12.4: The Eigenvalue Equation and operators

12.4.1 Eigenfunctions of different operators

There is one last, very important lesson about operators and eigenfunctions which is one of the most complicated things about quantum mechanics. And that is the fact that the eigenfunctions of one operator may, or may not, be the eigenfunctions of another operator. This is shown by the Venn diagram in Figure 12.4, and as an example let’s go back to the example of a Hamiltonian operator with no potential energy, i.e. \(\hat{H}=\frac{{-\hslash }^2}{2m}\frac{{\partial }^2}{\partial x^2}\). A wavefunction of this Hamiltonian is \(\psi =cos\left(\frac{2\pi }{\lambda }x\right)\), and has an energy as we showed on the previous page. Now, if we apply the momentum operator to the same state:

\[\hat{p}\psi =\frac\hslash{i}\frac{\partial }{\partial x}cos\left(\frac{2\pi }{\lambda }x\right)=\frac{-2\pi \hslash }{i\lambda }\cdot sin\left(\frac{2\pi }{\lambda }x\right)\neq \omega \psi \nonumber \]

Then you should know that the momentum of the state most definitely is not \(\frac{-2\pi \hslash }{i\lambda }\). The next section will discuss in great detail how we deal with this uncomfortable situation.

12.4.2 Practice with the Eigenvalue Equation and Complex Wave Equations.

We have already shown that wave equations, when squared, provides a measure of probability that a quantum mechanical particle is at a particular position. We have also shown how a wave equation can provide additional information, that being what is returned when it is operated on by, oddly, operators. We will make this more concrete with examples here. Let’s say that the normalized wavefunction for an electron is: \(\psi =N\cdot cos\left(kx\right)\) where N is the normalization constant and \(k=\frac{2\pi }{\lambda }\) is the wavevector. We know how to square this function, which then tells us the probability that the electron is at a position x that we are curious about (for whatever reason). What about the energy of this electron? Just like in the previous examples we apply the potential energy free (i.e. \(\hat{V}=0\)) Hamiltonian:

\[\hat{H}\psi =-\frac{\hslash^2}{2m}\frac{{\partial }^2}{\partial x^2}\left\{N\cdot cos\left(kx\right)\right\}=-\frac{\hslash^2}{2m}\frac{\partial }{\partial x}\left\{N\cdot k\cdot sin\left(kx\right)\right\}=\frac{\hslash^2k^2}{2m}\left\{N\cdot cos\left(kx\right)\right\} \nonumber \]

Comparison to the eigenvalue equation \(\hat{H}\psi =E\psi\) reveals that the above is in the proper form, so we can be sure that the energy is: \(\frac{\hslash^2k^2}{2m}\).

Now let’s repeat the above using the complex mathematical version of the wavefunction, i.e. \(\psi =N\cdot cos\left(kx\right)=N\cdot \left(\frac{1}{2}e^{ikx}+\frac{1}{2}e^{-ikx}\right)\):

\[\hat{H}\psi =-\frac{\hslash^2}{2m}\frac{{\partial }^2}{\partial x^2}\left\{N\cdot \left(\frac{1}{2}e^{ikx}+\frac{1}{2}e^{-ikx}\right)\right\}=-\frac{\hslash^2}{2m}\frac{\partial }{\partial x}\left\{N\cdot \left(\frac{ik}{2}e^{ikx}+\frac{-ik}{2}e^{-ikx}\right)\right\}=-\frac{\hslash^2}{2m}\left\{N\cdot \left(\frac{{-k}^2}{2}e^{ikx}+\frac{{-k}^2}{2}e^{-ikx}\right)\right\} \nonumber \]

The next step is to factor out \({-k}^2\) which gives us:

\[\hat{H}\psi =\frac{\hslash^2k^2}{2m}\left\{N\cdot \left(\frac{1}{2}e^{ikx}+\frac{1}{2}e^{-ikx}\right)\right\}=E\psi \nonumber \]

where again we see that \(E=\frac{\hslash^2k^2}{2m}\). So, everything seems fine, but why are we using this approach? While solving \(\hat{H}\psi\) using the complex representation of \(\psi =N\cdot cos\left(kx\right)\) seems more difficult, there are going to be many examples coming up where the complex representation is far easier to work with. For example, the electron’s wavefunction could have been \(\psi =N\cdot e^{ikx}\). In this case, which do you think is harder to solve:

\[\hat{H}\psi =-\frac{\hslash^2}{2m}\frac{{\partial }^2}{\partial x^2}\left\{N\cdot e^{ikx}\right\} \nonumber \]

or:

\[\hat{H}\psi =-\frac{\hslash^2}{2m}\frac{{\partial }^2}{\partial x^2}\left\{N\cdot cos\left(kx\right)+i\cdot sin\left(kx\right)\right\} \nonumber \]

