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10: Expectation values, 2D-PIB and Heisenberg Uncertainty Principle (Lecture)

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    93435
  • Recap of Lecture 9:

    Last lecture focused on gaining an intuition of wavefunctions with an emphasis on the energies of the particle in the box. We continued the discussion of the PIB and the intuition we want from the model system. We revised the time-dependent solutions to the model system (which is always there). We emphasized that the total wavefunction must be oscillating in time (although we often ignore that in this class) and that it has both a real and imaginary component (we will revisit that again later on). We discussion symmetry of functions and integration over odd integrands. We used that fact to justify (in part) orthogonality, which is a powerful (blow your mind) fact that the eigenstates of an operator will be orthogonal.

    \[\int_{0}^{L}\psi _{m}(x)\psi _{n}(x)dx=0\: \: \: if\: \: \: m \neq n \label{3.5.19}\]

    This is really cool as we will see. So, ideally we want wavefunctions to be normalized and they are intrinsically orthogonal (this is an intrinsic property from solving the eigenvalue/vector problem).

    Expectation Value of Position of Particle in a Box (with n=1)

    For the ground-state particle in a box wavefunction with \(n=1\):

    \[\psi_{n=1} = \sqrt{\dfrac{2}{L}} \sin \left(\dfrac{\pi x}{L} \right) \label{PIB}\]

    This state has the following probability distribution:

    \[\psi^*_{n=1} \psi_{n=1} = \dfrac{2}{L} \sin^2 \left(\dfrac{\pi x}{L} \right)\]

    (left) The ground-state (n=1) wavefunction for a particle in a box. (right) The ground-state (n=1) probability for a particle in a box.

    In general the expectation value is the measurable for an operator

    \[ \langle o \rangle = \int _{-\infty}^{+\infty} \psi^* \hat{O} \psi \, dx \label{expect}\]

    The expectation value for position with the \(\hat{x} = x\) operation for this any wavefunction (Equation \(\ref{expect}\)) is

    \[ \langle x \rangle = \int _{-\infty}^{+\infty} \psi^* x \psi \, dx \]

    which for the ground-state wavefunction (Equation \(\ref{PIB}\)) shown in Figure \(\PageIndex{1}\) is

    \[\begin{align} \langle x \rangle &= \int _{-\infty}^{+\infty} \sqrt{\dfrac{2}{L}} \sin \left(\dfrac{\pi x}{L} \right) x \sqrt{\dfrac{2}{L}} \sin \left(\dfrac{\pi x}{L} \right) \, dx \\[4pt] &= \dfrac{2}{L} \int _{-\infty}^{+\infty} x \sin^2 \left(\dfrac{\pi x}{L} \right) \label{GSExpect} \end{align}\]

    You can do this equation, but you can see via symmetry that

    \[ \langle x \rangle = \dfrac{L}{2}\]

    limits of Integration

    A wavefunction gives you the probability of a particle to be seen somewhere. If it's not confined to be in some region, like a potential well or something, then it has a probability to be seen anywhere in the space. That's why the integration runs from \(−∞\) to \(+∞\) (remember that you could never pinpoint a quantum mechanical particle in space). That can be converted to a finite volume of space if the confinement is strong. For example, in the case of a particle in an infinite potential well, the boundary condition is that the wavefunction should vanish at the potential well boundaries.

    Expectation Value of Momentum of Particle in a Box

    What is the average momentum of a particle in the box? We start with Equation \(\ref{expect}\) and use the momentum operator

    \[\hat{p}_{x}=-i\hbar\dfrac{\partial}{\partial x}\label{3.2.3a}\]

    We note that the particle-in-a-box wavefunctions are not eigenfunctions of the momentum operator (Exercise \(\PageIndex{4}\)). However, this does not mean that Equation \(\ref{expect}\) is inapplicable as Example \(\PageIndex{2}\) demonstrates.

    Example \(\PageIndex{2}\): The Average Momentum of a Particle in a Box is Zero

    Even though the wavefunctions are not momentum eigenfunctions, we can calculate the expectation value for the momentum. Show that the expectation or average value for the momentum of an electron in the box is zero in every state (i.e., arbitrary values of \(n\)).

    Strategy:

    First write the expectation value integral (Equation \(\ref{expect}\)) with the momentum operator. Then insert the expression for the wavefunction and evaluate the integral as shown here.

