# Momentum operator action [duplicate]

I have a density matrix defined as

$$\rho(t) = \int_{-\infty}^{+\infty} \sigma(t,x) \otimes |x\rangle\langle x|$$

I want to show that $$\hat{P} \rho(t) \hat{P} - \frac{1}{2} (\hat{P^2} \rho(t) - \rho(t) \hat{P^2}) = 2\frac{\partial^2 }{\partial x^2} \sigma(t,x)$$ where $\hat{P}$ is the momentum operator.

The question is similar to Density Operator, Expectation Value, Coherent States, but this doesn't help me in getting the answer. $\hat{P} = -i \frac{\partial}{\partial x}$ in the position basis, but I am unable to get the final answer. Can someone help me?

## marked as duplicate by Emilio Pisanty, Jon Custer, ZeroTheHero, sammy gerbil, John Rennie quantum-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 11 '17 at 6:05

Your question is so malformed it is hard to rectify into something meaningful. In any case, this is my attempt. It is basically but an exercise of appreciating the bra-ket notation. I doubt there is anything special about a density matrix. The direct product symbol is misplaced and possibly obstructive.

You may mean $$\rho=\int dx ~ |x\rangle \sigma(x) \langle x| ~.$$ Recall what you expressed in elliptical language, $$P= \int dx ~ |x\rangle \frac{\partial_x}{i} \langle x| ~,$$ as well as $\langle x|y\rangle=\delta(x-y)$, and $i\langle x|P|y\rangle=\partial_x\delta(x-y)=-\partial_y \delta(x-y)$. (Use the narrowing Gaussian limit to remind yourself of this.)

It is then evident that $$[P,\rho]= \int dx dy ~ \left(|y\rangle \frac{\partial_y}{i} \langle y| x\rangle \sigma(x)\langle x| -|y\rangle \sigma(y) \langle y| x\rangle \frac{\partial_x}{i} \langle x|\right )=\\ \int dx dy ~ \left(|y\rangle \frac{\partial_y \delta(x-y)}{i} \sigma(x)\langle x| -|y\rangle \sigma(y) \delta(x-y) \frac{\partial_x}{i} \langle x|\right )=\\ \int dx ~ \left(|x\rangle \frac{\partial_x}{i} ~ \sigma(x)\langle x| -|x\rangle \sigma(x) \frac{\partial_x}{i} \langle x|\right )=\int dx ~ |x\rangle \frac{ \sigma'(x)}{i}\langle x| ~.$$ To get from the 2nd to the 3rd line, you've integrated out of $-\partial_x \delta(x-y)$ by parts. Appreciate the first derivative there (3rd line) acts on everything to its right, so its action on the bra is cancelled by the second term!

Iterating this move once again yields $$[P,[P,\rho]]=-\int dx ~ |x\rangle \sigma''(x)\langle x|= -2P\rho P+ P^2\rho+\rho P^2,$$ not quite your expression, but symmetry considerations may convince you that your original desideratum expression was/is wrong. (Hint: consider the classical limit after reinstating the $\hbar$s.)

Note added in response to comment: The above linchpin expression $i\langle x|P|y\rangle=\partial_x\delta(x-y)=-\partial_y \delta(x-y)$ is a consequence of the characteristic Hermitian P action, namely $\langle x| P= -i\partial_x \langle x|$, hence $P|x\rangle=i\partial_x|x\rangle$.

Note in response to further comment by @Jyothi : This is at the heart of the bra-ket notation, which I assume you have appreciated--I can't quite give you a tutorial on it (competing with Dirac's breathtakingly superb text...). If you can start from $P \psi(x)=-i\partial_x \psi(x)$, choosing $|\psi \rangle=|y\rangle$, you simply have the above matrix elements $i\langle x| P| y\rangle=\partial_x\delta(x-y)=-\partial_y \delta(x-y)$, as stated.

Now multiply this on the left by $|x\rangle$ and on the right by $\langle y|$, and integrate over x and y, and then integrate by parts to get $$iP=\int dx dy |x\rangle \langle x|i P | y\rangle \langle y| =\int dx dy ~ |x\rangle (-\partial_y \delta (x-y) ) \langle y| \\=\int dx dy ~ |x\rangle \delta (x-y)\partial_y \langle y| = \int dx ~ |x\rangle \partial_x \langle x| ~.$$ This is essentially a mere explanation of what the notation actually means: it transitions from the matrix bra-ket representation to functions-of-x representation, and vice versa. The reason of its popularity is that it is so compact and intuitive, once you learn the language.

• @CosmasZachos, how do you write P as \int dx |x>-i\partial_x <x| ? – ABCD Aug 10 '17 at 9:32
• Thanks a lot. Given the density matrix as $$\rho(t) = \int_{-\infty}^{+\infty} \sigma(t,x) \otimes |x\rangle\langle x|$$ and $\frac{d}{dt}\rho(t) = i\gamma (\hat{P}\rho\sigma_{x} - \sigma_{x}\rho\hat{P})$ where $\hat{P}$ is the momentum operator and $\sigma_x$ is the Paauli matrix, how can one show that $\frac{\partial}{\partial t}\sigma(t,x) = -\gamma(\sigma_x\frac{\partial \sigma(t,x) }{\partial x}+\frac{\partial \sigma(t,x) }{\partial x} \sigma_x)?$ Following the above prescription doesn't lead to this. Could you please help with this? – ABCD Aug 11 '17 at 7:28
• I have no clue what you are talking about... That's why I invited the OP to rewrite his question clearly. Where did this Pauli matrix thing come from? It appears to me σ(x,t) is just a scalar function representing the operator ρ in coordinate space. You want to bypass the OP's intent and interpret it as a noncommutative matrix? Where did this bit on time derivatives pop up? Are we talking about the same question, or another one you wish to ask? – Cosmas Zachos Aug 11 '17 at 15:01
• The question is based on a density matrix in eq.3 in the paper arxiv.org/abs/1402.3253. $\rho_i$ in this paper is written as $\sigma(x,t)$ and $|x\rangle$ takes the role of $|i\rangle$ in the paper. Now, the equation of motion of density matrix is written down and thats how Pauli matrices enter the scene. Then taking the density matrix as with the above structure, how to write the equation of motion for $\sigma(x,t)$ is the question. – ABCD Aug 12 '17 at 10:06
• But that's a completely different question, so far unasked. Rather than expecting an answer in the comments of another question, why don't you ask it? But without a hamiltonian and clear indication of what on earth you are talking about, I don't see much that could be thought about. – Cosmas Zachos Aug 12 '17 at 13:17