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I'm trying to self study quantum mechanics and am having a little trouble manipulating commutators. I get two different answers below, depending on the method I'm using. The second method gives me the correct answer for the more complicated textbook problem, but I'm just trying to understand why the first method is wrong.

So here $D$ is just the partial derivative with respect to $x$. $A$ is the $x$-component of the electromagnetic vector potential. I'm trying to calculate $[D^2, A]$.

Method 1: Use a test function. $$\begin{array}{rcl} [D^2, A]f & = & D^2(Af) - A(D^2f) \\ & = & D(A'f + Af') - Af''\\ & = & A''f + A'f' + A'f' +Af'' - Af'' \\ & = & A''f+2A'f'.\end{array}$$ So from this version, we end up with $[D^2, A] = A'' + 2A'D$.

Method 2: Use the formula $[AB, C] = A[B,C] + [A,C]B$.

Using this formula, $$\begin{array}{rcl} [D^2, A] & = & D[D, A] + [D, A]D \\ & = & D(A') + A'D = A'' +A'D. \end{array}$$ So in this method, $[D^2, A] = A'' + A'D$, which is different from the first method! (It doesn't have that factor of 2.) Can you please help me see what I'm doing wrong!

Qmechanic
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TKT
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2 Answers2

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The problem comes with your evaluation of $D[D,A]$ according to the second method. You are correct that this equals $DA'$, but this is not the same as $A''$. The differential operators $D$ still acts on everything to its right. If you include a test function $f$ again and evaluate $DA'f$, you find $$DA'f=(DA')f+A'(Df)=A''f+A'Df.$$ Thus, $DA'=A''+A'D$, and the second term on the right-hand side is exactly what you were missing; include it, and your answer agrees with the first method.

Buzz
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In agreement with Buzz, the moment we introduce any power $n$ of the differential operator $D$, it kinds of loses its identity and transforms into a new operator $B = D^n$. This creates a chain rule in the operation with the new operator $B = D^n$, which differentiates other operators and wavefunction which lie to its right side.

So, your first answer is correct, second is not as far as the problem is dealing with differential operator.

Buzz
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Tanay Gupta
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