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PHY356F: Quantum Mechanics I. Lecture 8 — Making Sense of Quantum Mechanics

Posted by peeterjoot on November 3, 2010

My notes from Lecture 8, November 2, 2010. Taught by Prof. Vatche Deyirmenjian.

[Click here for a PDF of this post with nicer formatting]


Desai: “Quantum Theory is a linear theory …”

We can discuss SHM without using sines and cosines or complex exponentials, say, only using polynomials, but it would be HARD to do so, and much more work. We want the framework of Hilbert space, linear operators and all the rest to make our life easier.

Dirac: “Mathematics is only a tool, and one should learn the … (FIXME: LOOKUP)”

You have to be able to understand the concepts and apply the concepts as well as the mathematics.

Deyirmenjian: “Think before you compute.”

Joke: With his name included it is the 3Ds. There’s a lot of information included in the question, so read it carefully.

Q: The equation A {\lvert {a_n} \rangle} = a_n {\lvert {a_n} \rangle} for operator A, eigenvalue a_n, n = 1,2 and eigenvector {\lvert {a_n} \rangle} that is identified by the eigenvalue a_n says that

\item (a) measuring the physical quantity associated with A gives result a_n
\item (b) A acting on the state {\lvert {a_n} \rangle} gives outcome a_n
\item (c) the possible outcomes of measuring the physical quantity associated with A are the eigenvalues a_n
\item (d) Quantum mechanics is hard.

{\lvert {a_n} \rangle} is a vector in a vector space or Hilbert space identified by some quantum number a_n, n \in 1,2, \cdots.

The a_n values could be expressions. Example, Angular momentum is describe by states {\lvert {lm} \rangle}, l = 0,1,2,\cdots and m = 0, \pm 1, \pm 2

Recall that the problem is

\begin{aligned}\mathbf{L}^2 {\lvert {lm} \rangle} &= l(l+1) \hbar^2 {\lvert {lm} \rangle} \\ L_z {\lvert {lm} \rangle} &= m \hbar {\lvert {lm} \rangle}\end{aligned} \hspace{\stretch{1}}(4.72)

We have respectively eigenvalues l(l+1)\hbar^2, and m \hbar.

A: Answer is (c). a_n isn’t a measurement itself. These represent possibilities. Contrast this to classical mechanics where time evolution is given without probabilities

\begin{aligned}\mathbf{F}_{\text{net}} &= m \mathbf{a} \\ \mathbf{x}(0), \mathbf{x}'(0) &\implies \mathbf{x}(t), \mathbf{x}'(t)\end{aligned} \hspace{\stretch{1}}(4.74)

The eigenvalues are the possible outcomes, but we only know statistically that these are the possibilities.

(a),(b) are incorrect because we do not know what the initial state is, nor what the final outcome is. We also can’t say “gives result a_n”. That statement is too strong!

Q: We wouldn’t say that A acts on pure state {\lvert {a_n} \rangle}, instead. If the state of the system is {\lvert {\psi} \rangle} = {\lvert {a_5} \rangle}, the probability of measuring outcome a_5 is
\item (a) a_5
\item (b) a_5^2
\item (c) \left\langle{{a_5}} \vert {{\psi}}\right\rangle = \left\langle{{a_5}} \vert {{a_5}}\right\rangle = 1.
\item (d) {\left\lvert{\left\langle{{a_5}} \vert {{\psi}}\right\rangle}\right\rvert}^2 = {\left\lvert{\left\langle{{a_5}} \vert {{a_5}}\right\rangle}\right\rvert}^2 = {\left\lvert{1}\right\rvert}^2 = 1.

A: (d) The eigenformula equation doesn’t say anything about any specific outcome. We want to talk about probability amplitudes. When the system is prepared in a particular pure eigenstate, then we have a guarentee that the probability of measuring that state is unity. We wouldn’t say (c) because the probability amplitudes are the absolute square of the complex number \left\langle{{a_n}} \vert {{a_n}}\right\rangle.

The probability of outcome a_n, given initial state {\lvert {\Psi} \rangle} is {\left\lvert{\left\langle{{a_n}} \vert {{\Psi}}\right\rangle}\right\rvert}^2.

Wave function collapse: When you make a measurement of the physical quantity associated with A, then the state of the system will be the value {\lvert {a_5} \rangle}. The state is not the number (eigenvalue) a_5.

Example: SGZ. With a “spin-up” measurement in the z-direction, the state of the system is {\lvert {z+} \rangle}. The state before the measurement, by the magnet, was {\lvert {\Psi} \rangle}. After the measurement, the state describing the system is {\lvert {\phi} \rangle} = {\lvert {z+} \rangle}. The measurement outcome is +\frac{\hbar}{2} for the spin angular momentum along the z-direction.

FIXME: SGZ picture here.

There is an interaction between the magnet and the silver atoms coming out of the oven. Before that interaction we have a state described by {\lvert {\Psi} \rangle}. After the measurement, we have a new state {\lvert {\phi} \rangle}. We call this the collapse of the wave function. In a future course (QM interpretations) the language used and interpretations associated with this language can be discussed.

Q: Express Hermitian operator A in terms of its eigenvectors.
Q: The above question is vague because

\item (a) The eigenvectors may form a descrete set.
\item (b) The eigenvectors may form a continuous set.
\item (c) The eigenvectors may not form a complete set.
\item (d) The eigenvectors are not given.

