Stationary States

Consider separable solutions to Schrödinger's equation of the form

$$\psi(x,t) = \psi(x) {\rm e}^{-{\rm i} \omega t}.$$ (C.63)

According to Equation (C.20), such solutions have definite energies, $E=\hbar \omega$ . For this reason, they are usually written

$$\psi(x,t) = \psi(x) {\rm e}^{-{\rm i} (E/\hbar) t}.$$ (C.64)

The probability of finding the particle between $x$ and $x+dx$ at time $t$ is

$$P(x,t) = \vert\psi(x,t)\vert^{ 2} dx = \vert\psi(x)\vert^{ 2} dx.$$ (C.65)

This probability is time independent. For this reason, states whose wavefunctions are of the form (C.64) are known as stationary states. Moreover, $\psi(x)$ is called a stationary wavefunction. Substituting (C.64) into Schrödinger's equation, (C.24), we obtain the following differential equation for $\psi(x)$ :

$$-\frac{\hbar^{ 2}}{2 m} \frac{d^{ 2}\psi}{d x^{ 2}} + U(x) \psi = E \psi.$$ (C.66)

This equation is called the time-independent Schrödinger equation. Of course, the most general form of this equation is

$$H \psi = E \psi,$$ (C.67)

where $H$ is the Hamiltonian. (See Section C.5.)

Consider a particle trapped in a one-dimensional square potential well, of infinite depth, which is such that

$$U(x) = \left\{ \begin{array}{lll} 0&\mbox{\hspace{0.5cm}}&0\leq x \leq a\ [0.5ex] \infty &&\mbox{otherwise} \end{array}\right..$$ (C.68)

The particle is excluded from the region $x<0$ or $x>a$ , so $\psi=0$ in this region (i.e., there is zero probability of finding the particle outside the well). Within the well, a particle of definite energy $E$ has a stationary wavefunction, $\psi(x)$ , that satisfies

$$-\frac{\hbar^{ 2}}{2 m} \frac{d^{ 2}\psi}{d x^{ 2}} = E \psi.$$ (C.69)

The boundary conditions are

$$\psi(0) = \psi(a) = 0.$$ (C.70)

This follows because $\psi=0$ in the region $x<0$ or $x>a$ , and $\psi(x)$ must be continuous [because a discontinuous wavefunction would generate a singular term (i.e., the term involving $d^{ 2}\psi/dx^{ 2}$ ) in the time-independent Schrödinger equation, (C.66), that could not be balanced, even by an infinite potential].

Let us search for solutions to Equation (C.69) of the form

$$\psi(x) = \psi_0 \sin(k x),$$ (C.71)

where $\psi_0$ is a constant. It follows that

$$\frac{\hbar^{ 2} k^{ 2}}{2 m} = E.$$ (C.72)

The solution (C.71) automatically satisfies the boundary condition $\psi(0)=0$ . The second boundary condition, $\psi(a)=0$ , leads to a quantization of the wavenumber: that is,

$$k= n \frac{\pi}{a},$$ (C.73)

where $n=1, 2, 3,$ et cetera. (A ``quantized'' quantity is one that can only take certain discrete values.) Here, the integer $n$ is known as a quantum number. According to Equation (C.72), the energy is also quantized. In fact, $E=E_n$ , where

$$E_n = n^{ 2} \frac{\hbar^{ 2} \pi^{ 2}}{2 m a^{ 2}}.$$ (C.74)

Thus, the allowed wavefunctions for a particle trapped in a one-dimensional square potential well of infinite depth are

$$\psi_n(x,t) = A_n \sin\left(n \pi \frac{x}{a}\right) \exp\left(-{\rm i} n^{ 2} \frac{E_1}{\hbar} t\right),$$ (C.75)

where $n$ is a positive integer, and $A_n$ a constant. We cannot have $n=0$ , because, in this case, we obtain a null wavefunction: that is, $\psi=0$ , everywhere. Furthermore, if $n$ takes a negative integer value then it generates exactly the same wavefunction as the corresponding positive integer value (assuming $A_{-n}=-A_n$ ).

The constant $A_n$ , appearing in the previous wavefunction, can be determined from the constraint that the wavefunction be properly normalized. For the case under consideration, the normalization condition (C.32) reduces to

$$\int_0^a \vert\psi(x)\vert^{ 2} dx = 1.$$ (C.76)

It follows from Equation (C.75) that $\vert A_n\vert^{ 2}=2/a$ . Hence, the properly normalized version of the wavefunction (C.75) is

$$\psi_n(x,t) = \left(\frac{2}{a}\right)^{1/2} \sin\left(n \pi \frac{x}{a}\right) \exp\left(-{\rm i} n^{ 2} \frac{E_1}{\hbar} t\right).$$ (C.77)

At first sight, it seems rather strange that the lowest possible energy for a particle trapped in a one-dimensional potential well is not zero, as would be the case in classical mechanics, but rather $E_1= \hbar^{ 2} \pi^{ 2}/(2 m a^{ 2})$ . In fact, as explained in the following, this residual energy is a direct consequence of Heisenberg's uncertainty principle. A particle trapped in a one-dimensional well of width $a$ is likely to be found anywhere inside the well. Thus, the uncertainty in the particle's position is ${\mit\Delta} x\sim a$ . It follows from the uncertainty principle, (C.60), that

$${\mit\Delta} p \gtrsim \frac{\hbar}{2 {\mit\Delta} x}\sim \frac{\hbar}{a}.$$ (C.78)

In other words, the particle cannot have zero momentum. In fact, the particle's momentum must be at least $p\sim \hbar/a$ . However, for a free particle, $E=p^{ 2}/2 m$ . Hence, the residual energy associated with the particle's residual momentum is

$$E \sim \frac{p^{ 2}}{m}\sim \frac{\hbar^{ 2}}{m a^{ 2}}\sim E_1.$$ (C.79)

This type of residual energy, which often occurs in quantum mechanical systems, and has no equivalent in classical mechanics, is called zero point energy.