# Time Derivative of Expectation Value of Position

• Matt Chu
In summary: If you can't help the person you're talking to, you're not helping them, are you?I think the problem is that you are trying to integrate by parts twice. The second time you should not integrate the second term by parts, but instead use the fact that ##\frac{\partial \Psi}{\partial x}## is a real function, so that you can multiply it to get rid of the complex conjugate.
Matt Chu

## 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?

Last edited:
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.

kuruman said:
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.

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.
Matt Chu said:
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.

Matt Chu said:
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.

Matt Chu said:

## 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.

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

bobob said:
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##.

PeroK said:
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.

Matt Chu said:

## 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.

## 1. What is the meaning of the "time derivative" in the context of the expectation value of position?

The time derivative is a measure of how quickly the expectation value of position is changing over time. It tells us the rate of change of the average position of a particle in a quantum mechanical system.

## 2. How is the time derivative of the expectation value of position calculated?

The time derivative of the expectation value of position is calculated using the Schrödinger equation, which describes how the wavefunction of a quantum system evolves over time. The expectation value is given by the integral of the position operator multiplied by the wavefunction, and the time derivative is then found by differentiating this expression with respect to time.

## 3. Can the time derivative of the expectation value of position be negative?

Yes, the time derivative of the expectation value of position can be negative. This means that the particle is moving in the opposite direction to the positive direction of the coordinate axis, which is often the x-axis.

## 4. How does the time derivative of the expectation value of position relate to the uncertainty principle?

The time derivative of the expectation value of position is related to the uncertainty principle through the Heisenberg uncertainty principle. This principle states that the product of the uncertainty in position and the uncertainty in momentum is always greater than or equal to the reduced Planck constant divided by 2. This means that as the time derivative of the expectation value of position becomes more well-defined, the uncertainty in momentum must increase.

## 5. What implications does the time derivative of the expectation value of position have for quantum mechanical systems?

The time derivative of the expectation value of position has important implications for quantum mechanical systems. It allows us to predict the movement of particles in a system and understand how their positions change over time. It also reveals information about the momentum and energy of the particles, which are key quantities in quantum mechanics.

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