### The Wave Function 04

** — 1.5. Momentum and other Dynamical quantities — **

Let us suppose that we have a particle that is described by the wave function $ {\Psi}$ then the expectation value of its position is (as we saw in The Wave Function 02):

$ \displaystyle <x>=\int_{-\infty}^{+\infty}x|\Psi(x,t)|^2\, dx $

Neophytes interpret the previous equations as if it was saying that the expectation value coincides with the average of various measurements of the position of a particle that is described by $ {\Psi}$. This interpretation is wrong since the first measurement will make the wave function collapse to the value that is actually obtained and if the following measurements of the position are done right away they'll just be of the same value of the first measurement.

Actually $ {<x>}$ is the average of position measurements of particles that are all described by the state $ {\Psi}$. That is to say that we have two ways of actually accomplishing what is implied by the previous interpretation of $ {<x>}$:

- We have a single particle. Then after a position measurement is made we have to able to make the particle to return to its $ {\Psi}$ state before we make a new measurement.
- We have a collection - a statistical ensemble is a more respectable name - of a great number of particles (in order for it to be statistically significant) and we arrange them all to be in state $ {\Psi}$. If we perform the measurement of the position of all this particles, then average of the measurements should be $ {<x>}$.

To put it more succinctly:

The expectation value is the average of repeated measurements on an ensemble of identically prepared systems.

Since $ {\Psi}$ is a time dependent mathematical object it is obvious that $ {<x>}$ also is a time dependent quantity:

$ {\begin{aligned} \dfrac{d<x>}{dt}&= \int_{-\infty}^{+\infty}x\dfrac{\partial}{\partial t}|\Psi|^2\, dx \\ &= \dfrac{i\hbar}{2m}\int_{-\infty}^{+\infty}x\dfrac{\partial}{\partial x}\left( \Psi^*\dfrac{\partial \Psi}{\partial x}-\dfrac{\partial \Psi^*}{\partial x}\Psi \right)\, dx \\ &= -\dfrac{i\hbar}{2m}\int_{-\infty}^{+\infty}\left( \Psi^*\dfrac{\partial \Psi}{\partial x}-\dfrac{\partial \Psi^*}{\partial x}\Psi \right)\,dx \\ &= -\dfrac{i\hbar}{m}\int_{-\infty}^{+\infty}\left( \Psi^*\dfrac{\partial \Psi}{\partial x}\right)\,dx \end{aligned}}$

where we have used integration by parts and the fact that the wave function has to be square integrable which is to say that the function is vanishingly small as $ {x}$ approaches infinity.

(Allow me to go on a tangent here but I just want to say that rigorously speaking the Hilbert space isn't the best mathematical space to construct the mathematical formalism of quantum mechanics. The problem with the Hilbert space approach to quantum mechanics is two fold:

- the functions that are in Hilbert space are necessarily square integrable. The problem is that many times we need to calculate quantities that depend not on a given function but on its derivative (for example), but just because a function is square integrable it doesn't mean that its derivative also is. Hence we don't have any mathematical guarantee that most of the integrals that we are computing actually converge.
- The second problem is that when we are dealing with continuous spectra (later on we'll see what this means) the eigenfunctions (we'll see what this means) are divergent

The proper way of doing quantum mechanics is by using rigged Hilbert spaces. A good first introduction to rigged Hilbert spaces and their use in Quantum Mechanics is given by Rafael de la Madrid in the article The role of the rigged Hilbert space in Quantum Mechanics )

The previous equation doesn't express the average velocity of a quantum particle. In our construction of quantum mechanic nothing allows us to talk about the velocity of particle. In fact we don't even know what the meaning of

velocity of a particle

is in quantum mechanics!

Since a particle doesn't have a definitive position prior to is measurement it also can't have a well defined velocity. Later on we'll see how how to construct the probability density for velocity in the state $ {\Psi}$.

For the purposes of the present section we'll just postulate that the expectation value of the velocity is equal to the time derivative of the expectation value of position.

$ \displaystyle <v>=\dfrac{d<x>}{dt} \ \ \ \ \ (24)$

As we saw in the lagrangian formalism and in the hamiltonian formalism posts of our blog it is more customary (since it is more powerful) to work with momentum instead of velocity. Since $ {p=mv}$ the relevant equation for momentum is;

Since $ {x}$ represents the position operator operator we can say in an analogous way that

$ \displaystyle \frac{\hbar}{i}\frac{\partial}{\partial x}$

represents the momentum operator. A way to see why this definition makes sense is to rewrite the definition of the expectation value of the position

$ \displaystyle <x>=\int \Psi^* x \Psi \, dx$

and to rewrite equation 25 in a more compelling way

$ \displaystyle <p> = \int \Psi^*\left( \frac{\hbar}{i}\frac{\partial}{\partial x} \right) \Psi \, dx$

After knowing how to calculate the expectation value of these two dynamical quantities the question now is how can one calculate the expectation value of other dynamical quantities of interest?

The thing is that all dynamical quantities can be expressed as functions of of $ {x}$ and $ {p}$. Taking this into account one just has to write the appropriate function of the quantity of interest in terms of $ {p}$ and $ {x}$ and then calculate the expectation value.

