San José State University

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Thayer Watkins
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 The Ehrenfest Theorem: Its Nature and Proof

In classical physics a particle may have a location and a speed. In quantum physics a particle instead has a probability density function. In physics that probability density function is expressed in a unique way; i.e., in terms of a wave function. A wave function is a complex number valued function over space φ(X), where X is the three dimensional coordinates of a point in space. The expected value E of some quantity, say potential energy, which is a function of space coordinates, V(X) is then given by

#### E{V} = ∫φ*(X)V(X)φ(X)dx³

where φ* is the complex conjugate of φ.

Because φ*φ is a probability density function it follows that

#### ∫φ*φdx³ = 1

In physics there is a special notation for the expected value of a quantity V; i.e., <V>.

## The Schrödinger Equation for the Wave Function

The reason for expressing the probability density function in the esoteric fashion of a wave function is that the wave function is determined by a special differential equation, the Schrödinger equation; i.e.,

#### ∂φ/∂t = (1/ih)Hφ

where t is time, i is the imaginary complex unit (−1)½, h is Planck's constant divided by 2π and H is the Hamiltonian operator for the system. The Hamiltonian operator is derived from the Hamiltonian function for the system. The Hamiltonian function for a system is its total energy expressed in terms of its momenta and state variables. The Hamiltonian operator is derived from the function by expressing a momentum in terms of the partial derivative with respect to its state variable. More precisely a momentum p is replaced with ih(∂/∂x) where x is the state variable correponding to the momentum p. Furthermore the exponential power of a momentum becomes the order of the partial derivative. For example, a free particle in space has only kinetic energy. That kinetic energy can be expressed as p²/(2m), where p is the linear momentum of the particle and m is its mass. Thus the Hamiltonian function is p²/(2m) and p=m∂x/∂t, the Hamiltonian operator is

#### H = (ih)²(∂²/∂x²)/(2m) which reduces to H = −h²(∂²/∂x²)/(2m)

The Schrödinger equation for a free particle is then

## The Ehrenfest Theorem

Let V now stand for any quantity for a quantum system. The value of V and the value of the wave function at points in space may change with time. The total derivative of the expected value of V with respect to time is the change in due to both changes in V(X,t) and changes in the wave function φ(X,t) with respect to time. This is expressed as d/dt.

There is a special function for the quantity V and the Hamiltonian operator H of the system. It is called the commutator of V and is defined as

#### [V, H]φ = VHφ − H(Vφ) which is usually expressed as [V, H] = VH − HV

For example, for the free particle H=ih(∂²/∂x²)/(2m) and thus

#### [V, H]φ = ihV(∂²φ/∂x²)/(2m) − ih(∂²(Vφ)/∂x²)/(2m) which reduces to [V, H]φ = −ih{(∂²V/∂x²)φ/(2m) + 2(∂V/∂x)(∂φ/∂x)/(2m)}

The Ehrenfest Theorem is then

#### d <V>/dt = <(∂V/∂t)> + (1/ih)<[V, H]>

This says that the time rate of change of the expected value of V is the expected value of the time rate of change of V plus the expected value of the commutator of V and H divided by ih.

Proof:

By definition

#### d <V>/dt = d/dt ∫φ*Vφdx³ = ∫((∂φ*/∂t))Vφdx³ + ∫φ*((∂V/∂t))dx³ + ∫φ*V((∂φ/∂t))dx³ which reduces to = ∫((∂φ*/∂t))Vφdx³ + <(∂V/∂t)> + ∫φ*V((∂φ/∂t))dx³

By the Schrödinger equation ∂φ/∂t = (1/ih)Hφ and therefore

#### ∂φ*/∂t = (−1/ih)(Hφ)* which reduces to ∂φ*/∂t = (−1/ih)φ*H*

Since H involves no imaginary terms H*=H. Thus the previous relation reduces further to

Thus

#### ∫((∂φ*/∂t))Vφdx³ = (−1/ih)∫φ*HVφdx³ and ∫φ*V((∂φ/∂t))dx³ = (1/ih)∫φ*VHφdx³

When these two expressions are substituted into the equation for d <V>/dt the result is

## Illustration of the Theorem

Consider a particle in a vertical gravitational field such that the potential energy is gz, where z is vertical distance and is a parameter giving the strength of the field. The Hamiltonian function H is then

#### H = p²/(2m) + gz

where m is the mass of the particle and p is the momentum. The linear momentum p is given by m(dz/dt). Thus the Hamiltonian operator H is

#### H = ih(∂²/∂z²)/(2m) + gz

Suppose the quantity of interest is z. Then the commutator of z and H is given by

#### zHφ − H(zφ) = z(ih(∂²φ/∂z²)/(2m) + gzφ) − ih(∂²/∂z²)/(2m) + gz)(zφ) which reduces to 2ih(∂φ/∂z) + gz²φ − gz²φ and obviously further to 2ih(∂φ/∂z)

Thus the commutator of z and H is 2(∂/∂z). The expected value of the commutator is

#### <[z, H]> = ∫φ*(∂φ/∂z)dz

(To be continued.)