Operators#

What you need to know

  • For every experimental observable there is a corresponding operator in quantum mechanics.

  • Operators must be linear. Becasue they are dervied from Schrodinger equation which itself is linear.

  • Operatprs must be Hermitian becasue only Hermitian operators produce real eigenvalues.

  • Operatprs must produce real eigenvalues. Becasue eigenvalue are the only possible values that are measured in experiments.

  • Operators commutations show weather two experimental observables can be measured simulateneously. E.g can one simulatenoulsy and precisely determine the values of position and momentum of an electron.

  • Commuting operators share eignefunctions, non-commuiting operators have different eigenfunctions.

Operators: A reminder#

We have already seen examples of operators. For short, operators are deonted by a hat symbol which implies that they encode a mathematical operations that can be carried out on functions. For example, the quantum mechanical momentum operator is:

\[{\hat{p}_x = -i\hbar\frac{d}{dx}}\]

  • When this operates on a function, it does the following: differentiate the function with respect to \(x\) and then multiply the result from by \(-i\hbar\).

  • The quantum mechanical momentum operator of position is:

\[\hat{x} = x\]

  • When this operates on a function, it does the following: multiply function by x.

Linearity of operators#

Operators in quantum mechanics are linear, which means that they fulfill the following rules:

\[{\hat{A}\left(\psi_1 + \psi_2\right) = \hat{A}\psi_1 + \hat{A}\psi_2}\]
\[{\hat{A}\left(c\psi\right) = c\hat{A}\psi\textnormal{ where \textit{c} is a constant}}\]

and \(\hat{A}\) is a linear operator. Operator algebra defines how operators are added, multiplied, etc. For example, adding two operators is equivalent to \(\hat{A}_1 + \hat{A}_2\). Multiplication corresponds to them operating one after another.

Example Apply the following operators on the given functions:

  1. Operator \(\hat{A} = d/dx\) and function \(x^2\)

  2. Operator \(\hat{A} = d^2/dx^2\) and function \(4x^2\)

  3. Operator \(\hat{A} = \left(\partial / \partial y\right)_x\) and function \(xy^2\)

  4. Operator \(\hat{A} = -i\hbar d/dx\) and function \(\exp(-ikx)\)

  5. Operator \(\hat{A} = -\hbar^2 d^2/dx^2\) and function \(\exp(-ikx)\)

Solution

  • \(\hat{A}\left(x^2\right) = \frac{d}{dx}x^2 = 2x\).

  • \(\hat{A}\left(4x^2\right) = \frac{d^2}{dx^2}\left(4x^2\right) = 8\).

  • \(\hat{A}\left(xy^2\right) = \left(\frac{\partial}{\partial y}\left(xy^2\right)\right)_x = 2xy\). Note that \(x\) is a constant.

  • \(\hat{A}\left(e^{-ikx}\right) = -i\hbar\frac{d}{dx}\left(e^{-ikx}\right) = -\hbar ke^{-ikx}\).

  • \(\hat{A}\left(e^{-ikx}\right) = -\hbar^2\frac{d^2}{dx^2}e^{-ikx} = i\hbar^2k\frac{d}{dx}e^{-ikx} = \hbar^2k^2e^{-ikx}\)

Eigenvalue problem#

\[{\hat{H}\psi_i(x,y,z) = E_i\psi_i(x,y,z)}\]

  • This is an eigenvalue problem where one needs to determine the eigenfunctions \(\psi_i\) and the eigenvalues \(E_i\). If \(\psi_i\) is an eigenfunction of \(\hat{H}\), operating with \(\hat{H}\) on it must yield a constant times \(\psi_i\).

