Schrödinger's wave-equation

Let us now consider the motion of a particle in three dimensions in the Schrödinger picture. The fixed dynamical variables of the system are the position operators ${\bf x}\equiv (x_1, x_2, x_3)$, and the momentum operators ${\bf p}\equiv (p_1, p_2, p_3)$. The state of the system is represented as some time evolving ket $\vert At\rangle$.

Let $\vert{\bf x'}\rangle$ represent a simultaneous eigenket of the position operators belonging to the eigenvalues ${\bf x'} \equiv (x_1', x_2', x_3')$. Note that, since the position operators are fixed in the Schrödinger picture, we do not expect the $\vert{\bf x}'\rangle$ to evolve in time. The wave-function of the system at time $t$ is defined

$$\psi({\bf x'}, t) = \langle {\bf x'}\vert At\rangle.$$ (266)

The Hamiltonian of the system is taken to be

$$H = \frac{{\bf p}^2}{2 m} + V({\bf x}).$$ (267)

Schrödinger's equation of motion (228) yields

$${\rm i} \hbar \frac{\partial \langle {\bf x'}\vert At\rangle}{\partial t} = \langle {\bf x'}\vert H\vert At\rangle,$$ (268)

where use has been made of the time independence of the $\vert{\bf x'}\rangle$. We adopt Schrödinger's representation in which the momentum conjugate to the position operator $x_i$ is written [see Eq. (164)]

$$p_i = -{\rm i} \hbar\frac{\partial}{\partial x_i}.$$ (269)

Thus,

$$\left\langle {\bf x}'\left\vert\frac{{\bf p}^2}{2 m}\right\... ...bar^2}{2 m}\right) \nabla'^2\langle {\bf x'} \vert At\rangle,$$ (270)

where use has been made of Eq. (168). Here, $\nabla'\equiv(\partial/\partial x', \partial/\partial y', \partial/\partial z')$ denotes the gradient operator written in terms of the position eigenvalues. We can also write

$$\langle {\bf x'}\vert V({\bf x}) = V({\bf x'})\langle {\bf x'} \vert,$$ (271)

where $V({\bf x'})$ is a scalar function of the position eigenvalues. Combining Eqs. (267), (268), (270), and (271), we obtain

$${\rm i} \hbar \frac{\partial \langle {\bf x}'\vert At\ran... ...\vert At\rangle + V({\bf x}') \langle {\bf x}'\vert At\rangle,$$ (272)

which can also be written

$${\rm i} \hbar \frac{\partial \psi({\bf x}', t)}{\partial t... ... \nabla'^2\psi({\bf x}', t) + V({\bf x'}) \psi({\bf x'}, t).$$ (273)

This is Schrödinger's famous wave-equation, and is the basis of wave mechanics. Note, however, that the wave-equation is just one of many possible representations of quantum mechanics. It just happens to give a type of equation which we know how to solve. In deriving the wave-equation, we have chosen to represent the system in terms of the eigenkets of the position operators, instead of those of the momentum operators. We have also fixed the relative phases of the $\vert{\bf x'}\rangle$ according to Schrödinger's representation, so that Eq. (269) is valid. Finally, we have chosen to work in the Schrödinger picture, in which state kets evolve and dynamical variables are fixed, instead of the Heisenberg picture, in which the opposite is true.

Suppose that the ket $\vert At\rangle$ is an eigenket of the Hamiltonian belonging to the eigenvalue $H'$:

$$H\vert At\rangle = H'\vert At\rangle.$$ (274)

Schrödinger's equation of motion (228) yields

$${\rm i} \hbar \frac{d \vert At\rangle}{dt} = H' \vert At\rangle.$$ (275)

This can be integrated to give

$$\vert At\rangle = \exp[ -{\rm i} H'(t-t_0)/\hbar] \vert At_0\rangle.$$ (276)

Note that $\vert At\rangle$ only differs from $\vert A t_0\rangle$ by a phase-factor. The direction of the vector remains fixed in ket space. This suggests that if the system is initially in an eigenstate of the Hamiltonian then it remains in this state for ever, as long as the system is undisturbed. Such a state is called a stationary state. The wave-function of a stationary state satisfies

$$\psi({\bf x}', t) = \psi({\bf x'}, t_0) \exp[ -{\rm i} H' (t-t_0)/\hbar].$$ (277)

Substituting the above relation into Schrödinger's wave equation (273), we obtain

$$-\left(\frac{\hbar^2}{2 m}\right)\nabla'^2\psi_0({\bf x'}) + (V({\bf x'})-E) \psi_0({\bf x'}) =0,$$ (278)

where $\psi_0({\bf x'}) \equiv \psi({\bf x'}, t_0)$, and $E= H'$ is the energy of the system. This is Schrödinger's time-independent wave-equation. A bound state solution of the above equation, in which the particle is confined within a finite region of space, satisfies the boundary condition

$$\psi_0({\bf x}') \rightarrow 0 \mbox{\hspace{1cm}as $\vert{\bf x}'\vert\rightarrow \infty$}.$$ (279)

Such a solution is only possible if

$$E < \lim_{\vert{\bf x}'\vert\rightarrow \infty} V({\bf x'}).$$ (280)

Since it is conventional to set the potential at infinity equal to zero, the above relation implies that bound states are equivalent to negative energy states. The boundary condition (279) is sufficient to uniquely specify the solution of Eq. (278).

