MHB Calculation of the commutator of the hamiltonian and position

[Fantini]

The book calculates the commutator $[H,x_i]$ as
$$[H,x_i] = \left[ \sum_j \frac{p_j^2}{2m}, x_i \right] = \frac{2}{2m} \sum_j p_j \frac{\hbar}{i} \delta_{ij} = - \frac{i \hbar p_i}{m},$$
where the hamiltonian operator $H$ is
$$H = \sum_j \frac{{\mathbf p}_j^2}{2m_j} + V({\mathbf x}).$$
The book claims to use the property of commutators that
$$[AB,C] = A[B,C] + [A,C]B,$$
but I don't see how that applies.

 

[GJA]

Hey Fantini,

The author is using the identity

$$[AB,C] = A[B,C] + [A,C]B$$

to write the commutator $$[p_{j}^{2},x_{i}]$$ in terms of the (negative) canonical commutation relation

$$[p_{j},x_{i}] = -ih\delta_{ji}$$

Do this by taking $$A = B = p_{j}$$ and $$C = x_{i}$$.

Let me know if anything is unclear/not quite right!

 

[Fantini]

Hey Fantini,

The author is using the identity

$$[AB,C] = A[B,C] + [A,C]B$$

to write the commutator $$[p_{j}^{2},x_{i}]$$ in terms of the (negative) canonical commutation relation

$$[p_{j},x_{i}] = -ih\delta_{ji}$$

Do this by taking $$A = B = p_{j}$$ and $$C = x_{i}$$.

Let me know if anything is unclear/not quite right!

This is unclear: $$[p_j,x_i] = -i \hbar \delta_{ji}.$$ How do you show this? When I do it by the definition all I get is $$[p_j, x_i] = \frac{\hbar}{i} \left(\delta_{ij} - x_i \frac{\partial}{\partial x_j} \right).$$

 

[GJA]

Hi Again Fantini,

Knowing the canonical commutation relation is essential in Quantum Mechanics, so it's good to understand where it comes from. I'll give two answers - the first is probably what you're looking for, the second is a more "advanced" method (I don't mean more difficult, I just mean it's probably not something you've encountered yet).

1) There are some subtleties to using what you called "the definition" in your computation. In no particular order the following is all going on behind the scenes:

• In Quantum Mechanics position and momentum are operators.
  • There is a slight error in your computation of the commutator and it stems from not thinking of $$x_{i}$$ as an operator.
• You are implicitly using what is called the position basis/representation of quantum mechanics when you write $$p_{j} "=" \frac{h}{i}\frac{\partial}{\partial x_{j}}$$. Furthermore, when working in the position space basis, the operator $$x_{i}$$ is a multiplication operator (it is not simply a real coordinate); essentially $$x_{i}$$ sitting alone is meaningless, what does have meaning is something of the form $$x_{i}f({\bf x})$$, because $$x_{i}$$ is acting as an operator on $$f$$ via multiplication.
  • Note: There is no compulsory need to work in the position basis. In fact, there is something called the momentum basis/representation, and in the momentum basis $$p_{j}\neq \frac{\partial}{\partial x_{j}}$$. I only bring this up to emphasize that you are making a choice when you write $$p_{j} "="\frac{h}{i}\frac{\partial}{\partial x_{j}}$$ (i.e. choosing between position or momentum space).
• This is how we put what I've mentioned above to work to correct the commutator identity: we take an arbitrary test function $$f({\bf x})$$ and compute

$$\begin{align*}

[p_{j},x_{i}]f &= p_{j}x_{i}f-x_{i}p_{j}f\\
&=\frac{h}{i}\frac{\partial(x_{i}f)}{\partial x_{j}}-x_{i}\frac{h}{i}\frac{\partial f}{\partial x_{j}}\\
&=\delta_{ij}\frac{h}{i}f+x_{i}\frac{h}{i}\frac{\partial f}{\partial x_{j}}-x_{i}\frac{h}{i}\frac{\partial f}{\partial x_{j}}\\
&=\frac{h}{i}\delta_{ji}f\end{align*}$$Since $$f$$ was arbitrary, the commutator operator satisfies

$$[p_{j},x_{i}]=-ih\delta_{ji}$$

I could have just said that a product rule was missing from your calculation, but that would not have been any fun :p.

2) The second explanation can be found in most Intermediate Quantum Mechanics texts, and stems from the fact that momentum is what is known as the "generator of space translations."

Let me know if you're still unclear on this.

 

[Fantini]

Thank you! It's all clear now. :)

 

[Ackbach]

Hi Again Fantini,

Knowing the canonical commutation relation is essential in Quantum Mechanics, so it's good to understand where it comes from. I'll give two answers - the first is probably what you're looking for, the second is a more "advanced" method (I don't mean more difficult, I just mean it's probably not something you've encountered yet).

1) There are some subtleties to using what you called "the definition" in your computation. In no particular order the following is all going on behind the scenes:

• In Quantum Mechanics position and momentum are operators.
  • There is a slight error in your computation of the commutator and it stems from not thinking of $$x_{i}$$ as an operator.
• You are implicitly using what is called the position basis/representation of quantum mechanics when you write $$p_{j} "=" \frac{h}{i}\frac{\partial}{\partial x_{j}}$$. Furthermore, when working in the position space basis, the operator $$x_{i}$$ is a multiplication operator (it is not simply a real coordinate); essentially $$x_{i}$$ sitting alone is meaningless, what does have meaning is something of the form $$x_{i}f({\bf x})$$, because $$x_{i}$$ is acting as an operator on $$f$$ via multiplication.
  • Note: There is no compulsory need to work in the position basis. In fact, there is something called the momentum basis/representation, and in the momentum basis $$p_{j}\neq \frac{\partial}{\partial x_{j}}$$. I only bring this up to emphasize that you are making a choice when you write $$p_{j} "="\frac{h}{i}\frac{\partial}{\partial x_{j}}$$ (i.e. choosing between position or momentum space).
• This is how we put what I've mentioned above to work to correct the commutator identity: we take an arbitrary test function $$f({\bf x})$$ and compute

$$\begin{align*}

[p_{j},x_{i}]f &= p_{j}x_{i}f-x_{i}p_{j}f\\
&=\frac{h}{i}\frac{\partial(x_{i}f)}{\partial x_{j}}-x_{i}\frac{h}{i}\frac{\partial f}{\partial x_{j}}\\
&=\delta_{ij}\frac{h}{i}f+x_{i}\frac{h}{i}\frac{\partial f}{\partial x_{j}}-x_{i}\frac{h}{i}\frac{\partial f}{\partial x_{j}}\\
&=\frac{h}{i}\delta_{ji}f\end{align*}$$Since $$f$$ was arbitrary, the commutator operator satisfies

$$[p_{j},x_{i}]=-ih\delta_{ji}$$

I could have just said that a product rule was missing from your calculation, but that would not have been any fun :p.

2) The second explanation can be found in most Intermediate Quantum Mechanics texts, and stems from the fact that momentum is what is known as the "generator of space translations."

Let me know if you're still unclear on this.

Normally, I wouldn't clutter up a thread with "useless" posts, but I just have to comment on this post: this is an incredibly lucid, well-thought-out presentation. Excellent work, GJA!