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Section 3.5 Schrödinger's Wave Equation

We have talked about the idea of a wavefunction \(\psi(x)\) that describes the wave-nature of a particle, and we have introduced the idea that the wave represents probability amplitudes, with the probability density \(P(x)\) for locating a particle being given by \(P(x) = |\psi(x)|^2\text{.}\) But how can we determine what that wavefunction is for a specific problem?

In 1926, Erwin Schrödinger proposed a differential equation whose solutions give the wavefunctions corresponding to matter waves. For a particle of mass \(m\) and total energy \(E\text{,}\) Schrödinger's equation written in one spatial dimension is

\begin{equation} -\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} + U(x)\psi(x) = E\psi(x)\text{,}\label{eq_schrodinger}\tag{3.18} \end{equation}

where \(U(x)\) is the potential energy function of the particle. Schrödinger's equation is similar in form to the types of equations that describe other wave phenomena, such as sound waves and electromagnetic waves. The basic idea here is that, given a particle with mass \(m\) and a potential energy \(U(x)\text{,}\) Schrödinger's equation will allow us to determine the possible energies \(E\) and associated wavefunction solutions \(\psi(x)\text{.}\)

Subsection 3.5.1 Testing Solutions to an Equation

There are advanced techniques from the theory of differential equations that we could use to start finding solutions to Schrödinger's equation from scratch. But for this course, we use only the “guess and check” method, which is also a standard technique in solving differential equations. (In lab, you will solve Schrödinger's equation using a numerical iteration method.) The approach is quite simple: you guess a particular function, you put it into the equation and see if it works. If it works for all values of \(x\text{,}\) then you say that it is a solution. If it doesn't, then you say that it isn't a solution.

We will illustrate the “guess and check” approach with an example:

Some equations have solutions which are functions, instead of just a number. Let's say you are trying to determine the function \(f(x)\) that solves the following equation:

\begin{equation} \left[f(x)\right]^2 - x^2 + 8x - 16 = 0\text{.}\label{eq_TestEx1}\tag{3.19} \end{equation}

You can probably solve this with algebra techniques, but let's pretend that you don't know how to solve it. Instead, we are going to test trial solutions.

  1. (a).

    Test the solution \(f(x) = ax+b\) to see if it satisfies (3.19). If it does, determine the values of the constants \(a\) and \(b\) for the solution.

  2. (b).

    Test the solution \(f(x) = ax\) to see if it satisfies (3.19). If it does, determine the value of the constant \(a\) for the solution.

  3. (c).

    Test the solution \(f(x) = ax^2+b\) to see if it satisfies (3.19). If it does, determine the values of the constants \(a\) and \(b\) for the solution.

Solution: (a) Substituting \(f(x) = ax+b\) into (3.19), we get

\begin{align*} \left[f(x)\right]^2-x^2+8x-16 =\mathstrut \amp (ax+b)^2-x^2+8x-16\\ =\mathstrut \amp a^2x^2+2abx+b^2-x^2+8x-16\\ =\mathstrut \amp (a^2-1)x^2+x(2ab+8)+(b^2-16) \end{align*}

The question is whether this can all add up to zero for all values of \(x\text{.}\) The answer to that question is yes only if the constants, the \(x\)-terms and the \(x^2\) terms all separately add to zero. That means that all of the terms in parentheses above need to add to zero separately. (You can't satisfy this by making \(x=0\text{,}\) because then the function would only work for that particular value of \(x\)). So

\begin{gather*} b^2-16 = 0 \rightarrow b = \pm 4\\ 2ab+8 = 0 \rightarrow 2ab=-8 \rightarrow a=-4/b\\ a^2-1=0 \rightarrow a= \pm1 \end{gather*}

All three of these relations work if \(b=4\) and \(a=-1\) or if \(b=-4\) and \(a=1\text{.}\) So, we can say that \(f(x)=ax+b\) is a solution to (3.19) if \(b=4\) and \(a=-1\) or if \(b=-4\) and \(a=1\text{,}\) i.e., \(f(x) = -x+4\) and \(f(x)=x-4\) are both solutions.

(b) Now let's substitute \(f(x) = ax\) into (3.19):

\begin{align*} \left[f(x)\right]^2-x^2+8x-16 =\mathstrut \amp (ax)^2-x^2+8x-16\\ =\mathstrut \amp a^2x^2-x^2+8x-16\\ =\mathstrut \amp (a^2-1)x^2+8x-16 \end{align*}

We could get rid of the \(x^2\) term by making \(a = \pm 1\text{,}\) but we would still be left with \(8x - 16 = 0\text{.}\) The only way to make that work would be to say \(x\) must be equal to 2. But that means that the test solution \(f(x) = ax\) (with \(a = 1\)) would only work at \(x = 2\text{.}\) If \(f(x)\) only works at one particular value of \(x\text{,}\) then it isn't a solution of the equation. (Solutions must work for every value of \(x\text{.}\))