Just for the heck of it let’s solve the former:

\[\hat{H}\psi =-\frac{\hslash^2}{2m}\frac{{\partial }^2}{\partial x^2}\left\{N\cdot e^{ikx}\right\}=-\frac{\hslash^2}{2m}\frac{\partial }{\partial x}\left\{ik\cdot N\cdot e^{ikx}\right\}=-\frac{\hslash^2}{2m}\left\{i^2k^2\cdot N\cdot e^{ikx}\right\}=\frac{\hslash^2k^2}{2m}\left\{N\cdot e^{ikx}\right\} \nonumber \]

Taking the derivative of an exponential is easy, and just like the previous example, we see that \(E=\frac{\hslash^2k^2}{2m}\). This wasn’t nearly as hard as taking the double derivative of two trig functions!

12.4.2 Expectation Values

How do we figure out the momentum of a particle with a wavefunction of the form\(psi =N\cdot cos(kx)\)? Give up? Sometimes! After all quantum mechanics is all about probability, and you cannot know everything. In this case, instead of giving up you can often solve these types of problems using the following approach. If we write out:

\[\psi =N\cdot

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you notice that particle’s wavefunction is composed of two equal momentum eigenfunctions, one that is moving to the right (\(e^{ikx}\)) and the other to the left (\(e^{-ikx}\)). Now you can guess that the total momentum is 0. Good intuition, but quantum class is sort of a math class, so how do we prove it? Here we introduce a new expression that is called the “expectation value” for an operator \(\widehat{\mathit{\Omega}}\):

\[\langle \widehat{\mathit{\Omega}}\rangle =\int^{upper\ limit}_{lower\ limit}{}{\psi }^*\widehat{\mathit{\Omega}}\psi \cdot \partial \tau \nonumber \]

where \(\psi\) may, or may not, be the eigenfunction of the operator \(\widehat{\mathit{\Omega}}\). What is great about expectation values is that it doesn’t matter- in either case you will get the right answer. Let’s apply this to our current problem with determining the momentum of \(\psi =N\cdot

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\[\langle \hat{p}\rangle =\int^{upper\ limit}_{lower\ limit}{}{\left\{N\cdot

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\[\frac{-\hslash }{i}N^2\int^{\infty }_{-\infty }{}

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When we look up this integral off the internet, we find \(\int^{\infty }_{-\infty }{}

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The expectation value approach also works with functions that are eigenfunctions. Let’s do an example using the normalized “right wave” \(\mathit{\Phi}=N\cdot e^{ikx}\) eigenfunction of momentum, that being the:

\[\langle \hat{p}\rangle =\int^{upper\ limit}_{lower\ limit}{}{\left\{N\cdot e^{ikx}\right\}}^*\frac\hslash{i}\frac{\partial }{\partial x}\left\{N\cdot e^{ikx}\right\}\cdot \partial x= \nonumber \]

\[\langle \hat{p}\rangle =\frac\hslash{i}\int^{\infty }_{-\infty }{}N^*\cdot e^{-ikx}{\cdot ik\cdot N\cdot e}^{ikx}\cdot \partial x= \nonumber \]

\[\langle \hat{p}\rangle =\frac{\hslash ik}{i}\int^{\infty }_{-\infty }{}N^*\cdot e^{-ikx}\cdot N{\cdot e}^{ikx}\cdot \partial x=\hslash k\int^{\infty }_{-\infty }{}{\mathit{\Phi}}^*\mathit{\Phi}\cdot \partial x=\hslash k \nonumber \]

