Linear Algebra and Quantum Mechanics Crash Course Conceptual Solutions

Quantum Mechanics (9 pts.) In your own words, define the following:
1. Wave-particle duality (3pts.) Wave-particle duality means that light and small particles have both properties of particles and properties of waves. This can be proven experimentally where both light and particles will behave as a wave in certain experiments but a particle in others
2. Superposition (3pts.) Superposition means that a quantum mechanical system can be in a combination of different states at the same time and only must chose one state to be in upon measurement.
3. Entanglement (3pts.) When quantum mechanical systems become entangled it means that they are no longer behaving independently but are now dependent on the states of the other systems.
Statistics (13 pts.) Consider a group of 10 rooms in an old building. Three of the rooms have no spiders, six of the rooms have one spider, and one room has 5 spiders.
- (1 pt.) Is the number of spiders in a room a discrete or continuous variable? Explain. The number of spiders is a discrete variable since it can only come as a whole number (integer)
- (2 pts.) What is the probability that you randomly select a room with 1 spider? Show your work. Number of rooms with one spider: 6 = \(N_1\) Total number of rooms: 3 + 6 + 1 = 10 = N Probability of selecting a room with one spider: \(P_1 = \frac{N_1}{N} = \frac{6}{10} = 0.6\)
- (4 pts.) Show that the total probability of this data set is 1. \[P_1 = 0.6\ (above)\] \[P_0 = \frac{3}{10} = 0.3\] \[P_5 = \frac{1}{10} = 0.1\] \[P_0 + P_1 + P_5 = 0.3 + 0.6 + 0.1 = 1\]
- (2 pts.) What is the average number of spiders in a room? \[\langle s \rangle = \sum_s sP(s) = 0P_0 + 1P_1 + 5P_5 = 0(0.3) + 1(0.6) + 5(0.1) = 1.1 spiders\]
- (4 pts.) If s is the number of spiders in a room, then the number of flies in the room, f, can be described by the equation \(f(s) = 10-2s\). What is the expectation value for the number of flies in the building? \[\langle f \rangle = \sum_s f(s)P(s) = f(0)P_0 + f(1)P_1 + f(5)P_5 = (10-2(0))(0.3) + (10-2(1))(0.6) + (10-2(5))(0.1) \] \[10(0.3) + 8(0.6) + 0(0.1) = 7.8 flies\]
Wavefunctions (4pts.). Consider a quantum mechanical state that can be in one of three states, \(|1\rangle\), \(|2\rangle\), and \(|3\rangle\). State \(|1\rangle\) has an energy of a, state \(|2\rangle\) has an energy of b, and state \(|3\rangle\) has an energy of c. The quantum mechanical state being studied is described by the following wavefunction. \[|\psi\rangle = \frac{2i}{\sqrt{30}}|1\rangle -\frac{5}{\sqrt{30}}|2\rangle +\frac{1}{\sqrt{30}}|3\rangle\] If you measure the energy of state \(|\psi\rangle\), what is the most likely result? Explain your answer. The most likely energy to be measured is b since the wavefunction has the highest probability of collapsing into state \(|2\rangle\) \[P(|2\rangle) = |\frac{5}{\sqrt{30}}|^2 = \frac{25}{30}\]
**Linear Algebra for Quantum Mechanics (24 pts.)

(6 pts.) Consider the two vectors \(|x\rangle\) and \(|y\rangle\) below. Show that \(\langle x | y \rangle ^* = \langle y | x \rangle\), where the \(*\) denotes the complex conjugate. Note that this is a general rule that we will be using many times throughout the course.** \[|x\rangle = \begin{bmatrix}-1 \\ 2i \\ 1 \end{bmatrix}\] \[|y\rangle = \begin{bmatrix} 1 \\ 0 \\ i \end{bmatrix}\]

First find the bra states that correspond to the above kets \[\langle x | = \begin{bmatrix}-1 & -2i & 1 \end{bmatrix}\] \[\langle y | = \begin{bmatrix} 1 & 0 & -i \end{bmatrix}\]