(c) Now let's substitute \(f(x) = ax^2+b\) into (3.19):

\begin{align*} \left[f(x)\right]^2-x^2+8x-16 =\mathstrut \amp (ax^2+b)^2-x^2+8x-16\\ =\mathstrut \amp a^2x^4+2abx^2+b^2-x^2+8x-16\\ =\mathstrut \amp a^2x^4+(2ab^2-1)x^2+ (8)x +(b^2-16) \end{align*}

This would work only if the coefficients of the \(x^4\text{,}\) \(x^2\text{,}\) \(x\) and constant terms were all zero. We can make the \(x^4\) term disappear by making \(a=0\) and we can make the constant term disappear by making \(b=\pm 4\text{,}\) but that would leave non-zero coefficients for the \(x^2\) and \(x\) terms. There is no combination of \(a\) and \(b\) that can make all four terms disappear, so there is no function of the form \(f(x) = ax^2+b\) that can satisfy (3.19).

The same approach works for any equation, including differential equations (i.e., equations that have derivatives in them).

Subsection 3.5.2 Testing Solutions of Schrödinger's Equation

Okay, let's look at Schrödinger's equation now. We won't “solve” this equation (that requires mathematics beyond the scope of this course), but we will test possible solutions to see if they work. First, since Schrödinger's equation is a generalization of de Broglie's result, we should get the same result for a free particle (i.e., if \(U(x)=0\)). De Broglie says that a free particle with a momentum \(p\) can be described as a sine wave with a wavelength \(\lambda=h/p\text{.}\) So, let's try a test solution \(\psi_1(x)=A\sin(kx)\) and see if this satisfies Schrödinger's equation.

With \(U(x) = 0\text{,}\) Schrödinger's equation becomes

\begin{equation} -\frac{\hbar^2}{2m}\frac{d^2\psi_1(x)}{dx^2} + 0 = E\,\psi_1(x)\text{.}\label{eq_schroed_with_0pe}\tag{3.20} \end{equation}

The Schrödinger equation has the second derivative of the wavefunction in it, so let's calculate that derivative for our trial solution:

\begin{equation} \frac{d\psi_1}{dx} = kA\cos(kx) \hspace{0.5in} \mbox{and} \hspace{0.5in} \frac{d^2\psi_1}{dx^2} = -k^2A\sin(kx)\text{.}\tag{3.21} \end{equation}

Next, substitute the second derivative and the wavefunction into (3.20)

\begin{equation} -\frac{\hbar^2}{2m}\left[-k^2A\sin(kx)\right] + 0 = E\left[A\sin(kx)\right]\tag{3.22} \end{equation}

and combine the \(\sin(kx)\) terms

\begin{equation} A\sin(kx)\left[\frac{\hbar^2k^2}{2m}-E\right] = 0\text{.}\tag{3.23} \end{equation}

This works for \(A=0\text{,}\) but that isn't interesting because then \(\psi(x)\) would just be \(0\text{.}\) But \(\psi_1(x)\) also works if the terms in brackets add up to zero, i.e., if

\begin{equation} \frac{\hbar^2k^2}{2m} - E = 0 \longrightarrow E = \frac{\hbar^2k^2}{2m}\tag{3.24} \end{equation}

and

\begin{equation} k=\pm\frac{\sqrt{2mE}}{\hbar}\text{,}\tag{3.25} \end{equation}

where \(k\) is the wavenumber \(k=2\pi/\lambda\text{.}\) The mechanical energy \(E\) is equal to the kinetic energy \(K\) in this case (since \(U = 0\) everywhere), so

\begin{equation} \frac{2\pi}{\lambda}=\pm \frac{\sqrt{2m\left(\frac{1}{2}mv^2\right)}}{h/2\pi} = \frac{2\pi\sqrt{m^2v^2}}{h} \ = \frac{2\pi p}{h}\text{.}\tag{3.26} \end{equation}

We have dropped the \(\pm\) since a negative wavelength doesn't have any physical relevance.) So, we can say that a test solution \(\psi_1(x) = A\sin(kx)\) does satisfy Schrödinger's equation for a free particle (\(U = 0\)), assuming the wavenumber \(k\) has a value that corresponds to a wavelength \(\lambda = h/p\text{,}\) as expected by de Broglie's relation.

(Note that this process has not determined the value of \(A\) in our test solution. The trial function \(\psi_1(x) = A\sin(kx)\) works for any value of \(A\) (assuming the correct \(k\)). The value of \(A\) needs to be determined by normalization conditions. We'll discuss this in the next chapter.)

For contrast, let's try a test solution that does not satisfy Schrödinger's equation.

Test the trial solution \(\psi_1(x) = Bx^2\) to see whether or not it satisfies Schrödinger's equation for a free particle.