where the complex conjugate \({\left\{N\cdot e^{ikx}\right\}}^*\) is: \({N^*\cdot e}^{-ikx}\), and we used the formula \(\int{}{\mathit{\Phi}}^*\mathit{\Phi}\partial x=\int{}{\left|\mathit{\Phi}\right|}^2\partial x=1\) in the last step which is the definition of normalization. While we determined the corrent momentum, you might ask, why not stick with \(\widehat{\mathit{\Omega}}\mathit{\Phi}=\omega \cdot \mathit{\Phi}\) given that \(\langle \hat{p}\rangle\) was seemingly much more complicated to work with? You are correct, it is generally mathematically far simpler to work with the eigenvalue equation over the expectation value expression. However, the expectation value method always works, and also gives us a “clean” answer because we don’t have to try to untangle the eigenvalue from the eigenfunction.

Now you might ask, why does the expectation value method work? For one, if we are working on eigenfunctions of the operator \(\widehat{\mathit{\Omega}}\), the answer is seen in a simple derivation:

\[\langle \widehat{\mathit{\Omega}}\rangle =\int^{\infty }_{-\infty }{}{\mathit{\Phi}}^*\widehat{\mathit{\Omega}}\mathit{\Phi}\cdot \partial \tau =\int^{\infty }_{-\infty }{}{\mathit{\Phi}}^*\cdot \omega \cdot \mathit{\Phi}\cdot \partial \tau =\omega \cdot \int^{\infty }_{-\infty }{}{\mathit{\Phi}}^*\cdot \mathit{\Phi}\cdot \partial \tau =\omega \nonumber \]

where we assume that \(\mathit{\Phi}\) is normalized. However, this proof breaks down when we are not applying an eigenfunction of the operator, i.e. when \(\widehat{\mathit{\Omega}}\psi \neq \omega \cdot \psi\). What do we do in this case? Here is another important lesson, which is that wavefunctions can always be written as linear combinations of other wavefunctions. For example, let’s say that the operator \(\widehat{\mathit{\Omega}}\) has two eigenfunctions \({\mathit{\Phi}}_1\) and \({\mathit{\Phi}}_2\), but \(\psi\) is not an eigenfunction of \(\widehat{\mathit{\Omega}}\). Upon further analysis you realize that \(\psi\) is a linear combination of the \(\mathit{\Phi}'s\):

\[\psi =c_1\cdot {\mathit{\Phi}}_1+c_2\cdot {\mathit{\Phi}}_2 \nonumber \]

where \(c_1\) and \(c_2\) are constants. When you apply the above to the expectation value expression:

\[\langle \widehat{\mathit{\Omega}}\rangle =\int^{\infty }_{-\infty }{}{\psi }^*\widehat{\mathit{\Omega}}\psi \cdot \partial \tau =\int^{\infty }_{-\infty }{}{\left\{c_1\cdot {\mathit{\Phi}}_1+c_2\cdot {\mathit{\Phi}}_2\right\}}^*\widehat{\mathit{\Omega}}\left\{c_1\cdot {\mathit{\Phi}}_1+c_2\cdot {\mathit{\Phi}}_2\right\}\cdot \partial \tau \nonumber \]

This problem has now turned into an algebraic mess which is a common occurrence. Fortunately, algebra is a middle school level of mathematics and as such we can deal with it:

\[\langle \widehat{\mathit{\Omega}}\rangle =\int^{\infty }_{-\infty }{}{\left\{c_1\cdot {\mathit{\Phi}}_1+c_2\cdot {\mathit{\Phi}}_2\right\}}^*\widehat{\mathit{\Omega}}\left\{c_1\cdot {\mathit{\Phi}}_1+c_2\cdot {\mathit{\Phi}}_2\right\}\cdot \partial \tau =\int^{\infty }_{-\infty }{}\left\{c^*_1{\mathit{\Phi}}^*_1\widehat{\mathit{\Omega}}c_1{\mathit{\Phi}}_1+c^*_2{\mathit{\Phi}}^*_2\widehat{\mathit{\Omega}}c_2{\mathit{\Phi}}_2+c^*_1{\mathit{\Phi}}^*_1\widehat{\mathit{\Omega}}c_2{\mathit{\Phi}}_2+c^*_2{\mathit{\Phi}}^*_2\widehat{\mathit{\Omega}}c_1{\mathit{\Phi}}_1\right\}\cdot \partial \tau \nonumber \]

