# Time Derivative of Expectation Value of Position

## Homework Statement

I want to prove that ##\frac{\partial \langle x \rangle}{\partial t} = \frac{\langle p_x \rangle}{m}##.

## Homework Equations

$$i\hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2} + V \Psi$$

## The Attempt at a Solution

[/B]
So given the expectation value of position,

$$\langle x \rangle = \int_{-\infty}^\infty \Psi^* x \Psi \ dx$$

I'm trying to show that the time derivative of this is equal to ##\frac{ \langle p_x \rangle}{m}##.

I started by using the product rule, which gave:

$$\frac{\partial \langle x \rangle}{\partial t} = \int_{-\infty}^\infty \left[ \Psi^* x \frac{\partial \Psi}{\partial t} + \frac{\partial \Psi^*}{\partial t} x \Psi \right] dx$$

Then, using the time-dependent Schrodinger equation:

$$\frac{\partial \langle x \rangle}{\partial t} = \frac{1}{i\hbar} \int_{-\infty}^\infty \Psi^* x (H \Psi) dx - \frac{1}{i\hbar} \int_{-\infty}^\infty (H \Psi)^* x \Psi dx$$
$$= \frac{1}{i\hbar} \int_{-\infty}^\infty \Psi^* x \left( - \frac{\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2} + V \Psi \right) dx - \frac{1}{i\hbar} \int_{-\infty}^\infty \left( - \frac{\hbar^2}{2m} \frac{\partial^2 \Psi^*}{\partial x^2} + V \Psi^* \right) x \Psi dx$$

(The ##V## components cancel out.)

$$= -\frac{i \hbar}{2m} \int_{\infty}^\infty \left[ -\Psi^* x \frac{\partial^2 \Psi}{\partial x^2} \right] dx + \frac{i\hbar}{2m} \int_{-\infty}^\infty \left[\frac{\partial^2 \Psi^*}{\partial x^2} x \Psi \right] dx$$

I then tried integrating by parts, which gives:

$$= \frac{i\hbar}{2m} \left( \left. -x\Psi^* \frac{\partial \Psi}{\partial x} \right|_{-\infty}^\infty + \int_{-\infty}^\infty \Psi^* \frac{\partial \Psi}{\partial x} dx + \int_{\infty}^\infty x \left| \frac{\partial \Psi}{\partial x} \right|^2 dx \right) + \frac{i\hbar}{2m} \left( \left. x\Psi^* \frac{\partial \Psi}{\partial x} \right|_{-\infty}^\infty - \int_{-\infty}^\infty \Psi^* \frac{\partial \Psi}{\partial x} dx - \int_{-\infty}^\infty x \left| \frac{\partial \Psi}{\partial x} \right|^2 dx \right)$$

However, at this point, every part in the above equation is real, so the equation ends up equalling zero. Where did I go wrong?

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## Answers and Replies

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kuruman
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As I recall, this is done by integrating by parts twice working on one of terms only. You will see better what's going on if you write $$x \frac{\partial \Psi}{\partial x}\frac{\partial \Psi^*}{\partial x} dx$$ after the fist parts integration. Then after the second parts integration you get a term $$x \Psi^* \frac{\partial \Psi}{\partial x} dx.$$Also, please preview and fix (if necessary) your LaTeX code before posting.

As I recall, this is done by integrating by parts twice working on one of terms only. You will see better what's going on if you write $$x \frac{\partial \Psi}{\partial x}\frac{\partial \Psi^*}{\partial x} dx$$ after the fist parts integration. Then after the second parts integration you get a term $$x \Psi^* \frac{\partial \Psi}{\partial x} dx.$$Also, please preview and fix (if necessary) your LaTeX code before posting.
I'm not exactly sure why I would integrate by parts again. Integrating by parts once already gives me zero, which is definitely not correct. There must be some mathematical or conceptual mistake that I've made because I don't think it's possible to proceed with what I've done.

kuruman
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In this expression that you have posted
$$-\frac{i \hbar}{2m} \int_{\infty}^\infty \left[ -\Psi^* x \frac{\partial^2 \Psi}{\partial x^2} \right] dx + \frac{i\hbar}{2m} \int_{-\infty}^\infty \left[\frac{\partial^2 \Psi^*}{\partial x^2} x \Psi \right] dx$$the second integral must have a negative sign in front of it. Look at the line above. The second integral has an overall positive sign in front of the ##\hbar^2## term. When you bring ##i## to the numerator you get an overall negative sign.
I'm not exactly sure why I would integrate by parts again.
You don't. I was thinking of the solution to another problem. Sorry for the confusion.

PeroK
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I then tried integrating by parts, which gives:

$$= \frac{i\hbar}{2m} \left( \left. -x\Psi^* \frac{\partial \Psi}{\partial x} \right|_{-\infty}^\infty + \int_{-\infty}^\infty \Psi^* \frac{\partial \Psi}{\partial x} dx + \int_{\infty}^\infty x \left| \frac{\partial \Psi}{\partial x} \right|^2 dx \right) + \frac{i\hbar}{2m} \left( \left. x\Psi^* \frac{\partial \Psi}{\partial x} \right|_{-\infty}^\infty - \int_{-\infty}^\infty \Psi^* \frac{\partial \Psi}{\partial x} dx - \int_{-\infty}^\infty x \left| \frac{\partial \Psi}{\partial x} \right|^2 dx \right)$$