Note

Example

What are the eigenfunctions and eigenvalues of an operator \(\hat{A} = d/dx\)

Note

Solution

Start with the eigenvalue equation:

\[{\frac{d}{dx}f = kf(x) \Rightarrow \frac{df(x)}{f(x)} = kdx { (integrate both sides)}}\]
\[\ln(f) = kx + c\]
\[{f_k = e^ce^{kx} = c'e^{kx}}\]

Exepctation#

  • The eigenfunctions are \(f_k(x)\) with the corresponding eigenvalue given by \(k\). In general, for operator \(\hat{A}\), the expectation value (quantum mechanical average) is defined as:

\[ {\left < {A}\right> = \int\psi^*\hat{A}\psi d\tau} \]

  • If \(\psi\) is an eigenfunction of \(\hat{A}\) then the expectation value is equal to the corresponding eigenvalue (\(a\)):

\[ {\hat{A}\psi = a\psi \Rightarrow \left<{A}\right>=\int\psi^*\underbrace{\hat{A}\psi}_{a\psi} d\tau = a\underbrace{\int\psi^*\psi d\tau}_{=1} = a} \]

  • Note that operators and eigenfunctions may be complex valued; however, eigenvalues of quantum mechanical operators must be real because they correspond to values obtained from measurements.

  • By allowing wavefunctions to be complex, it is merely possible to store more information in it (i.e., both the real and imaginary parts or ``density and velocity’’)

  • When computing experimental quantities complex conjugate pair of wavefunctions must be combined to yield real values.

Hermitian property#

Real eigenvalues#

  • Operator \(\hat{A}\) is Hermitian if it fulfills the following condition for all well-behaved functions \(\psi_j\) and \(\psi_k\):

\[ {\int {\color{blue} \psi^*_j} {\color{green}\hat{A} \psi_k} d\tau = \int { \color{green} \psi_k} {\color{blue} (\hat{A}\psi_j)^{*} } d\tau} \]

  • Note the symmetry between complex conjugate pair of wavefunctions: The expression remains the same wether the same operator acts on wavefunction or its complex conjugate pair.

  • In general most operators are not hermitian. Meaning you get different result when you feed complex conjugate function to the same operator. Some examples are below

  • This symmetry implies that the eigenvalues are real: Let \(\psi\) be an eigenfunction of \(\hat{A}\) with eigenvalue \(a\). Choose \(\psi_j = \psi_k = \psi\). Then we can write the result of the left and right hand side of hermitian condition:

\[\int\psi^*\hat{A}\psi d\tau = a\]
\[\int\psi\left(\hat{A}\psi\right)^*d\tau = a^*\]

Hence \(a = a^*\), which means that \(a\) must be real!

Example

Prove that the momentum operator (in one dimension) is Hermitian.

Solution

\({\int\limits_{-\infty}^{\infty}\psi_j^*(x)\left(-i\hbar\frac{d\psi_k(x)}{dx}\right)dx} = -i\hbar\int\limits_{-\infty}^{\infty}\psi_j^*(x)\frac{d\psi_k(x)}{dx}dx = \\ \overbrace{\int\limits_{-\infty}^{\infty}\psi_k(x)\left(i\hbar\frac{d\psi_j^*(x)}{dx}\right)dx}^{{integration\, by\, parts}}\) \( = {\int\limits_{-\infty}^{\infty}\psi_k(x)\left(-i\hbar\frac{d\psi_j(x)}{dx}\right)^*dx} \Rightarrow \hat{p}_x\textnormal{ is Hermitian}\).

Note that the wavefunctions approach zero at infinity and thus the boundary term in the integration by parts does not contribute. In 3-D, one would have to use the Green identities.

Orthogonality of eigenfunctions#

The Hermitian property can also be used to show that the eigenfunctions (\(\psi_j\) and \(\psi_k\)), which have different eigenvalues (i.e., \(a_j\) and \(a_k\) with \(a_j \ne a_k\); ``non-degenerate’’), are orthogonal to each other:

\[ {\textnormal{LHS: }\int\psi_j^*\hat{A}\psi_kd\tau = \int\psi_j^*a_k\psi_kd\tau = a_k\int\psi_j^*\psi_kd\tau} \]
\[ {\textnormal{RHS: }\int\psi_k\left(\hat{A}\psi_j\right)^*d\tau = \int\psi_k\left(a_j\psi_j\right)^*d\tau = a_j\int\psi_j^*\psi_kd\tau} \]

Here Hermiticity requires LHS = RHS. If \(a_j \ne a_k\), then we are dealing with:

\[ {{\left(a_k - a_j\right)}{\ne 0}\int\psi^*_j\psi_kd\tau = 0} \]

Note that if \(a_j = a_k\), meaning that the values are degenerate, this result does not hold.