This can be broken up into four smaller integrals which is less scary.

\[\langle \widehat{\mathit{\Omega}}\rangle =\int^{\infty }_{-\infty }{}c^*_1{\mathit{\Phi}}^*_1\widehat{\mathit{\Omega}}c_1{\mathit{\Phi}}_1\cdot \partial \tau +\int^{\infty }_{-\infty }{}c^*_2{\mathit{\Phi}}^*_2\widehat{\mathit{\Omega}}c_2{\mathit{\Phi}}_2\cdot \partial \tau +\int^{\infty }_{-\infty }{}c^*_1{\mathit{\Phi}}^*_1\widehat{\mathit{\Omega}}c_2{\mathit{\Phi}}_2\cdot \partial \tau +\int^{\infty }_{-\infty }{}c^*_2{\mathit{\Phi}}^*_2\widehat{\mathit{\Omega}}c_1{\mathit{\Phi}}_1\cdot \partial \tau \nonumber \]

We can simplify this further using the relationships:

\[\widehat{\mathit{\Omega}}c_1{\mathit{\Phi}}_1={\omega }_1\cdot c_1{\mathit{\Phi}}_1, \widehat{\mathit{\Omega}}c_2{\mathit{\Phi}}_2={\omega }_2\cdot c_2{\mathit{\Phi}}_2, c^*_1c_1={\left|c_1\right|}^2, and c^*_2c_2={\left|c_2\right|}^2 \nonumber \]

to yield:

\[\langle \widehat{\mathit{\Omega}}\rangle ={\left|c_1\right|}^2\cdot {\omega }_1\cdot \int^{\infty }_{-\infty }{}{\left|{\mathit{\Phi}}_1\right|}^2\cdot \partial \tau +{\left|c_2\right|}^2\cdot {\omega }_2\cdot \int^{\infty }_{-\infty }{}{\left|{\mathit{\Phi}}_2\right|}^2\cdot \partial \tau +c^*_1c_2\cdot {\omega }_2\int^{\infty }_{-\infty }{}{\mathit{\Phi}}^*_1{\mathit{\Phi}}_2\cdot \partial \tau +c^*_2c_1\cdot {\omega }_1\int^{\infty }_{-\infty }{}{\mathit{\Phi}}^*_2{\mathit{\Phi}}_1\cdot \partial \tau \nonumber \]

Now the above monster can be solved using something that we know already, which is that eigenfunctions are normalized:

\[\int^{\infty }_{-\infty }{}{\left|{\mathit{\Phi}}_1\right|}^2\cdot \partial \tau =\int^{\infty }_{-\infty }{}{\left|{\mathit{\Phi}}_2\right|}^2\cdot \partial \tau =1 \nonumber \]

Now we also must introduce a new concept called “orthonormality” for the 3\({}^{rd}\) and 4\({}^{th}\) expression above:

\[\int^{\infty }_{-\infty }{}{\mathit{\Phi}}^*_1{\mathit{\Phi}}_2\cdot \partial \tau =\int^{\infty }_{-\infty }{}{\mathit{\Phi}}^*_2{\mathit{\Phi}}_1\cdot \partial \tau =0 \nonumber \]

What this means is that, for two eigenfunctions of the same operator, when you integrate them together you get 0. The proper language is that “they do not overlap”, and we will explain this further in the next section on Hermitian operators. Regardless, the remainder of the proof is:

\[\langle \widehat{\mathit{\Omega}}\rangle ={\left|c_1\right|}^2\cdot {\omega }_1+{\left|c_2\right|}^2\cdot {\omega }_2 \nonumber \]

Now you may have said to yourself, “I can’t imagine when would I every run into an equation like: \(\psi =c_1\cdot {\mathit{\Phi}}_1+c_2\cdot {\mathit{\Phi}}_2\).” Actually, you already have, with:

\[N\cdot

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Here, \(\psi =N\cdot

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\[\langle \hat{p}\rangle ={\left|\frac{1}{2}\right|}^2\cdot \hslash k+{\left|\frac{1}{2}\right|}^2\cdot \left(-\hslash k\right)=0 \nonumber \]

Consequently, we not only see once again how the expectation value can allow us to figure out observables from difficult functions (ones that are not eigenfunctions), we also see how it works. We also see that \(\psi =N\cdot

