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I believe it's standard to place the operator in between the conjugate of the wavefunction and the wavefunction itself. For instance,

$$\langle p\rangle = \int_{-\infty}^{\infty}\Psi * \frac{\hbar}{i}\frac{d}{dx}\Psi dx$$

Is it wrong to do this?

$$\langle p\rangle = \int_{-\infty}^{\infty}\frac{\hbar}{i}\frac{d}{dx}|\Psi|^2 dx$$

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So I believe it's standard to place the operator inbetween the conjugate of the wavefunction and the wavefunction itself. For instance,

$$\langle p\rangle = \int_{-\infty}^{\infty}\Psi * \frac{\hbar}{i}\frac{d}{dx}\Psi dx$$

Yes, that is correct, and

Is it wrong to do this?

$$\langle p\rangle = \int_{-\infty}^{\infty}\frac{\hbar}{i}\frac{d}{dx}|\Psi|^2 dx$$

yes, that is wrong.

One easy way to see why the latter must be wrong is that the integral of an exact derivative is always doable:

$$\int_{-\infty}^{\infty}\frac{d}{dx}|\Psi|^2 dx=\left.|\Psi(x)|^2\right|_{-\infty}^{\infty}=0,$$ for all $\Psi$, which can't be right, not least because you require $⟨p⟩$ to depend on $\Psi$ in some meaningful fashion.

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  • $\begingroup$ I see. I also see why it works in the homework problem that I'm doing. I was finding $\langle x\rangle$ for a simple harmonic oscillator. I wrote $x$ in terms of the raising and lowering operators, but these operators, when they act on a wavefunction don't require derivatives. $a_+ \Psi_n = \sqrt{n+1}\Psi_{n+1}$. Since they don't have any derivatives, I can use the incorrect way? $\endgroup$ – DWade64 Sep 29 '14 at 11:56
  • $\begingroup$ No, you can't. The operators you are using are linear operators acting on Hilbert spaces $T:\text{dom}\, T \subset \mathcal{H}\rightarrow \mathcal{H}$. What you want to calculate is the inner product $\langle \psi, T\psi\rangle$, which is given in the $L^2(\Bbb{R})$ space by $\int_\Bbb{R} \psi(x)^* (T \psi)(x) dx$. It may be the case that $|\psi(x)|^2$ is not even in $L^2(\Bbb{R})$, so the action of the operator may not even make sense. $\endgroup$ – Mateus Sampaio Sep 29 '14 at 12:09
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    $\begingroup$ The raising and lowering operators are linear combinations of $x$ and $p$, and therefore do have derivatives in them. But you shouldn't have to take an integral to calculate that expectation value anyway. $\endgroup$ – ZachMcDargh Sep 29 '14 at 14:17
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The 2nd way

$$\langle p\rangle = \int_{-\infty}^{\infty}\frac{\hbar}{i}\frac{d}{dx}|\Psi|^2 dx$$

will produce a complex result in general (in the example above it is will simply be zero), not having a physical measurement analog. The operator operates on some vector (either $\Psi$ or $\bar{\Psi}$), whereas the $|\Psi|^2$ is a simple real number.

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