    Answer:

    \[\left \langle P \right \rangle = \int \limits ^L_0 \psi ^*_n (x) \left ( -i\hbar \dfrac {d}{dx} \right ) \psi _n (x) dx\]

    \[ = \int \limits ^L_0 \left (\dfrac {2}{L} \right )^{1/2} \sin (\dfrac {n \pi x}{L}) \left ( -i\hbar \dfrac {d}{dx} \right ) \left (\dfrac {2}{L} \right )^{1/2} \sin (\dfrac {n \pi x }{L} ) dx \]

    \[ = -i \hbar \left (\dfrac {2}{L} \right ) \int \limits ^L_0 \sin (\dfrac {n \pi x}{L}) \left ( \dfrac {d}{dx} \right ) \sin (\dfrac {n \pi x}{L}) dx\]

    \[ = -i \hbar \left (\dfrac {2}{L} \right ) \left ( \dfrac {n \pi}{L} \right ) \int \limits ^L_0 \sin (\dfrac {n \pi x}{L}) \cos (\dfrac {n \pi x}{L}) dx\]

    \[= 0\]

    Note that this makes sense since the particles spends an equal amount of time traveling in the \(+x\) and \(–x\) direction.

    Interpretation:

    It may seem that this means the particle in a box does not have any momentum, which is incorrect because we know the energy is never zero. In fact, the energy that we obtained for the particle-in-a-box is entirely kinetic energy because we set the potential energy at 0. Since the kinetic energy is the momentum squared divided by twice the mass, it is easy to understand how the average momentum can be zero and the kinetic energy finite

    Time to Bring in the Heavy Guns: Dirac Notation

    Kets

    In Dirac’s notation what is known is put in a ket, \(| \, \rangle\). So, for example, \(| p \rangle\) expresses the fact that a particle has momentum \(p\). It could also be more explicit: \(| p=2 \rangle\), the particle has momentum equal to 2; \(| x=1.23 \rangle\), the particle has position 1.23. \(| \Psi \rangle\) represents a system in the state \( \Psi \) and is therefore called the state vector. The ket can also be interpreted as the initial state in some transition or event.

    Bras

    The bra \(\langle \, | \) represents the final state or the language in which you wish to express the content of the ket \(| \, \rangle\). For example,\( \langle 0.25 | \Psi \rangle\) is the probability amplitude that a particle in state \( \Psi\) will be found at position \(x = 0.25\). In conventional notation we write this as \( \Psi(x=0.25) \), the value of the function \( \Psi \) at \(x\)=0.25. The absolute square of the probability amplitude, \( \left| \langle x=0.25| \Psi \rangle \right|^2\), is the probability density that a particle in state \( \Psi \) will be found at \(x\) = 0.25. Thus, we see that a bra-ket pair can represent an event, the result of an experiment. In quantum mechanics an experiment consists of two sequential observations - one that establishes the initial state (ket) and one that establishes the final state (bra).

    Bra-Ket Pairs

    If we write \( \langle x| \Psi \rangle \), we are expressing \( \Psi \) in coordinate space without being explicit about the actual value of \(x\). \(\langle 0.25 | \Psi \rangle \) is a number, but the more general expression \(\rangle x | \Psi \rangle \) is a mathematical function, a mathematical function of \(x\), or we could say a mathematical algorithm for generating all possible values of \( \langle x| \Psi \rangle \), the probability amplitude that a system in state \( | \Psi \rangle \) has position \(x\).

    Heisenberg's Uncertainty Principle Redux

    This Heisenberg Uncertainty Principles was originally introduces as

    \[ \begin{align} \Delta{p_x}\Delta{x} &\ge \dfrac{h}{4\pi} \\[4pt] &\ge \dfrac{\hbar}{2} \label{1.9.5} \end{align}\]

    with \(\hbar = \frac{h}{2\pi}= 1.0545718 \times 10^{-34}\; m^2 \cdot kg / s\).

    reality_wavepacket.gif
    Figure \(\PageIndex{1}\): The animation shows the relevant spreads in the uncertainty for position and momentum of light/photons (light wave's corresponding photon particle). From the result of de Broglie, we know that for a particle with known momentum, \(p\) will have a precise value for its de Broglie wavelength can be determined (and hence a specific color of the light).

