For verifying I set the inner product relation for hermitian conjugate operators: $$<\psi|L_-\psi>=<L_+\psi |\psi>$$ But I get: $<\psi|L_-\psi>=-<L_-\psi|\psi>$

Since the lowering/raising operator acts on the x and y basis specificaly by definition ($L_\pm=L_x\pm iL_y$)

I tried to evaluate over volume $d^3 \textbf{r} = dxdydz$. Then, by considering $L_x=y\hat{p}_z-z \hat{p}_y$, and $\hat{p}_{x,y,z}=\frac{\hbar}{i}\frac{\partial }{\partial (x,y,z)}$, I get: $$<\psi|L_-\psi>=\int\psi^*(L_x- iL_y)\psi d^3 \textbf{r}$$ $$\int\psi^*(L_x- iL_y)\psi d^3 \textbf{r}=\int\psi^* \left [y\frac{\hbar}{i}\frac{\partial }{\partial z}\psi-z\frac{\hbar}{i}\frac{\partial }{\partial y}\psi-i(z\frac{\hbar}{i}\frac{\partial }{\partial x}\psi-x\frac{\hbar}{i}\frac{\partial }{\partial z}\psi)\right ]d^3\textbf{r}$$

$$=\int\psi^*y\frac{\hbar}{i}\frac{\partial }{\partial z}\psi-\psi^*z\frac{\hbar}{i}\frac{\partial }{\partial y}\psi-\psi^*z\hbar\frac{\partial }{\partial x}\psi+\psi^*x\hbar\frac{\partial }{\partial z}\psi d^3\textbf{r}$$

Now noting that the terms are very similar, in general the integration by parts for each term will be of the form:

$$\int \psi^*C\frac{\partial }{\partial (x,y,z)}\psi d^3\textbf{r}=\psi^*\psi|^{\infty}_{-\infty}-\int\psi\frac{\partial }{\partial (x,y,z)}\psi^* d^3\textbf{r}=-\int\psi\frac{\partial }{\partial (x,y,z)}\psi^* d^3\textbf{r}$$

Then, after integrating each term by parts we get:

$$<\psi|L_-\psi>=\int-\psi y\frac{\hbar}{i}\frac{\partial \psi^*}{\partial z}+\psi z\frac{\hbar}{i}\frac{\partial \psi^*}{\partial y}+\psi z\hbar\frac{\partial \psi^*}{\partial x}-\psi x\hbar\frac{\partial \psi^*}{\partial z} d^3\textbf{r}$$

$$=\int-\psi y\hat{p}_z\psi^*+\psi z\hat{p}_y\psi^*+i\left (\psi z\frac{\hbar}{i}\frac{\partial \psi^*}{\partial x}-\psi x\frac{\hbar}{i}\frac{\partial \psi^*}{\partial z}\right ) d^3\textbf{r}$$

$$=\int-\psi y\hat{p}_z\psi^*+\psi z\hat{p}_y\psi^*+i\left (\psi z\hat{p}_x\psi^*-\psi x\hat{p}_z\psi^*\right ) d^3\textbf{r}$$

$$=\int \psi\left [ -y\hat{p}_z+ z\hat{p}_y+i\left ( z\hat{p}_x- x\hat{p}_z\right ) \right ]\psi^* d^3\textbf{r} $$

$$=\int \psi\left [ -L_x +i L_y \right ]\psi^* d^3\textbf{r} $$

$$=-\int \psi\left [ L_x -i L_y \right ]\psi^* d^3\textbf{r} $$

Here is the thing, the last integral isn't equal to?:

$$-\int \psi\left [ L_x -i L_y \right ]\psi^* d^3\textbf{r}=-<L_-\psi|\psi> $$

But for the verification it has to be: $<L_+\psi|\psi>$

Maybe I'm missing something trivial...


