# When and why can ∂p/∂t=0 in position space?

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• George Keeling
George Keeling
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TL;DR Summary
When and why can I avoid expanding momentum as derivative operator (in position space) and treat it as an independent variable? So ∂p/∂t=0
I'm doing Griffiths & Schroeter Problem 3.37 and I would like to prove that the expectation value of a rather peculiar observable vanishes:
$$\left\langle\frac{\partial}{\partial t}\left(\hat{x}\hat{p}\right)\right\rangle=0$$My cheat sheet expands the LHS to an integral and continues …
$$\int{\Psi^\ast\frac{\partial}{\partial t}\left(xp\right)\Psi d x}=\int{\Psi^\ast\left(0\right)\Psi d x}=0$$but, along with ##\Psi\left(x,t\right)##, we are in position space so should be setting
$$\hat{p}=-i\hbar\frac{\partial}{\partial x}$$so the interesting term becomes
$$-i\hbar\frac{\partial}{\partial t}\left(x\frac{\partial\Psi}{\partial x}\right)=-i\hbar\left(\frac{\partial x}{\partial t}\frac{\partial\Psi}{\partial x}+x\frac{\partial^2\Psi}{\partial t\partial x}\right)$$
The first part vanishes because ##\frac{\partial x}{\partial t}=0## but the second doesn't.

It's a bit tempting to say that ##x,p,t## are all independent variables so that
$$\frac{\partial}{\partial t}\left(xp\right)=\frac{\partial x}{\partial t}p+x\frac{\partial p}{\partial t}=0+0$$but I don't see why you can do that in this case. Or can you? If so why?

Well, it is a weird problem setup to begin with when you are working in the Schrodinger picture. There is no time dependence in the operators so the derivative with respect to time is clearly zero. On the mathematical operations side of things, there has to be some care about what the time derivative is action on.

##\frac{\partial}{\partial t}\left(\hat{A}\left(\hat{B}f(x,t)\right)\right)##

is different than

## \left(\frac{\partial}{\partial t}\left(\hat{A}\hat{B}\right)\right)f(x,t)##

I interpret your problem as using the latter, whereas your calculation assumes the former.

Is there a Hamiltonian
George Keeling said:
TL;DR Summary: When and why can I avoid expanding momentum as derivative operator (in position space) and treat it as an independent variable? So ∂p/∂t=0

I'm doing Griffiths & Schroeter Problem 3.37 and I would like to prove that the expectation value of a rather peculiar observable vanishes:
$$\left\langle\frac{\partial}{\partial t}\left(\hat{x}\hat{p}\right)\right\rangle=0$$My cheat sheet expands the LHS to an integral and continues …
$$\int{\Psi^\ast\frac{\partial}{\partial t}\left(xp\right)\Psi d x}=\int{\Psi^\ast\left(0\right)\Psi d x}=0$$but, along with ##\Psi\left(x,t\right)##, we are in position space so should be setting
$$\hat{p}=-i\hbar\frac{\partial}{\partial x}$$so the interesting term becomes
$$-i\hbar\frac{\partial}{\partial t}\left(x\frac{\partial\Psi}{\partial x}\right)=-i\hbar\left(\frac{\partial x}{\partial t}\frac{\partial\Psi}{\partial x}+x\frac{\partial^2\Psi}{\partial t\partial x}\right)$$
The first part vanishes because ##\frac{\partial x}{\partial t}=0## but the second doesn't.

It's a bit tempting to say that ##x,p,t## are all independent variables so that
$$\frac{\partial}{\partial t}\left(xp\right)=\frac{\partial x}{\partial t}p+x\frac{\partial p}{\partial t}=0+0$$but I don't see why you can do that in this case. Or can you? If so why?
Which edition is this? I could not find it. In the 3ed there is a similar problem in 3.37 but the ##\frac{\mathrm d}{\mathrm d t}## is outside of the angle brackets and leads to the virial theorem.

pines-demon said:
Is there a Hamiltonian

Which edition is this? I could not find it. In the 3ed there is a similar problem in 3.37 but the ##\frac{\mathrm d}{\mathrm d t}## is outside of the angle brackets and leads to the virial theorem.
Mine is the third edition. We get to ##\left\langle\frac{\partial}{\partial t}\left(\hat{x}\hat{p}\right)\right\rangle=0## after using equation 3.73. Hamiltonian comes into it.

Haborix said:
I interpret your problem as using the latter, whereas your calculation assumes the former.
So I think what you are saying is that the 'interesting term' should be
$$-i\hbar\left[\frac{\partial}{\partial t}\left(x\frac{\partial}{\partial x}\right)\right]\Psi$$and so, as the part in square brackets contains no ##t##s, it vanishes when differentiated. I disagree. I would expand the square brackets as
$$\frac{\partial}{\partial t}\left(x\frac{\partial}{\partial x}\right)=\frac{\partial x}{\partial t}\frac{\partial}{\partial x}+x\frac{\partial^2}{\partial t\partial x}=x\frac{\partial^2}{\partial t\partial x}$$so I end up with the same result as I had before. Bracketing the operators is really immaterial. Just like matrix operators, they are associative.

It is fine to disagree, but it puts you at odds with standard conventions.

George Keeling
George Keeling said:
Mine is the third edition.
This is problem 3.37 from the 3ed:

The time derivative is outside the angle brackets in Eq. (3.112).
Note that it says that for a stationary state the ##\mathrm d \langle \hat x\hat p\rangle /\mathrm d t=0##, not what you wrote.
George Keeling said:
We get to ##\left\langle\frac{\partial}{\partial t}\left(\hat{x}\hat{p}\right)\right\rangle=0## after using equation 3.73. Hamiltonian comes into it.
Equation 3.73 is Ehrenfest's theorem:
$$\frac{d}{dt}\langle \hat Q\rangle = \frac{i}{\hbar}\langle [\hat H,\hat Q] \rangle+ \left\langle \frac{\partial \hat Q}{\partial t}\right\rangle$$
if you want to replace operator ##\hat Q## with ##\hat x\hat p## you have to take into account both terms in the right hand side of 3.73. If the question is "do ##\hat x## and ##\hat p## are independent of time (in Schrödinger's picture)?" the answer is yes.