Solution: With \(U=0\text{,}\) Schrödinger's equation is

\begin{equation} -\frac{\hbar^2}{2m}\frac{d^2\psi_1(x)}{dx^2} + 0 = E\,\psi_1(x)\text{.}\tag{3.27} \end{equation}

Calculate the second derivative for our trial solution:

\begin{equation} \frac{d\psi_1}{dx} = 2Bx \hspace{0.5in} \mbox{and} \hspace{0.5in} \frac{d^2\psi_1}{dx^2} = 2B\text{.}\tag{3.28} \end{equation}

Next, substitute the second derivative and the wavefunction into Schrödinger's equation:

\begin{equation} -\frac{\hbar^2}{2m}\left[2B\right] + 0 = E\left[Bx^2\right]\text{,}\tag{3.29} \end{equation}

and cancel common factors and simplify, giving

\begin{equation} E B x^2 + \frac{\hbar^2 B}{m} = 0\text{.}\tag{3.30} \end{equation}

In this equation there is no physically-meaningful choice of \(E\) and \(B\) that could make this true for all \(x\text{,}\) 1  so \(\psi_1(x)\) is not a solution.

Finally, let's illustrate a harder problem, the quantum harmonic oscillator.

Recall that the spring potential energy is given by \(U(x) = \frac{1}{2}k_{ sp}x^2 = \frac{1}{2}m\omega^2x^2\text{,}\) where \(\omega = \sqrt{k_{ sp}/m}\text{.}\) Inserting this into the Schrödinger equation, we find

\begin{equation} \frac{-\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} + \frac{1}{2}k_\text{ sp } x^2\,\psi(x) = E\,\psi(x)\text{.}\label{eq_schroed_spring}\tag{3.31} \end{equation}

Try the test function \(\psi_2(x)=xe^{-ax^2}\) to see if this is a solution to Schrödinger's equation for a harmonic oscillator. If it is, determine the unknown constant \(a\) in the solution and the unknown energy \(E\text{.}\)

Solution.

Again, we must calculate the second derivative of this function. We will leave the calculation of the second derivative for you to do (remember to be careful to correctly use the product rule and the chain rule). The final result for the second derivative is

\begin{equation} \frac{d^2 \psi_2(x)}{dx^2} = \left(4 a^2 x^3 - 6 a x \right) e^{- a x^2}\text{.}\label{eq_2ndDerivative}\tag{3.32} \end{equation}

Now substitute this and \(\psi_2(x)\) into Schrödinger's equation, (3.31):

\begin{equation} -\frac{\hbar^2}{2m}\left(4a^2x^3 - 6ax\right) e^{-ax^2} +\frac{1}{2}k_\text{ sp } x^2\left(x \,e^{-ax^2}\right) = E\left(x \,e^{-ax^2}\right)\text{.}\tag{3.33} \end{equation}

Every term has the same exponential factor in it, so we cancel those to get

\begin{equation} \frac{-2\hbar^2a^2}{m}x^3 + 3\frac{a\hbar^2}{m}x + \frac{1}{2}k_\text{ sp } x^3 = Ex\text{.}\tag{3.34} \end{equation}

Now to finish, move \(Ex\) to the left hand side and collect common powers of \(x\text{:}\)

\begin{equation} x^3\left(\frac{1}{2}k_\text{ sp } - \frac{2\hbar^2a^2}{m}\right) + x\left(3\frac{a\hbar^2}{m} - E\right) = 0\text{.}\tag{3.35} \end{equation}

For this equation to hold for all \(x\text{,}\) we must make the coefficient of each power of \(x\) vanish separately. So both the coefficient of \(x^3\) and the coefficient of \(x\) are set to zero:

\begin{gather*} \frac{1}{2}k_\text{ sp } - \frac{2\hbar^2a^2}{m} = 0 \text{ and } 3\frac{a\hbar^2}{m} - E = 0\text{.} \end{gather*}

Solving these for \(a\) and \(E\text{,}\) we find

\begin{equation} a = \frac{m\omega}{2\hbar} \mathbf{ and} E = 3\frac{a\hbar^2}{m} = 3\left(\frac{m\omega}{2\hbar}\right)\frac{\hbar^2}{m} = \frac{3}{2}\hbar\omega\text{.}\tag{3.36} \end{equation}

We see that \(\psi_2(x) = x e^{-ax^2}\) is a solution to the Schrödinger equation if the constants \(a\) and \(E\) are as given in the previous equation. We see that Schrödinger's equation gives us the energy \(E\) for this state, and also determines the unknown constant \(a\) in our trial wavefunction, \(\psi_2\text{.}\) This wavefunction solution is actually the first excited state of the quantum oscillator: \(\psi_2(x) = x\,e^{-(m\omega/2\hbar)x^2}\) with definite energy \(E_2 = \frac{3}{2}\hbar\omega\text{.}\) In one of the assigned problems at the end of the chapter, you will test the wavefunction solution corresponding to the ground state of the harmonic oscillator.

We'll say a lot more about quantization of energy states in the next chapter.

\(B=0\) would work, but then you'd have \(\psi_1(x) = 0\text{,}\) which isn't a physically relevant solution, because \(\psi_1(x) = 0\) means that there is no probability of finding the particle anywhere.