The quantity $\rho({\bf x'}, t)$, defined by

$$\rho({\bf x'}, t) = \vert\psi({\bf x'}, t)\vert^2,$$ (281)

is termed the probability density. Recall, from Eq. (120), that the probability of observing the particle in some volume element $d^3 x'$ around position ${\bf x'}$ is proportional to $\rho({\bf x'}, t) d^3 x'$. The probability is equal to $\rho({\bf x'}, t) d^3 x'$ if the wave-function is properly normalized, so that

$$\int \rho({\bf x'}, t) d^3 x' = 1.$$ (282)

Schrödinger's time-dependent wave-equation, (273), can easily be written in the form of a conservation equation for the probability density:

$$\frac{\partial \rho}{\partial t} + \nabla'\cdot {\bf j} = 0.$$ (283)

The probability current ${\bf j}$ takes the form

$${\bf j}({\bf x}', t) = - \left(\frac{{\rm i} \hbar}{2 m}\... ...left(\frac{\hbar}{m} \right){\rm Im} (\psi^\ast \nabla' \psi).$$ (284)

We can integrate Eq. (283) over all space, using the divergence theorem, and the boundary condition $\rho\rightarrow 0$ as $\vert{\bf x'}\vert\rightarrow \infty$, to obtain

$$\frac{\partial}{\partial t} \int \rho({\bf x'}, t) d^3 x' = 0.$$ (285)

Thus, Schrödinger's wave-equation conserves probability. In particular, if the wave-function starts off properly normalized, according to Eq. (282), then it remains properly normalized at all subsequent times. It is easily demonstrated that

$$\int {\bf j}({\bf x}', t) d^3 x' = \frac{\langle {\bf p }\rangle_t}{m},$$ (286)

where $\langle {\bf p} \rangle_t$ denotes the expectation value of the momentum evaluated at time $t$. Clearly, the probability current is indirectly related to the particle momentum.

In deriving Eq. (283) we have, naturally, assumed that the potential $V({\bf x}')$ is real. Suppose, however, that the potential has an imaginary component. In this case, Eq. (283) generalizes to

$$\frac{\partial \rho}{\partial t} + \nabla'\cdot {\bf j} = \frac{2 {\rm Im}(V)}{\hbar} \rho,$$ (287)

giving

$$\frac{\partial}{\partial t} \int \rho({\bf x'}, t) d^3 x' ... ...ar} {\rm Im}\!\int V({\bf x}') \rho({\bf x'}, t) d^3 x'.$$ (288)

Thus, if ${\rm Im}(V)<0$ then the total probability of observing the particle anywhere in space decreases monotonically with time. Thus, an imaginary potential can be used to account for the disappearance of a particle. Such a potential is often employed to model nuclear reactions in which incident particles can be absorbed by nuclei.

The wave-function can always be written in the form

$$\psi({\bf x}', t) = \sqrt{\rho({\bf x}', t)} \exp\left[\frac{{\rm i} S({\bf x}',t)}{ \hbar}\right],$$ (289)

where $\rho$ and $S$ are both real functions. The interpretation of $\rho$ as a probability density has already been given. What is the interpretation of $S$? Note that

$$\psi^\ast \nabla'\psi = \sqrt{\rho} \nabla'(\sqrt{\rho}) + \left(\frac{{\rm i}} {\hbar} \right) \rho \nabla' S.$$ (290)

It follows from Eq. (284) that

$${\bf j} = \frac{\rho \nabla' S}{m}.$$ (291)

Thus, the gradient of the phase of the wave-function determines the direction of the probability current. In particular, the probability current is locally normal to the contours of the phase-function $S$.

Let us substitute Eq. (289) into Schrödinger's time-dependent wave-equation. We obtain

$$-\frac{1}{2 m}\left[ \hbar^2 \nabla'^2 \sqrt{\rho} + 2{\rm i} \... ...bla' S\vert^2 +{\rm i} \hbar \sqrt{\rho} \nabla'^2 S\right] + \sqrt{\rho} V$$

$$= \left[ {\rm i} \hbar \frac{\partial \sqrt{\rho}}{\partial t} - \sqrt{\rho} \frac{\partial S}{\partial t} \right].$$ (292)

Let us treat $\hbar$ as a small quantity. To lowest order, Eq. (292) yields

$$-\frac{\partial S({\bf x}', t)}{\partial t} = \frac{1}{2 m}... ...x}', t) \vert^2 + V ({\bf x}', t) = H({\bf x}', \nabla' S, t),$$ (293)

where $H({\bf x}, {\bf p}, t)$ is the Hamiltonian operator. The above equation is known as the Hamilton-Jacobi equation, and is one of the many forms in which we can write the equations of classical mechanics. In classical mechanics, $S$ is the action (i.e., the path-integral of the Lagrangian). Thus, in the limit $\hbar\rightarrow 0$, wave mechanics reduces to classical mechanics. It is a good approximation to neglect the terms involving $\hbar$ in Eq. (292) provided that

$$\hbar \vert\nabla'^2 S\vert \ll \vert\nabla' S\vert^2.$$ (294)

Note that, according to Eq. (289),

$${\mathchar'26\mskip-10mu\lambda}= \frac{\hbar}{\vert\nabla' S\vert},$$ (295)

where ${\mathchar'26\mskip-10mu\lambda}$ is the de Broglie wave-length divided by $2\pi$. The inequality (294) is equivalent to

$$\vert\nabla' {\mathchar'26\mskip-10mu\lambda}\vert \ll 1.$$ (296)

In other words, quantum mechanics reduces to classical mechanics whenever the de Broglie wave-length is small compared to the characteristic distance over which things (other than the quantum phase) vary. This distance is usually set by the variation scale-length of the potential.