Sakura page 85: how to do this proof?

[omoplata]

From "Modern Quantum Mechanics, revised edition" by J. J. Sakurai, page 85.

Eqn (2.2.23a),$$[x_i,F(\mathbf{p})] = i \hbar \frac{\partial F}{\partial p_i}$$Eqn(2.2.23b)$$[p_i,G(\mathbf{x})] = -i \hbar \frac{\partial G}{\partial x_i}$$
It says "We can easily prove both formulas by repeatedly applying (1.6.50e)."

(1.6.50e) is,$$[A,BC]=[A,B]C+B[A,C]$$

I can't figure it out. How do I do this?

 

[jeppetrost]

Say you taylor expand the function. Then you just need to figure out what is e.g. [x,p^n]

 

[Leandro Tabak]

Hi, in the case of

\begin{bmatrix}x,F(p)\end{bmatrix}

for the unidimensional variable x for example, you must write $\textit{F(p)}$ as an expansion.

Then,

F(P) = sum Cn.p^n

Then apply the derivate to the expansion and replace in the commutator.

Keep trying and let us know if you could.

Regards

 

[omoplata]

Thanks for the help.

Like this?
$$F(\mathbf{p}) = F(\mathbf{0}) + \Sigma_{n=1}^\infty \left( \frac{\partial^n F(\mathbf{p})}{\partial p^n} \right)_{\mathbf{p} = \mathbf{0}} \frac{\mathbf{p}^n}{n!}$$
Then,$$\left[ x_i, F(\mathbf{p}) \right] = \left[ x_i, F(\mathbf{0}) + \Sigma_{n=1}^\infty \left( \frac{\partial^n F(\mathbf{p})}{\partial p^n} \right)_{\mathbf{p} = \mathbf{0}} \frac{\mathbf{p}^n}{n!} \right] = \left[ x_i, \Sigma_{n=1}^\infty \left( \frac{\partial^n F(\mathbf{p})}{\partial p^n} \right)_{\mathbf{p} = \mathbf{0}} \frac{\mathbf{p}^n}{n!} \right]$$, and ignore second order and higher terms, and take $\left( \frac{\partial F(\mathbf{p})}{\partial p} \right)_{\mathbf{p} = \mathbf{0}}$ to be a constant?

 

[Fightfish]

and ignore second order and higher terms, and take $\left( \frac{\partial F(\mathbf{p})}{\partial p} \right)_{\mathbf{p} = \mathbf{0}}$ to be a constant?

Why would you ignore the higher order terms? You will need to work out $[x,p^{n}]$, as jeppetrost mentioned earlier.
Isn't $\left( \frac{\partial F(\mathbf{p})}{\partial p} \right)_{\mathbf{p} = \mathbf{0}}$ a constant to begin with?

 

[Fredrik]

You use (1.6.50e) to find $[x,p^n]$. The proper way is to do it explicitly for the first few values of n, then guess the formula for arbitrary n, and finally prove by induction that your guess is correct.

 

[omoplata]

OK, I almost got it this time.
$$\begin{eqnarray} \left[ x_i, F(\mathbf{p}) \right] & = & \left[ x_i, F(\mathbf{0}) + \Sigma_{n=1}^\infty \left( \frac{\partial^n F(\mathbf{p})}{\partial p^n} \right)_{\mathbf{p} = \mathbf{0}} \frac{\mathbf{p}^n}{n!} \right]\\ & = & \left[ x_i, \Sigma_{n=1}^\infty \left( \frac{\partial^n F(\mathbf{p})}{\partial p^n} \right)_{\mathbf{p} = \mathbf{0}} \frac{\mathbf{p}^n}{n!} \right]\\ & = & \Sigma_{n=1}^\infty \left( \frac{\partial^n F(\mathbf{p})}{\partial p^n} \right)_{\mathbf{p} = \mathbf{0}} \frac{1}{n!} \left[x_i, \mathbf{p}^n \right] \end{eqnarray}$$
Then,
$$\left[x_i,\mathbf{p}^n \right] = \left[x_i,\mathbf{p} \cdot \mathbf{p}^{n-1} \right] = \left[x_i,\mathbf{p} \right] \mathbf{p}^{n-1} + x_i \left[\mathbf{p}, \mathbf{p}^{n-1} \right] = i \hbar \mathbf{p}^{n-1}$$
Plugging this back in,
$$\begin{eqnarray} \left[ x_i, F(\mathbf{p}) \right] & = & i \hbar \Sigma_{n=1}^\infty \left( \frac{\partial^n F(\mathbf{p})}{\partial p^n} \right)_{\mathbf{p} = \mathbf{0}} \frac{\mathbf{p}^{n-1}}{n!}\\ & = & i \hbar \Sigma_{n=1}^\infty \left( \frac{\partial^{n-1} \left( \frac{ \partial F(\mathbf{p})}{\partial p} \right) } {\partial p^{n-1}} \right)_{\mathbf{p} = \mathbf{0}} \frac{\mathbf{p}^{n-1}}{n!}\\ & = & i \hbar \Sigma_{n=0}^\infty \left( \frac{\partial^{n} \left( \frac{ \partial F(\mathbf{p})}{\partial p} \right) } {\partial p^{n}} \right)_{\mathbf{p} = \mathbf{0}} \frac{\mathbf{p}^{n}}{(n+1)!} \end{eqnarray}$$
If I somehow changed that (n+1) to an n, that would be the answer. Where did I go wrong?