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12.4.3 Schrödinger’s Cat

In the previous example, it was found that a quantum particle described by the wavefunction \(\psi =N\cdot cos\left(kx\right)\) or \(\psi =N\cdot sin\left(kx\right)\) have \(\langle \hat{p}\rangle =0\). However, does this mean that \(E=\frac{p^2}{2m}=0\)? Afterall, this is a particle with no potential energy. And now we ”””””””have a conundrum because we already demonstrated that \(E=\frac{p^2}{2m}=\frac{\hslash^2k^2}{2m}\)! The resolution lies in the fact that these trigonometric wavefunctions are composed of equal left- and right-travelling waves. Meaning that the particle has momentum, and thus kinetic energy, but since the momentum can be pointed either left or right they cancel each other out to yield a net \(\langle \hat{p}\rangle =0\)! This is an example of quantum superposition, as exemplified by the famous Schrödinger’s cat who is both alive and dead at the same time as explained in the box. It is important to note that the superposition is not because the particle is moving right and left at the same time, rather, the wavefunction is used because the experimentalist was not able (or wasn’t told) about which direction the particle is travelling. Although it is true that the particle is moving either right or left, and an experimentalist was able to guess which direction and use the corresponding wavefunction for further analysis, the results would turn out to be incorrect. This seems like absolute nonsense, but this is true and has been confirmed multiple times. In fact, quantum superposition is the basis for quantum computing.

12.4.4 Expectation Value examples: Position

We have already discussed how the position operator \(\hat{x}\) is simply x. Consequently, let’s apply the operator to our favorite wavefunction \(\psi =N\cdot

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\[\hat{x}\psi =x\cdot N\cdot

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Whups- this is a fail, the wavefunction on the right is supposed to be a number (\(\omega\)) multiplying the original wavefunction. However, if \(\left(kx\right)) and \(\left(kx\right)), then clearly \(f(x)\neq g(x)\) since “x” is not a finite value like 5 or \(\piup\). To be more plainspoken, you need to see \(\omega =5\) or \(\omega =\pi\), not \(\omega =x\). The example above is undoubtedly confusing; we have two explanations. For one, the application of an operator is akin to asking a question. The position operator is asking, “Where are you at?” However, this question is nonsensical when applied to \(\psi =N\cdot

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12.4.5 Hermitian operators

One of the most important relationships in quantum mechanics is called orthonormality. This means that, if you have a few functions \({\psi }_n\) that are eigenfunctions of the operator \(\widehat{\mathit{\Omega}}\), then the following is true:

\[\int^{upper\ limit}_{lower\ limit}{}{{\psi }^*_{n^'}\psi }_n\cdot \partial \tau ={\delta }_{n^',n} \nonumber \]

Where d is the Kronecker delta function:

\[{\delta }_{n^',n}={\{}^{1,\ if\ n^'=n\ }_{0,\ if\ n^'\neq n} \nonumber \]

We saw this previously in our discussion on the proof of the expectation value equation. Where does this come from? It is assumed that the wavefunctions are eigenfunctions of an operator that is Hermitian. The definition of a Hermitian operator is:

\[\int{}{\psi }^*_{n^'}\widehat{\mathit{\Omega}}{\psi }_n\cdot \partial \tau =\int{}{\psi }_n{\left({\widehat{\mathit{\Omega}}\psi }_{n^'}\right)}^*\cdot \partial \tau \nonumber \]

Now while this seems very abstract, you’re right, it is. However, it turns out that nearly all quantum mechanical operators (and most important the Hamiltonian operator) has this mathematical trait. The fact that the operator behaves this way has implications for the solutions to the operator, i.e. the wavefunctions. To see what we mean, first assume that the wavefunctions \({\psi }_n\) and \({\psi }_{n^'}\) are actually the exact same thing, meaning \(n=n^'\). Also \(\widehat{\mathit{\Omega}}{\psi }_n={\omega }_n{\psi }_n\). As a result:

\[\int{}{\psi }^*_n\widehat{\mathit{\Omega}}{\psi }_n\cdot \partial \tau =\int{}{\psi }^*_n{\omega }_n{\psi }_n\cdot \partial \tau ={\omega }_n\cdot \int{}{\psi }^*_n{\psi }_n\cdot \partial \tau \nonumber \]