In a more formal (hence more respectable) way the equation for the expectation value of the dynamical quantity $ {Q=Q(x,p)}$ is

As an example let us look into what would be the relevant expression for the kinetic energy the relevant definition can be found at Newtonian Mechanics 01. Henceforth we'll use $ {T}$ to denote the kinetic energy instead of $ {K}$ in order to use the same notation that is used in Introduction to Quantum Mechanics (2nd Edition).

$ \displaystyle T=\frac{1}{2}mv^2=\frac{p^2}{2m} $

Hence the expectation value is

Exercise 3 Why can't you do integration by parts directly on
$ \displaystyle \frac{d<x>}{dt}=\int x\frac{\partial}{\partial t}|\Psi|^2 \, dx$ pull the time derivative over onto $ {x}$, note that $ {\partial x/\partial t=0}$ and conclude that $ {d<x>/dt=0}$? Because integration by parts can only be used when the differentiation and integration are done with the same variable. |

Exercise 4 Calculate
$ \displaystyle \frac{d<p>}{dt}$ First lets us remember the the Schroedinger equation: And its complex conjugate for the time evolution of the expectation value of momentum is $ {\begin{aligned} \dfrac{d<p>}{dt} &= \dfrac{d}{dt}\int\Psi ^* \dfrac{\hbar}{i}\dfrac{\partial \Psi}{\partial x}\, dx\\ &= \dfrac{\hbar}{i}\int \dfrac{\partial}{\partial t}\left( \Psi ^* \dfrac{\partial \Psi}{\partial x}\right)\, dx\\ &= \dfrac{\hbar}{i}\int\left( \dfrac{\partial \Psi^*}{\partial t}\dfrac{\partial \Psi}{\partial x}+\Psi^* \dfrac{\partial}{\partial x}\dfrac{\partial \Psi}{\partial t} \right) \, dx\\ &= \dfrac{\hbar}{i}\int \left[ \left( -\dfrac{i\hbar}{2m}\dfrac{\partial^2\Psi^*}{\partial x^2}+\dfrac{i}{\hbar}V\Psi^* \right)\dfrac{\partial \Psi}{\partial x} + \Psi^*\dfrac{\partial}{\partial x}\left( \dfrac{i\hbar}{2m}\dfrac{\partial^2\Psi}{\partial x^2}-\dfrac{i}{\hbar}V\Psi \right)\right]\, dx\\\ &= \dfrac{\hbar}{i}\int \left[ -\dfrac{i\hbar}{2m}\left(\dfrac{\partial^2\Psi^*}{\partial x^2}\dfrac{\partial\Psi}{\partial x}-\Psi^*\dfrac{\partial ^3 \Psi}{\partial x^3} \right)+\dfrac{i}{\hbar}\left( V\Psi ^*\dfrac{\partial\Psi}{\partial x}-\Psi ^*\dfrac{\partial (V\Psi)}{\partial x}\right)\right]\, dx \end{aligned}}$ First we'll calculate the first term of the integral (ignoring the constant factors) doing integration by parts (remember that the boundary terms are vanishing) two times $ {\begin{aligned} \int \left(\dfrac{\partial^2\Psi^*}{\partial x^2}\dfrac{\partial\Psi}{\partial x}-\Psi^*\dfrac{\partial ^3 \Psi}{\partial x^3}\right)\, dx &= \left[ \dfrac{\partial \Psi^*}{\partial x^2} \dfrac{\partial \Psi}{\partial x}\right]-\int\dfrac{\partial \Psi^*}{\partial x}\dfrac{\partial ^2 \Psi}{\partial x^2}\, dx- \int \Psi^*\dfrac{\partial ^3 \Psi}{\partial x^3}\, dx \\ &=-\left[ \Psi ^*\dfrac{\partial ^2 \Psi}{\partial x^2} \right]+\int \Psi^*\dfrac{\partial ^3 \Psi}{\partial x^3}\, dx - \int \Psi^*\dfrac{\partial ^3 \Psi}{\partial x^3}\, dx \\ &= 0 \end{aligned}}$ Then we'll calculate the second term of the integral $ {\begin{aligned} \int \left( V\Psi ^*\dfrac{\partial\Psi}{\partial x}-\Psi ^*\dfrac{\partial (V\Psi)}{\partial x} \right)\, dx &= \int \left( V\Psi ^*\dfrac{\partial\Psi}{\partial x}-\Psi ^* \dfrac{\partial V}{\partial x}\Psi-\Psi ^*V\dfrac{\partial \Psi}{\partial x} \right)\, dx\\ &= -\int\Psi ^* \dfrac{\partial V}{\partial x}\Psi\, dx\\ &=<-\dfrac{\partial V}{\partial x}> \end{aligned}}$ In conclusion it is $ \displaystyle \frac{d<p>}{dt}=<-\dfrac{\partial V}{\partial x}> \ \ \ \ \ (30)$ Hence the expectation value of the momentum operator obeys Newton's Second Axiom. The previous result can be generalized and its generalization is known in the Quantum Mechanics literature as Ehrenfest's theorem |

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