    Calculating Uncertainties

    Equation \(\ref{1.9.5}\) relates the uncertainty of momentum and position. An immediate questions that arise is if \(\Delta x\) represents the full range of possible \(x\) values or if it is half (e.g., \(\langle x \rangle \pm \Delta x\)). \(\Delta x\) is the standard deviation and is a statistic measure of the spread of \(x\) values. The use of half the possible range is more accurate estimate of \(\Delta x\). Later, we can show that once a wavefunction can be constructed to describe the system, then both \(x\) and \(\Delta x\) can be explicitly derived.

    In quantum mechanics,\(\langle x \rangle\) is the expected value of \(x\), which is the idea of a mean value in statistics. The expected value \(\langle x \rangle\) is equal to

    \[\langle x \rangle = \int_{-\infty}^{\infty} \psi^* x \psi \,dx\]

    where the wavefunctions are normalized (to satisfy the Born interpretation of wavefunctions)

    \[\int_{-\infty}^{\infty} \psi^* \psi \,dx = 1 \]

    This can be generalized to any operator such that

    \[\langle \hat{A} \rangle= \int_{-\infty}^{\infty} \psi^* \hat{A} \psi \, dx \label{5}\]

    The expected value of \(x\) in the 1D Particle in a Box problem can be found, because we know \(\psi (x)\), and we are given the operator \(\hat{x}\). Thus,

    \[\langle x \rangle = \int_{-\infty}^{\infty} \psi^* (x) \hat{x} \psi (x) dx = \int_0^L \sqrt{\frac {2}{L}} \sin \left(\frac{n\pi x}{L}\right) x \sqrt{\frac {2}{L}} \sin \left(\frac{n\pi x}{L} \right)dx = \frac {2}{L} \int_0^a x\cdot \sin^2\left(\frac {n\pi x}{L}\right)dx \label{6}\]

    Following through with the integral ultimately results in \(\langle x \rangle = L/2\), for all \(n\) values. The variance of the expected value of \(x\), represented as \(\sigma^2_x\) can be found by taking the difference of \(\langle x^2 \rangle\) and \(\langle x \rangle^2\), which is a measure of the "spread of the distribution" (see Worksheet #1) and is given by

    \[ \langle \sigma_x^2 \rangle = \int_{a}^{b} (x-\langle x \rangle )^2 \psi^* \psi dx \label{5eq}\]

    By using the same method to find \(\langle x \rangle\), Equation \(\ref{6}\) is solved to equal

    \[\langle x^2 \rangle = \dfrac {L^2}{3} - \dfrac {L^2}{2n^2 \pi^2}\]

    Square rooting the variance gives the standard deviation

    \[\sigma_x = \sqrt{ \left(\dfrac {L}{2\pi n} \right)^2 \left(\dfrac{\pi^2 h^2}{3} -2\right)}.\]

    The standard deviation for momentum is found in a similar fashion. Multiplying the resulting \(\sigma_p\) and \(\sigma_x\) gives Heisenberg's uncertainty principle product:

    \[\sigma_x \sigma_p \geq \hbar/2\]

    We can substitute the standard deviation for the uncertainty in the Heisenberg expression

    \[\sigma_x = \Delta x\]

    which provides a quantitative mechanism to interpret the uncertainty relationship.

    Example \(\PageIndex{1}\):

    The uncertainty of a position of a particle (\(\Delta x\)) is defined as

    \[ \Delta x = \sqrt{ \langle x^2 \rangle - \langle x \rangle ^2} \]

    For the particle in the box at the ground eigenstate (\(n=1\)), what is the uncertainty in the value \( x \)?

    Step 1: expectation value of the position

    The expectation value of the position operator is:

    \[\langle \hat{x}\rangle = \int_{0}^{a}\psi_n^*\hat{x}\psi_n dx\label{1}\]

    or in Dirac notation

    \[\langle \hat{x}\rangle = \langle n | \hat{x} | n \rangle\]

    which expands to

    \[\langle \hat{x}\rangle = \dfrac{2}{a}\int_{0}^{a}x\sin^2\dfrac{\pi x}{a}dx = \dfrac{a}{2}\tag{2}\]

    Where we used the half angle formula and integration by parts to compute the integral.