Now to compute the dot products \[\langle x | y \rangle = \begin{bmatrix}-1 & -2i & 1 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \\ i \end{bmatrix}\] \[ = -1(1) + -2i(0) + 1(i) = i-1\]

\[\langle y | x \rangle = \begin{bmatrix} 1 & 0 & -i \end{bmatrix}\begin{bmatrix}-1 \\ 2i \\ 1 \end{bmatrix}\] \[= 1(-1) + 0(2i) + -i(1) = -i-1\]

Note that \(\langle x | y \rangle^* = -i-1 = \langle y | x \rangle\)

(6 pts.) Given below are the three Pauli matrices which we will be using many times in this course. Show that \(\sigma_x\sigma_x = \sigma_y\sigma_y = \sigma_z\sigma_z = \textbf{I}\) (the identity matrix) and that \([\sigma_x, \sigma_y] = 2i\sigma_z\). \[\sigma_x = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\] \[\sigma_y = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}\] \[\sigma_z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\]

The below matrix multiplication will be performed using Wolfram alpha: \[\sigma_x\sigma_x = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix} = \text{bf{I}\] \[\sigma_y\sigma_y = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}\begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix} = \textbf{I}\] \[\sigma_z\sigma_z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}\]

\[\sigma_x\sigma_y = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} = \begin{bmatrix}i&0\\0&-i\end{bmatrix}\] \[\sigma_y\sigma_x = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix}-i&0\\0&i\end{bmatrix}\]

\[[\sigma_x, \sigma_y] = \sigma_x\sigma_y - \sigma_y\sigma_x\] \[= \begin{bmatrix}i&0\\0&-i\end{bmatrix}-\begin{bmatrix}-i&0\\0&i\end{bmatrix} = \begin{bmatrix} 2i&0\\0&-2i\\\end{bmatrix} = 2i\begin{bmatrix} 1&0\\0&-1\\\end{bmatrix} = 2i\sigma_z\]

(6 pts.) Given N vectors of length N, these vectors are called a basis if any vector of length N can be constructed as a weighted linear sum of the original vectors (i.e. the original vectors multiplied by scalars and added or subtracted from each other). Show that \(|0\rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix}\) and \(|1\rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix}\) for a basis for all vectors of length 2.

Consider the generic vector of length 2 \(\begin{bmatrix}\alpha\\\beta\end{bmatrix}\):

\[\begin{bmatrix}\alpha\\\beta\end{bmatrix} = \begin{bmatrix}\alpha\\0\end{bmatrix} + \begin{bmatrix}0\\\beta\end{bmatrix}\] \[= \alpha\begin{bmatrix}1\\0\end{bmatrix} + \beta\begin{bmatrix}0\\1\end{bmatrix} = \alpha|0\rangle + \beta|1\rangle\]

Thus any vector of length 2 can be created using a linear combination of \(|0\rangle\) and \(|1\rangle\) \(\longrightarrow\) Thus \(|0\rangle\) and \(|1\rangle\) are a basis for vectors of length 2.

(6 pts.) Given \(|0\rangle\) and \(|1\rangle\) defined in the previous problem, show that \(\langle 0 | 0 \rangle = \langle 1 | 1 \rangle = 1\) and that \(\langle 0 | 1 \rangle = \langle 1 | 0 \rangle = 0\). Any basis which has these properties (\(\langle i | j \rangle = \delta_{ij}\), where \(\delta\) is the Kronecker delta) is called an orthonormal basis (it is both normalized and orthogonal).

First to define the bra states: \[\langle 0 | = \begin{bmatrix}1&0\end{bmatrix}\] \[\langle 1 | = \begin{bmatrix}0&1\end{bmatrix}\]

First, let’s do the dot products of each vector with itself:

\[\langle 0 | 0 \rangle = \begin{bmatrix}1&0\end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = 1(1) + 0(0) = 1\] \[\langle 1 | 1 \rangle = \begin{bmatrix}0&1\end{bmatrix}\begin{bmatrix} 0 \\ 1 \end{bmatrix} = 0(0) + 1(1) = 1\]

Now to show the relationship with the opposite state dot products:

\[\langle 0 | 1 \rangle = \begin{bmatrix}1&0\end{bmatrix}\begin{bmatrix} 0 \\ 1 \end{bmatrix} = 1(0) + 0(1) = 0\] \[\langle 1 | 0 \rangle = \begin{bmatrix}0&1\end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = 0(1) + 1(0) = 0\]