# Sakura page 85: how to do this proof?

1. Apr 29, 2013

### 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?

2. Apr 29, 2013

### jeppetrost

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

3. Apr 29, 2013

### 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

4. Apr 29, 2013

### 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?

5. Apr 29, 2013

### Fightfish

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?

6. Apr 29, 2013

### Fredrik

Staff Emeritus
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.

7. Apr 29, 2013

### 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?

8. Apr 29, 2013

### Fredrik

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

9. Apr 29, 2013

### 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 $-i\hbar\left.\frac{\partial G}{\partial x_i}\right|_{\mathbf{x}}$.

Last edited: Apr 29, 2013
10. Apr 29, 2013

### Fredrik

Staff Emeritus
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.

11. Apr 30, 2013

### omoplata

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.

12. Apr 30, 2013

### Fredrik

Staff Emeritus
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.

Last edited: Apr 30, 2013