1 Answer 1


First of all, the definition of the Hermitian conjugate actually uses different states, so the statement is $$\langle \alpha | A \beta \rangle = \langle A^\dagger \alpha | \beta\rangle $$

With respect to the lowering operator,

$$\langle \alpha | L_- \beta\rangle =\langle \alpha|(L_x-iL_y)\beta\rangle= \langle \alpha|L_x\beta\rangle -i\langle\alpha|L_y \beta\rangle$$

We know that both $L_x$ and $L_y$ are Hermitian, so $$\langle \alpha|L_x\beta\rangle -i\langle\alpha|L_y \beta\rangle = \langle L_x \alpha|\beta\rangle -i\langle L_y \alpha | \beta\rangle$$

However, remember that the inner product is not $\mathbb{C}$-linear, it is $\mathbb{C}$-sesquilinear. We can pull the factor of $i$ into the bra part, but it has to be complex-conjugated, so it picks up an additional minus sign and we find that

$$\langle \alpha|L_- \beta\rangle=\langle L_x \alpha|\beta\rangle -i\langle L_y \alpha | \beta\rangle = \langle(L_x+iL_y)\alpha|\beta\rangle=\langle L_+ \alpha | \beta\rangle$$

If you want to work things out at the level of the position basis, that's fine, but remember you need that extra conjugation at the end. For example, in the case of $L_x$,

$$\langle \alpha | L_x \beta\rangle =\int d^3r \ \bar\alpha(-i\hbar)\left(y\frac{\partial}{\partial z} - z\frac{\partial}{\partial y}\right)\beta = (-i\hbar)\int d^3r \ y\cdot \bar\alpha \frac{\partial \beta}{\partial z} - (-i\hbar)\int d^3r \ z \cdot \bar \alpha \frac{\partial \beta}{\partial y}$$

Integration by parts yields a minus sign, but pulling the $(-i\hbar)$'s into the conjugated part of the integrand yields another, and you find that

$$ \langle\alpha | L_x\beta\rangle = \int d^3r \ \left[\overline{(-i\hbar)\left(y\frac{\partial}{\partial z}-z \frac{\partial}{\partial y}\right)\alpha}\right] \beta = \langle L_x \alpha | \beta\rangle$$

To elaborate on the conjugation part, consider a 1D problem where you're supposed to show that the momentum operator is Hermitian. What you're trying to show is that $$\int dx \ \overline{\alpha} \left(\frac{\hbar}{i}\frac{d}{dx}\right) \beta=\int dx \ \overline{\left(\frac{\hbar}{i}\frac{d}{dx}\right)\alpha} \cdot \beta $$

You're getting an extra minus sign because you're finding that

$$\int dx \ \overline{\alpha} \left(\frac{\hbar}{i}\frac{d}{dx}\right) \beta=-\int dx \ \left(\frac{\hbar}{i}\frac{d}{dx}\right)\overline{\alpha} \cdot \beta $$

but that minus sign goes away when you extend the complex conjugation to the $\frac{\hbar}{i}$.

  • $\begingroup$ The $L_x$ isn't defined as $y\hat{p}_z-z\hat{p}_y$? $\endgroup$ Oct 15, 2017 at 3:46
  • 1
    $\begingroup$ @JoshuaSalazar Typo, fixed. $\endgroup$
    – J. Murray
    Oct 15, 2017 at 3:48
  • $\begingroup$ I'm not so used to the inner product conjugated/non conjugated parts. Can you see what mistake did I make in my step-by-step calculations above? I think is about the conjugated integrand part you talked about $\endgroup$ Oct 15, 2017 at 3:51
  • 1
    $\begingroup$ @JoshuaSalazar I added a simple example at the end. $\endgroup$
    – J. Murray
    Oct 15, 2017 at 4:03
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    $\begingroup$ The minus sign is because you didn't do the complex conjugation initially either. You see where you say "each term is of the form ..." ? You need to complex conjugate there as well. $\endgroup$
    – J. Murray
    Oct 15, 2017 at 4:36

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