However, at this point, every part in the above equation is real, so the equation ends up equalling zero. Where did I go wrong?
You have ended up with ##\Psi^*## in both integrals. It should be ##\Psi## in one of them. In any case, your expression should reduce to:

$$= \frac{i\hbar}{2m} \left( \int_{-\infty}^\infty \Psi \frac{\partial \Psi^*}{\partial x} dx - \int_{-\infty}^\infty \Psi^* \frac{\partial \Psi}{\partial x} dx \right)$$

And then you do hit one of the terms with integration by parts again.

bobob
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## Homework Statement

I want to prove that ##\frac{\partial \langle x \rangle}{\partial t} = \frac{\langle p_x \rangle}{m}##.
[...]
However, at this point, every part in the above equation is real, so the equation ends up equalling zero. Where did I go wrong?
First, you wrote down what you are supposed to prove, incorrectly. It should be:
##\frac{\partial \langle x \rangle}{\partial t} = \frac{\langle p \rangle}{m}##

And you needn't go through all of the integrations to do it. ##\langle x\rangle = \langle\psi|x|\psi\rangle##. Take the time derivative of both sides. You get three terms, two of which may be written as a commutator.

PeroK
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2020 Award
I want to prove that ##\frac{\partial \langle x \rangle}{\partial t} = \frac{\langle p_x \rangle}{m}##.
First, you wrote down what you are supposed to prove, incorrectly. It should be:
##\frac{\partial \langle x \rangle}{\partial t} = \frac{\langle p \rangle}{m}##

And you needn't go through all of the integrations to do it. ##\langle x\rangle = \langle\psi|x|\psi\rangle##. Take the time derivative of both sides. You get three terms, two of which may be written as a commutator.
The OP has merely emphasisd that it's ##p_x## (momentum in the x-direction). There's nothing wrong with that.

The OP is clearly looking for a wave-mechanical proof. Referring him to a linear algebra formalism that he may not yet have encountered is less than helpful.

If, however, you do want to be pedantic, then it should be an ordinary derivative ##\frac{d\langle x \rangle}{dt}##, as the expectation value is only a function of the one variable; namely, ##t##.

bobob
Gold Member
The OP has merely emphasisd that it's ##p_x## (momentum in the x-direction). There's nothing wrong with that.

The OP is clearly looking for a wave-mechanical proof. Referring him to a linear algebra formalism that he may not yet have encountered is less than helpful.

If, however, you do want to be pedantic, then it should be an ordinary derivative ##\frac{d\langle x \rangle}{dt}##, as the expectation value is only a function of the one variable; namely, ##t##.
Didn't look like a subscript when I saw it.

nrqed
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## Homework Statement

I want to prove that ##\frac{\partial \langle x \rangle}{\partial t} = \frac{\langle p_x \rangle}{m}##.

## Homework Equations

$$i\hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2} + V \Psi$$

## The Attempt at a Solution

[/B]
So given the expectation value of position,

$$\langle x \rangle = \int_{-\infty}^\infty \Psi^* x \Psi \ dx$$

I'm trying to show that the time derivative of this is equal to ##\frac{ \langle p_x \rangle}{m}##.

I started by using the product rule, which gave:

$$\frac{\partial \langle x \rangle}{\partial t} = \int_{-\infty}^\infty \left[ \Psi^* x \frac{\partial \Psi}{\partial t} + \frac{\partial \Psi^*}{\partial t} x \Psi \right] dx$$

Then, using the time-dependent Schrodinger equation:

$$\frac{\partial \langle x \rangle}{\partial t} = \frac{1}{i\hbar} \int_{-\infty}^\infty \Psi^* x (H \Psi) dx - \frac{1}{i\hbar} \int_{-\infty}^\infty (H \Psi)^* x \Psi dx$$
$$= \frac{1}{i\hbar} \int_{-\infty}^\infty \Psi^* x \left( - \frac{\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2} + V \Psi \right) dx - \frac{1}{i\hbar} \int_{-\infty}^\infty \left( - \frac{\hbar^2}{2m} \frac{\partial^2 \Psi^*}{\partial x^2} + V \Psi^* \right) x \Psi dx$$

(The ##V## components cancel out.)

$$= -\frac{i \hbar}{2m} \int_{\infty}^\infty \left[ -\Psi^* x \frac{\partial^2 \Psi}{\partial x^2} \right] dx + \frac{i\hbar}{2m} \int_{-\infty}^\infty \left[\frac{\partial^2 \Psi^*}{\partial x^2} x \Psi \right] dx$$

I then tried integrating by parts, which gives:

$$= \frac{i\hbar}{2m} \left( \left. -x\Psi^* \frac{\partial \Psi}{\partial x} \right|_{-\infty}^\infty + \int_{-\infty}^\infty \Psi^* \frac{\partial \Psi}{\partial x} dx + \int_{\infty}^\infty x \left| \frac{\partial \Psi}{\partial x} \right|^2 dx \right) + \frac{i\hbar}{2m} \left( \left. x\Psi^* \frac{\partial \Psi}{\partial x} \right|_{-\infty}^\infty - \int_{-\infty}^\infty \Psi^* \frac{\partial \Psi}{\partial x} dx - \int_{-\infty}^\infty x \left| \frac{\partial \Psi}{\partial x} \right|^2 dx \right)$$

However, at this point, every part in the above equation is real, so the equation ends up equalling zero. Where did I go wrong?
In your last expression, you have dropped the minus sign in front of the ##i \hbar/2m## in front of the first parenthesis. That's the source of the problem.