 

[Fredrik]

$$\left[x_i,\mathbf{p}^n \right] = \left[x_i,\mathbf{p} \cdot \mathbf{p}^{n-1} \right] = \left[x_i,\mathbf{p} \right] \mathbf{p}^{n-1} + x_i \left[\mathbf{p}, \mathbf{p}^{n-1} \right]$$

That last term is wrong. Look at (1.6.50e) again.

 

[MisterX]

I'm pretty sure there are other, easier, ways to solve this problem. For example one could consider the action of the commutator on the wavefunction in momentum space for Eqn (2.2.23a), and in position space for Eqn (2.2.23b).

$$[p_i,G(\mathbf{x})]\psi \leftrightarrow -i\hbar \frac{\partial }{\partial x_i} \left( G(\mathbf{x})\psi(\mathbf{x})\right) - G(\mathbf{x}) \left( -i\hbar \frac{\partial }{\partial x_i}\psi(\mathbf{x}) \right) = -i\hbar\left.\frac{\partial G}{\partial x_i}\right|_{\mathbf{x}} \psi(\mathbf{x})$$

There is no loss of rigor doing it this way. From the result above $[p_i,G(\mathbf{x})]$ clearly has the $\mathbf{x}$ eigenbasis with eigenvalues $-i\hbar\left.\frac{\partial G}{\partial x_i}\right|_{\mathbf{x}}$.

 

[Fredrik]

That's an interesting approach MisterX. I think it requires a deeper understanding of what we're allowed to do, so I'm not sure I can agree that it's easier (for a typical student of Sakurai), but your solution is certainly much shorter.

 

[omoplata]

That last term is wrong. Look at (1.6.50e) again.

Yeah, thanks. Correcting that,
$$\begin{eqnarray} \left[x_i,\mathbf{p}^n \right] & = & \left[x_i,\mathbf{p} \cdot \mathbf{p}^{n-1} \right] = \left[x_i,\mathbf{p} \right] \mathbf{p}^{n-1} + \mathbf{p} \left[x_i, \mathbf{p}^{n-1} \right]\\ \left[x_i,\mathbf{p}^{n-1} \right] & = & \left[x_i,\mathbf{p} \cdot \mathbf{p}^{n-2} \right] = \left[x_i,\mathbf{p} \right] \mathbf{p}^{n-2} + \mathbf{p} \left[x_i, \mathbf{p}^{n-2} \right]\\ & \dots & \\ \end{eqnarray}$$
So,
$$[x_i,\mathbf{p}^n] = n i \hbar \mathbf{p}^{n-1}$$
Therefore,
$$\begin{eqnarray} \left[ x_i, F(\mathbf{p}) \right] & = & i \hbar \Sigma_{n=1}^\infty \left( \frac{\partial^n F(\mathbf{p})}{\partial p^n} \right)_{\mathbf{p} = \mathbf{0}} \frac{\mathbf{p}^{n-1}}{(n-1)!}\\ & = & i \hbar \Sigma_{n=1}^\infty \left( \frac{\partial^{n-1} \left( \frac{ \partial F(\mathbf{p})}{\partial p} \right) } {\partial p^{n-1}} \right)_{\mathbf{p} = \mathbf{0}} \frac{\mathbf{p}^{n-1}}{(n-1)!}\\ & = & i \hbar \Sigma_{n=0}^\infty \left( \frac{\partial^{n} \left( \frac{ \partial F(\mathbf{p})}{\partial p} \right) } {\partial p^{n}} \right)_{\mathbf{p} = \mathbf{0}} \frac{\mathbf{p}^{n}}{n!}\\ & = & \frac{\partial F(\mathbf{p})}{\partial p}\\ \end{eqnarray}$$
,which is the solution.

One small final question. I think $\mathbf{p}$ here is a vector. In the above, I did the same as I would do for a scalar. Would the procedure be different for a vector.

 

[Fredrik]

One small final question. I think $\mathbf{p}$ here is a vector. In the above, I did the same as I would do for a scalar. Would the procedure be different for a vector.

Yes, the Taylor expansion would more complicated, and that makes the rest of it more complicated as well. You can still use the same idea, but the notation will be a pain. The multi-index notation used here can probably reduce that pain.