Commutations of operators#

The product \(\hat{A}\hat{B}\) of two operators \(\hat{A}\) and \(\hat{B}\) on some function f are defined as follows:

\[ {\hat{A}\hat{B}f = \hat{A}\left(\hat{B}f\right)} \]

In practice, this means that we first operate with \(\hat{B}\) and then with \(\hat{A}\). Note that the order of multiplication is important because they may not commute (\(\hat{A}\hat{B} \ne \hat{B}\hat{A}\); just like for matrices). The commutator of two operators \(\hat{A}\) and \(\hat{B}\) is defined as:

\[{\left[\hat{A},\hat{B}\right]f = \left(\hat{A}\hat{B} - \hat{B}\hat{A}\right)f}\]

If the commutator of \(\hat{A}\) and \(\hat{B}\) is zero, it means that their order in multiplication (or the operation order, in other words) may be changed. If the commutator is non-zero, the order may not be changed. Operator multiplication is associative:

\[{\hat{A}\hat{B}\hat{C} = \left(\hat{A}\hat{B}\right)\hat{C} = \hat{A}\left(\hat{B}\hat{C}\right)}\]

Example

Prove that operators \(\hat{A} = x\) and \(\hat{B} = d/dx\) do not commute (i.e., \(\left[\hat{A}, \hat{B}\right] \ne 0\)).

Solution

Let \(f\) be an arbitrary well-behaved function. We need to calculate both \(\hat{A}\hat{B}f\) and \(\hat{B}\hat{A}f\):

\[ \hat{A}\hat{B}f = xf'(x)\textnormal{ and } \hat{B}\hat{A}f = \frac{d}{dx}\left(xf(x)\right) = f(x) + xf'(x)\]
\[\left[\hat{A},\hat{B}\right]f = \hat{A}\hat{B}f - \hat{B}\hat{A}f = -f\]
\[\Rightarrow \left[\hat{A},\hat{B}\right] = -1\textnormal{ (this is non-zero and the operators do not commute)} \]

Simple rules for commutators#

\[{\left[A,A\right] = \left[A,A^n\right] = \left[A^n,A\right] = 0}\]
\[{\left[A,B\right] = -\left[B,A\right]}\]
\[{\left[A,B^2\right] = \left[A,B\right]B + B\left[A,B\right]}\]

Commutability measurements#

We have seen previously that operators may not always commute (i.e., \([A, B] \ne 0\)). An example of such operator pair is position \(\hat{x}\) and momentum \(\hat{p}_x\):

\[{\hat{p}_x\hat{x}\psi(x) = \hat{p}_x\left(x\psi(x)\right) = \left(\frac{\hbar}{i}\frac{d}{dx}\right)\left(x\psi(x)\right) = \frac{\hbar x}{i}\frac{d\psi(x)}{dx} + \frac{\hbar}{i}\psi(x)}\]
\[{\hat{x}\hat{p}_x\psi(x) = x\left(\frac{\hbar}{i}\frac{d\psi(x)}{dx}\right)}\]
\[{\Rightarrow \left[\hat{p}_x,\hat{x}\right]\psi(x) = \left(\hat{p}_x\hat{x} - \hat{x}\hat{p}_x\right)\psi(x) = \frac{\hbar}{i}\psi(x)}\]
\[{\Rightarrow \left[\hat{p}_x,\hat{x}\right] = \frac{\hbar}{i}}\]

