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The Schrödinger equation

Week 1, Lecture 3

So far we have been working with the time-dependent Schrödinger equation, which is a partial differential equation

\[i\hbar\frac{\partial\Psi}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\cdot\Psi\]

In general, the potential in which the particle sits will depend on time, i.e. \(V=V(x,t)\). Let's now assume for a moment, that the potential is stable in time \(V(x,t)=V(x)\). While this seems very limiting at first, it's actually quite a common case. We can then separate the variables by introducing two new functions \(\Psi(x,t)=\psi(x)\cdot\phi(t)\) and write the TDSE as

\[i\hbar\psi(x)\frac{\mathrm{d}\phi(t)}{\mathrm{d} t}=-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2\psi(x)}{\mathrm{d} x^2}\phi(t) + V(x)\cdot\psi(x)\phi(t)\]

We have no more partial derivates and we can re-arrange the equation such that the time dependent parts are on the left and the position dependent on the right

\[i\hbar\frac{1}{\phi(t)}\frac{\mathrm{d}\phi(t)}{\mathrm{d} t}=-\frac{\hbar^2}{2m}\frac{1}{\psi(x)}\frac{\mathrm{d}^2\psi(x)}{\mathrm{d} x^2} + V(x)=\mathrm{constant}=E\]

As the two sides depend on different variables and are equal to one another, the only possible solution is a constant, which we call \(E\). The position dependent part is called the time-independent Schrödinger equation (TISE):

\[-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2\psi(x)}{\mathrm{d} x^2} + V(x)\psi(x)=E\psi(x)\]

while the time-dependent part simplifies to

\[\frac{\mathrm{d}\phi(t)}{\mathrm{d}t}=-\frac{iE}{\hbar}\phi(t)\]

We ended up with two ordinary differential equations instead of one partial. As a next step, we will try to find a general solution:

  1. For the time-dependent part we guess the simple, general solution to be related to \(e^{ax}\), as it takes the right form \(\frac{\mathrm{d}}{\mathrm{d}x}e^{ax}=a\cdot e^{ax}\). Hence the general solution is \(\phi(t)=e^{-iEt/\hbar}\).

  2. In order to solve the TISE we will need to specify the potential \(V(x)\). This is exactly what we will do for the rest of the course: choose a specific \(V(x)\) and then solve the TISE.

Note

In order to make the distinction between variables and operators clear, we will from now on use "hats" on operators: \(\hat{x}\), \(\hat{p}\), etc.

The time-independent Schrödinger equation is an eigenvalue differential equation, with the set of solutions \(\psi_n(x)\) and eigenvalues \(E_n\). These seperable solutions have some interesting properties:

  1. They are stationary states, which means "nothing" except for the wavefunction changes in time. While \(\Psi(x,t)=\psi_n(x)e^{-iE_nt/\hbar}\) is time dependent, \(|\Psi(x,t)|^2\) is not. This is easy to see:

    \[|\Psi(x,t)|^2=\psi_n^*e^{iE_nt/\hbar}\psi_ne^{-iE_nt/\hbar}=|\psi_n(x)|^2\]

  2. The solutions have no uncertainty in their total energy! From classical mechanics you are familiar with the energy \(E=\frac{1}{2}mv^2+V=\frac{p^2}{2m}+V(x)=H(x,p)\), which is also called the Hamiltonian. In quantum mechanics we have something similar: remember the operator for momentum \(\hat{p}=\frac{\hbar}{i}\frac{\mathrm{d}}{\mathrm{d}x}\), which we can use to write the total energy as

    \[\hat{E}=\frac{1}{2m}\left(\frac{\hbar}{i}\frac{\mathrm{d}}{\mathrm{d}x}\right)\left(\frac{\hbar}{i}\frac{\mathrm{d}}{\mathrm{d}x}\right)+V(x)=\frac{-\hbar^2}{2m}\frac{\mathrm{d}^2}{\mathrm{d}x^2}+V(x)=\hat{H}\]

    Here \(\hat{H}\) is the Hamiltonian in QM and the TISE then simplifies to \(\hat{H}\psi_n=E_n\psi_n\). Now, in order to demonstrate that the solutions \(\psi_n\) have no uncertainty in energy, we need to calculate the standard deviation \(\sigma_E\) of the energy for \(\psi_n\): \(\sigma_E=\sqrt{\langle E^2\rangle-\langle E\rangle^2}\).

    \[\langle E\rangle=\int_{-\infty}^{+\infty} \psi_n^*(x)e^{iE_nt/\hbar}\hat{H}\psi_ne^{-iE_nt/\hbar} \,\mathrm{d}x = \int_{-\infty}^{+\infty} \psi_n^*(x)E_n\psi_n(x) \,\mathrm{d}x = \]
    \[=E_n \int_{-\infty}^{+\infty} \psi_n^*(x)\psi_n(x) \,\mathrm{d}x = E_n\]
    \[\langle E^2\rangle=\int_{-\infty}^{+\infty} \psi_n^*(x)\hat{H}^2\psi_n \,\mathrm{d}x=\int_{-\infty}^{+\infty} \psi_n^*(x)E_n^2\psi_n \,\mathrm{d}x=E_n^2\]
    \[\sigma_E^2=\langle E^2\rangle-\langle E\rangle^2=E_n^2-E_n^2=0\]

    Hence \(\psi_n\) have definite energy!