Also:

\[\int{}{\psi }_n{\left({\widehat{\mathit{\Omega}}\psi }_n\right)}^*\cdot \partial \tau =\int{}{\psi }_n{\left({\omega }_n{\psi }_n\right)}^*\cdot \partial \tau ={\omega }^*_n\cdot \int{}{\psi }_n{\psi }^*_n\cdot \partial \tau \nonumber \]

Since, for a Hermitian operator \(\int{}{\psi }^*_n\widehat{\mathit{\Omega}}{\psi }_n\cdot \partial \tau =\int{}{\psi }_n{\left({\widehat{\mathit{\Omega}}\psi }_n\right)}^*\cdot \partial \tau\), then:

\[{\omega }_n\cdot \int{}{\psi }^*_n{\psi }_n\cdot \partial \tau ={\omega }^*_n\cdot \int{}{\psi }_n{\psi }^*_n\cdot \partial \tau \nonumber \]

And thus:

\[{\omega }_n\cdot \int{}{\psi }^*_n{\psi }_n\cdot \partial \tau -{\omega }^*_n\cdot \int{}{\psi }_n{\psi }^*_n\cdot \partial \tau =\left({\omega }_n-{\omega }^*_n\right)\int{}{\psi }^*_n{\psi }_n\cdot \partial \tau =0 \nonumber \]

Where we used the fact that, through the associative axiom of multiplication: \(\int{}{\psi }^*_n{\psi }_n\cdot \partial \tau =\int{}{\psi }_n{\psi }^*_n\cdot \partial \tau\). Now, there are only two ways for \(\left({\omega }_n-{\omega }^*_n\right)\int{}{\psi }^*_n{\psi }_n\cdot \partial \tau =0\), either \(\int{}{\psi }^*_n{\psi }_n\cdot \partial \tau =0\) which we already know is false (its equal to 1) or \({\omega }_n-{\omega }^*_n=0\), which means that \({\omega }_n={\omega }^*_n\). When is a number equal to its complex conjugate? Only when that number is fully real. Thus, the eigenvalues of Hermitian operators have real eigenvalues.

Next assume that \(n\neq n^'\). The same analyses above yield:

\[\int{}{\psi }^*_{n^'}\widehat{\mathit{\Omega}}{\psi }_n\cdot \partial \tau =\int{}{\psi }^*_{n^'}{\omega }_n{\psi }_n\cdot \partial \tau ={\omega }_n\cdot \int{}{\psi }^*_{n^'}{\psi }_n\cdot \partial \tau \nonumber \]

Also:

\[\int{}{\psi }_n{\left({\widehat{\mathit{\Omega}}\psi }_{n^'}\right)}^*\cdot \partial \tau =\int{}{\psi }_n{\left({\omega }_{n^'}{\psi }_{n^'}\right)}^*\cdot \partial \tau ={\omega }^*_{n^'}\cdot \int{}{\psi }_n{\psi }^*_{n^'}\cdot \partial \tau \nonumber \]

Since \(\int{}{\psi }^*_{n^'}\widehat{\mathit{\Omega}}{\psi }_n\cdot \partial \tau =\int{}{\psi }_n{\left({\widehat{\mathit{\Omega}}\psi }_{n^'}\right)}^*\cdot \partial \tau\) then:

\[\left({\omega }_n-{\omega }^*_{n^'}\right)\int{}{\psi }^*_{n^'}{\psi }_n\cdot \partial \tau =0 \nonumber \]

And we now have to figure out whether \({\omega }_n-{\omega }^*_{n^'}=0\) or if \(\int{}{\psi }^*_{n^'}{\psi }_n\cdot \partial \tau =0\). Now, if \({\psi }_n\) and \({\psi }_{n^'}\) are different eigenfunctions of the operator then they must have different eigenvalues. If not, they would be the same. Thus, \({\omega }_n\neq {\omega }_{n^'}\), and we have to conclude that different eigenfunctions of the same operator are orthonormal:

\[\int{}{\psi }^*_{n^'}{\psi }_n\cdot \partial \tau ={\{}^{1,\ if\ n^'=n\ }_{0,\ if\ n^'\neq n} \nonumber \]