives the following explanation:
    "Note that the integral is a function of only t, so I use the total derivative (d/dt) in the first term, but the integrand is a function of x as well as t, so it's a partial derivative ([tex]\frac{\partial}{\partial t}[/tex]) in the second one."

    It is this explanation that I am having difficulty understanding. I just don't understand why he replaced the total derivative with a partial derivative, despite his explanation. Could someone please explain what he is trying to explain in greater detail? Thanks.
  2. jcsd
  3. Feb 15, 2007 #2


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    Suppose we integrate some function of x between limits, then after performing the integration, we are left with a number (since we have evaluated x in the resulting expression at the limits). Similarly, if we have a function of x and t, like above, then when integrating it with respect to x, we will be left with a function of only t (since, again, x has been evaluated at the limits). So, the derivative of this wrt t will be a total derivative.

    Now consider the second case above, where we are taking the time derivative into the integral. Now, the function inside the integral is a function of x and t (since it has not been integrated yet) and so the derivative wrt t will be partial.
  4. Feb 15, 2007 #3


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    In the LHS a function of both "x" and "t" is integrated DEFINTELY wrt "x" resulting in a function only of "t". The "x" dependent part has been replaced with the limits of the antiderivative at the points + and - infinity. Therefore, the resulting function doesn't depend on 2 variables anymore and depends on "t" wrt which is differentiated.

    Since "x" and "t" are independent variables, it doesn't matter whether you integrate wrt "x" first and differentiate wrt "t" afterwards (LHS) or viceversa (RHS). The reason we use the partial derivative in the RHS is because the probability density is a function of 2 independent variables and differentiation wrt one is denoted by the Jacobi symbol [itex] \partial [/itex].
  5. Feb 15, 2007 #4
    *thumbs up*

    Thanks guys, it's much clearer to me now :biggrin:
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