    Step 2: expectation value of the position operator squared

    The expectation value of the position operator squared is:

    \[\langle \hat{x^2}\rangle = \int_{0}^{a}\psi_n^*\hat{x^2}\psi_n dx\label{3}\]

    or in Dirac notation

    \[\langle \hat{x}\rangle = \langle n | \hat{x^2} | n \rangle\]

    which expands to

    \[\langle \hat{x^2}\rangle = \dfrac{2}{a}\int_{0}^{a}x^2\sin^2\dfrac{\pi x}{a}dx = \dfrac{a^2}{3} - \dfrac{a^2}{2\pi^2}\tag{4}\]

    Again we used the half angle formula and integration by parts to compute this integral.

    Answer:

    \[\Delta x = \sqrt{\langle \hat{x^2}\rangle - \langle \hat{x}\rangle^2 } \label{12}\]

    Now insert \(\langle \hat{x^2}\rangle\) and \(\langle \hat{x}\rangle^2\).

    \[\Delta x =\sqrt{\dfrac{a^2}{3} - \dfrac{a^2}{2\pi^2} - \dfrac{a^2}{4}}=a\sqrt{\dfrac{1}{12}-\dfrac{1}{2\pi^2} }=0.18a\]

    Exercise \(\PageIndex{3}\)

    Prove the following equality

    \[ \langle \sigma_x^2 \rangle = \langle x^2 \rangle - \langle x \rangle^2\]

    3D Boxes

    The 1D particle in the box problem can be expanded to consider a particle within a 3D box for three lengths \(a\), \(b\), and \(c\). When there is NO FORCE (i.e., no potential) acting on the particles inside the box.

    Figure used with permission (CC-SA-BY-NC; Information and Communication Technology)

    The potential for the particle inside the box

    \[V(\vec{r}) = 0\]

    • \(0 \leq x \leq a\)
    • \(0 \leq y \leq b\)
    • \(0 \leq z \leq c\)
    • \(a < x < 0\)
    • \(b < y < 0\)
    • \(c < z < 0\)

    \(\vec{r}\) is the vector with all 3 components along the 3 axis of the 3-D box: \(\vec{r} = a\hat{x} + b\hat{y} + c\hat{z}\). When the potential energy is infinite, then the wave function equals zero. When the potential energy is zero, then the wavefunction obeys the Schrödinger equation.

    \[\psi(r) = \sqrt{\dfrac{8}{v}}\sin \left( \dfrac{n_{x} \pi x}{a} \right) \sin \left(\dfrac{n_{y} \pi y}{b}\right) \sin \left(\dfrac{ n_{z} \pi z}{c} \right)\]

    \[v = a \times b \times c = volume \; of \; box\]

    with Total Energy levels:

    \[E_{n_x,n_y,x_z} = \dfrac{h^{2}}{8m}\left(\dfrac{n_{x}^{2}}{a^{2}} + \dfrac{n_{y}^{2}}{b^{2}} + \dfrac{n_{z}^{2}}{c^{2}}\right) \label{3.9.10}\]

    As with 1-D boxes, the quantum numbers \(n_x\), \(n_y\), and \(n_z\) go from 1 to infinity.

    The wavefunction of a 2D well with \(n_x=4\) and \(n_y=4\). image used with permission (Public Domain; Keenan pepper )

    Exercise \(\PageIndex{1}\)

    What is the energy of the ground state of a 3D box that is \(a\) by \(a\) by \(a\)?

    Answer:

    This is when \(n_x=1\), \(n_y=1\), and \(n_z=1\). So from Equation \(\ref{3.9.10}\), this is

    \[E_{1,1,1}= \dfrac{h^{2}}{8m}\left(\dfrac{1}{a^{2}} + \dfrac{1}{a^{2}} + \dfrac{1}{a^{2}}\right) = \dfrac{3h^{2}}{8ma^{2}} \]

    Exercise \(\PageIndex{2}\)

    What is the energy of the 1st excited state of a 3D box that is \(a\) by \(a\) by \(a\)?

    Answer:

    This is trickier, since there is a degeneracy in the system with three wavefunctions having the same energy:

    • \(\Psi_{2,1,1}\)
    • \(\Psi_{1,2,1}\)
    • \(\Psi_{1,1,2}\)

    Each of these wavefunctions have the same energy (via Equation \(\ref{3.9.10}\)) of

    \[E= \dfrac{6h^{2}}{8ma^{2}}\]