Last edited:
dextercioby, Demystifier and George Keeling
Haborix said:
It is fine to disagree, but it puts you at odds with standard conventions.
To begin with I thought you meant politeness conventions and I had offended you
But on further reflection I think you mean Schrödinger picture conventions which I am having difficulty getting used to. Thanks.
I was using that famous product rule convention in calculus.

pines-demon said:
if you want to replace operator ##\hat Q## with ##\hat x\hat p## you have to take into account both terms in the right hand side of 3.73. If the question is "do ##\hat x## and ##\hat p## are independent of time (in Schrödinger's picture)?" the answer is yes.
That's exactly what I was doing so that makes ##\left\langle\frac{\partial}{\partial t}\left(\hat{x}\hat{p}\right)\right\rangle=0##

I have also now found and read Schrödinger's picture. "operators (observables and others) are mostly constant with respect to time (an exception is the Hamiltonian which may change if the potential ##V## changes)". Well I never.

I'm a bit worried that operators are mostly constant with respect to time. How can I be sure about a particular operator, except by asking you guys?

George Keeling said:
I'm a bit worried that operators are mostly constant with respect to time. How can I be sure about a particular operator, except by asking you guys?
If there is not explicit time dependence, then the operator is independent of time.

pines-demon and George Keeling
George Keeling said:
To begin with I thought you meant politeness conventions and I had offended you
But on further reflection I think you mean Schrödinger picture conventions which I am having difficulty getting used to. Thanks.
I was using that famous product rule convention in calculus.
Ha! You are perfectly polite.

I think the Schrodinger picture only make the issue simple if the derivative with respect to time is just a derivative of operators, not a derivative of the result of operators applied to a state. For example, it is the difference between:

##\frac{\partial}{\partial t}\begin{pmatrix} a & b \\ c & d \end{pmatrix}=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}##

and

##\frac{\partial}{\partial t}\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x(t) \\ y(t)\end{pmatrix}\right) =\begin{pmatrix} ax'(t) + by'(t) \\ cx'(t) + dy'(t) \end{pmatrix} ##

To me, the issue seems to be about interpreting a mathematical expression. I think you interpret the expression in your original post as the latter, whereas it should be the former.

George Keeling and pines-demon
I think the problem is due to the hats in the expectation value expression. In my book one normally has no hats: ##\left\langle Q\right\rangle## not ##\left\langle\hat{Q}\right\rangle##. ##Q## is an observable and ##\hat{Q}## is its operator. Where I started from was
$$\left\langle\frac{\partial}{\partial t}\left(\hat{x}\hat{p}\right)\right\rangle=0\ ?$$If I had started with
$$\left\langle\frac{\partial}{\partial t}\left(xp\right)\right\rangle$$then I would have got
$$\left\langle\frac{\partial}{\partial t}\left(xp\right)\right\rangle=\int{\Psi^\ast\frac{\partial}{\partial t}\left[x\left(-i\hbar\frac{\partial}{\partial x}\right)\Psi\right]dx}$$which leads to the mess I got in.

But with hats I get (following @Haborix)
$$\left\langle\frac{\partial}{\partial t}\left(\hat{x}\hat{p}\right)\right\rangle=\int{\Psi^\ast\left[\frac{\partial}{\partial t}\left(x\left(-i\hbar\frac{\partial}{\partial x}\right)\right)\right]\Psi d x}=\int{\Psi^\ast\left[0\right]\Psi d x}$$and using @Claude's rule.

I'm not sure it's even proper to expand ##\left\langle\frac{\partial}{\partial t}\left(\hat{x}\hat{p}\right)\right\rangle## perhaps one should just say that
$$\frac{\partial}{\partial t}\left(\hat{x}\hat{p}\right)=0\Rightarrow\left\langle\frac{\partial}{\partial t}\left(\hat{x}\hat{p}\right)\right\rangle=\left\langle0\right\rangle=0$$

Thanks for your patience!

George Keeling said:
I think the problem is due to the hats in the expectation value expression.
I'm not sure I understand. The only mathematical quantity involved is the operator ##\hat{Q}##. The "observable" ##Q## is not something you can do math with, so I don't understand how you could even make sense of an expression like ##\partial / \partial t (x p)## (without the hats).

PeterDonis said:
The "observable" ##Q## is not something you can do math with
The problem from the book (post #7) asked to show something about
$$\frac{d}{dt}\left\langle x p\right\rangle$$so I suppose that means an observable that is two other observables multiplied together. So it looks like Griffiths can do math with observables. After that, observing its rate of change is not a stretch?

George Keeling said:
I suppose that means an observable that is two other observables multiplied together.
I would be more inclined to suppose that the operators representing those observables are what is meant, since we know how to multiply operators, but I have no idea what it would mean to multiply observables if that doesn't mean multiplying the operators that represent them. Unfortunately I don't have a copy of Griffiths handy to check the context. But since assuming that ##xp## somehow represents something other than operators is what got you into trouble, it seems to me that the simplest way out of trouble would be to treat them as operators.

For the record, ##xp## with or without hats is not an observable, because it can't be represented directly by a selfadjoint operator, but only through a symmetrization trick

$$xp = \frac{1}{2} (xp+px)$$ at classical level, then the operator for the RHS is a representation of an observable.

pines-demon

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