In contrast, the kinetic energy operator and the momentum operators commute:

\[ {\left[\hat{T},\hat{p}_x\right] = \left[\frac{\hat{p}_x^2}{2m},\hat{p}_x\right] = \frac{p_x^3}{2m} - \frac{p_x^3}{2m} = 0} \]

We had the uncertainty principle for the position and momentum operators:

\[ \Delta x\Delta p_x \ge \frac{\hbar}{2} \]

In general, it turns out that for operators \(\hat{A}\) and \(\hat{B}\) that do not commute, the uncertainty principle applies in the following form:

\[{\Delta A\Delta B \ge \frac{1}{2}\left|\left<\left[\hat{A},\hat{B}\right]\right>\right|}\]

Example

Obtain the position/momentum uncertainty principle

Solution

Denote \(\hat{A} = \hat{x}\) and \(\hat{B} = \hat{p}_x\).

\[\frac{1}{2}\left|\left<\left[\hat{A},\hat{B}\right]\right>\right| = \frac{1}{2}\left|\left<\left[\hat{x},\hat{p}_x\right]\right>\right| = \frac{1}{2}\left|\left<\frac{\hbar}{i}\right>\right| = \frac{1}{2}\left|\left<\psi\left|\frac{\hbar}{i}\right|\psi\right>\right| = \frac{1}{2}\left|\frac{\hbar}{i}\underbrace{\left<\psi\left|\psi\right.\right>}_{=1}\right| = \frac{\hbar}{2}\]
\[\Rightarrow \Delta x\Delta p_x \ge \frac{\hbar}{2}\]

Commuting operators share eigenfunctions#

Example

Show that if all eigenfunctions of operators \(\hat{A}\) and \(\hat{B}\) are identical, \(\hat{A}\) and \(\hat{B}\) commute with each other.

Solution

Denote the eigenvalues of \(\hat{A}\) and \(\hat{B}\) by \(a_i\) and \(b_i\) and the common eigenfunctions by \(\psi_i\). For both operators we have then:

\[\hat{A}\psi_i = a_i\psi_i\textnormal{ and }\hat{B}\psi_i = b_i\psi_i\]

By using these two equations and expressing the general wavefunction \(\psi\) as a linear combination of the eigenfunctions, the commutator can be evaluated as:

\[\hat{A}\hat{B}\psi = \hat{A}\left(\hat{B}\psi\right) = \hat{A}\overbrace{\left(\hat{B}\sum\limits_{i=1}^{\infty}c_i\psi_i\right)}^{\textnormal{complete basis}} = \hat{A}\overbrace{\left(\sum\limits_{i=1}^{\infty}c_i\hat{B}\psi_i\right)}^{\hat{B}\textnormal{ linear}} = \hat{A}\overbrace{\left(\sum\limits_{i=1}^{\infty}c_ib_i\psi_i\right)}^{\textnormal{eigenfunction of }\hat{B}}\]
\[= \overbrace{\sum\limits_{i=1}^{\infty}c_ib_i\hat{A}\psi_i}^{\hat{A}\textnormal{ linear}} = \overbrace{\sum\limits_{i=1}^{\infty}c_ib_ia_i\psi_i}^{\textnormal{eigenfunction of }\hat{A}} = \overbrace{\sum\limits_{i=1}^{\infty}c_ia_ib_i\psi_i}^{a_i\textnormal{ and }b_i\textnormal{ are constants}} = \sum\limits_{i=1}^{\infty}c_ia_i\hat{B}\psi_i\]
\[= \hat{B}\sum\limits_{i=1}^{\infty}c_ia_i\psi_i = \hat{B}\sum\limits_{i=1}^{\infty}c_i\hat{A}\psi_i = \hat{B}\hat{A}\sum\limits_{i=1}^{\infty}c_i\psi_i = \hat{B}\hat{A}\psi\]
\[\Rightarrow \left[\hat{A},\hat{B}\right] = 0\]

Note that the commutation relation must apply to all well-behaved functions and not just for some given subset of functions!