  3. The expectation value of any operator \(\hat{Q}\) is independent of time.

    \[\langle Q\rangle=\int_{-\infty}^{+\infty} \Psi*(x,t)\hat{Q}\left(x,\frac{\mathrm{d}}{\mathrm{d}x}\right)\Psi(x,t)\,\mathrm{d}x =\]
    \[=\int_{-\infty}^{+\infty} \psi_n^*(x)e^{iE_nt/\hbar}\hat{Q}\left(x,\frac{\mathrm{d}}{\mathrm{d}x}\right)\psi_n(x)e^{-iE_nt/\hbar}\,\mathrm{d}x =\]
    \[=\int_{-\infty}^{+\infty} \psi_n^*(x)\hat{Q}\left(x,\frac{\mathrm{d}}{\mathrm{d}x}\right)\psi_n(x)\,\mathrm{d}x\]

  4. The solutions \(\psi_n\) form a complete and orthonormal set:

    • We can perform a Fourier decomposition of any arbitrary function: \(f(x)=\sum_n c_n \psi_n(x)\)
    • \(\psi_n\) are normalized: \(\int_{-\infty}^{+\infty} \psi_n^*\psi_n \,\mathrm{d}x = 1\)
    • \(\int_{-\infty}^{+\infty} \psi_n^*\psi_m \,\mathrm{d}x = \delta_{n,m}\), where \(\delta_{n,m}\) is the Kronecker delta.

Note

Recipe for solving the TDSE with \(V=V(x)\)

  • Step 1: expand \(\Psi(x,0)=\sum_n c_n\psi_n(x)\) \(\int_{-\infty}^{+\infty} \psi_m^*\Psi(x,0) \,\mathrm{d}x = \int_{-\infty}^{+\infty} \psi_m^*\sum_n c_n\psi_n(x) \,\mathrm{d}x = \sum_n c_n\int_{-\infty}^{+\infty} \psi_m^*\psi_n(x) \,\mathrm{d}x = \sum_n c_n \delta_{m,n} = c_m\) \(\rightarrow c_n=\int_{-\infty}^{+\infty}\psi_n^*\Psi(x,0)\,\mathrm{d}x\)

  • Step 2: add the time-dependence of \(\psi_n\). If \(\Psi_n(x,t=0)=\psi_n(x) \rightarrow \Psi_n(x,t)=e^{-iE_nt/\hbar}\psi_n(x)\)

  • Step 3: use the superposition principle \(\Psi(x,t)=\sum_n c_n e^{-iE_nt/\hbar}\psi_n(x)\)

Example:

A particle is prepared in a superposition state (i.e. the linear combination of stationary states): \(\Psi(x,0)=c_1\psi_1+c_2\psi_2\). We know that \(\psi_1\) and \(\psi_2\) are stationary states with energies \(E_1\) and \(E_2\), respectively.

  • What is \(\Psi(x,t)\)? \(\Psi(x,t)=c_1e^{-iE_1t/\hbar}\psi_1+c_2e^{-iE_2t/\hbar}\psi_2\)

  • What is \(|\Psi(x,t)|^2\)?

\[|\Psi(x,t)|^2=\left(c_1^*e^{iE_1t/\hbar}\psi_1^*+c_2^*e^{iE_2t/\hbar}\psi_2^*\right)\left(c_1e^{-iE_1t/\hbar}\psi_1+c_2e^{-iE_2t/\hbar}\psi_2\right)=\]
\[=|c_1|^2|\psi_1|^2+|c_2|^2|\psi_2|^2+c_1c_2\left(e^{i(E_1-E_2)t/\hbar}+e^{-i(E_1-E_2)t/\hbar}\right)\psi_1\psi_2=\]
\[=c_1^2\psi_1^2+c_2^2\psi_2^2+2c_1c_2\cos\left(\frac{(E_1-E_2)t}{\hbar}\right)\psi_1\psi_2=\]

For simplicity, we have assumed \(c_1\), \(c_2\) and \(\psi_1\), \(\psi_2\) to be real.

The \(|c_n|^2\) gives the probability of measuring energy \(E_n\). This is immediately clear if \(\Psi=\psi_1\) and for a superposition state \(\Psi=c_1\psi_1+c_2\psi_2\) it implies that \(\sum_n|c_n|^2=1\), i.e. normalization. It turns out that for any \(\Psi=\sum c_n\psi_n\), the expectation value \(\langle E\rangle\) is independent of time. This is of course the familiar conservation of energy! Please note that a measurement does not preserve energy.