A: None of the above. A Hermitian operator is guarenteed to have a complete set of eigenvectors. The operator may also be both discrete and continuous (example: the complete spin wave function).


\begin{aligned}A &= A \mathbf{1} \\ &= A \left( \sum_n {\lvert {a_n} \rangle} {\langle {a_n} \rvert} \right) \\ &= \sum_n (A {\lvert {a_n} \rangle} ){\langle {a_n} \rvert} \\ &= \sum_n (a_n {\lvert {a_n} \rangle}) {\langle {a_n} \rvert} \\ &= \sum_n a_n {\lvert {a_n} \rangle} {\langle {a_n} \rvert}\end{aligned}


\begin{aligned}A &= A \mathbf{1} \\ &= A \left( \int d\alpha {\lvert {\alpha} \rangle} {\langle {\alpha} \rvert} \right) \\ &= \int d\alpha (A {\lvert {\alpha} \rangle} ){\langle {\alpha} \rvert} \\ &= \int d\alpha (\alpha {\lvert {\alpha} \rangle}) {\langle {\alpha} \rvert} \\ &= \int d\alpha \alpha {\lvert {\alpha} \rangle} {\langle {\alpha} \rvert}\end{aligned}

An example is the position eigenstate {\lvert {x} \rangle}, eigenstate of the Hermitian operator X. \alpha is a label indicating the summation.

general case with both discrete and continuous:

\begin{aligned}A &= A \mathbf{1} \\ &= A \left( \sum_n {\lvert {a_n} \rangle} {\langle {a_n} \rvert} + \int d\alpha {\lvert {\alpha} \rangle} {\langle {\alpha} \rvert} \right) \\ &= \sum_n \left(A {\lvert {a_n} \rangle} \right){\langle {a_n} \rvert} + \int d\alpha \left(A {\lvert {\alpha} \rangle} \right){\langle {\alpha} \rvert} \\ &= \sum_n \left(a_n {\lvert {a_n} \rangle}\right) {\langle {a_n} \rvert} + \int d\alpha \left(\alpha {\lvert {\alpha} \rangle}\right) {\langle {\alpha} \rvert} \\ &= \sum_n a_n {\lvert {a_n} \rangle} {\langle {a_n} \rvert} + \int d\alpha \alpha {\lvert {\alpha} \rangle} {\langle {\alpha} \rvert}\end{aligned}

Problem Solving

\item MODEL — Quantum, linear vector space
\item VISUALIZE — Operators can have discrete, continuous or both discrete and continuous eigenvectors.
\item SOLVE — Use the identity operator.
\item CHECK — Does the above expression give A {\lvert {a_n} \rangle} = a_n {\lvert {a_n} \rangle}.


\begin{aligned}A {\lvert {a_m} \rangle}&= \sum_n a_n {\lvert {a_n} \rangle} \left\langle{{a_n}} \vert {{a_m}}\right\rangle + \int d\alpha \alpha {\lvert {\alpha} \rangle} \left\langle{{\alpha}} \vert {{a_n}}\right\rangle \\ &= \sum_n a_n {\lvert {a_n} \rangle} \delta_{nm} \\ &= a_m {\lvert {a_m} \rangle}\end{aligned}

What remains to be shown, used above, is that the continous and discrete eigenvectors are orthonormal. He has an example vector space, not yet discussed.

Q: what is {\langle {\Psi_1} \rvert} A {\lvert {\Psi_1} \rangle}, where A is a Hermitian operator, and {\lvert {\Psi_1} \rangle} is a general state.

A: {\langle {\Psi_1} \rvert} A {\lvert {\Psi_1} \rangle} = average outcome for many measurements

of the physical quantity associated with A such that the system is prepared in state {\lvert {\Psi_1} \rangle} prior to each measurement.

Q: What if the preparation is {\lvert {\Psi_2} \rangle}. This isn’t neccessarily an eigenstate of A, it is some linear combination of eigenstates. It is a general state.
A: {\langle {\Psi_2} \rvert} A {\lvert {\Psi_2} \rangle} = average of the physical quantity associated with A, but the preparation is {\lvert {\Psi_2} \rangle}, not {\lvert {\Psi_1} \rangle}.

Q: What if our initial state is a little bit of {\lvert {\Psi_1} \rangle}, and a little bit of {\lvert {\Psi_2} \rangle}, and a little bit of {\lvert {\Psi_N} \rangle}. ie: how to describe what comes out of the oven in the SG experiment. That spin is a statistical mixture. We could understand this as only a statistical mix. This is a physical relavent problem.
A: To describe that statistical situtation we have the following.

\begin{aligned}\left\langle{{A}}\right\rangle_{\text{average}} = \sum_j w_j {\langle {\Psi_j} \rvert} A {\lvert {\Psi_j} \rangle}\end{aligned} \hspace{\stretch{1}}(4.76)

We sum up all the expectation values modified by statistical weighting factors. These w_j‘s are statistical weighting factors for a preparation associated with {\lvert {\Psi_j} \rangle}, real numbers (that sum to unity). Note that these states {\lvert {\Psi_j} \rangle} are not neccessarily orthonormal.

With insertion of the identity operator we have

\begin{aligned}\left\langle{{A}}\right\rangle_{\text{average}}&= \sum_j w_j {\langle {\Psi_j} \rvert} \mathbf{1} A {\lvert {\Psi_j} \rangle} \\ &= \sum_j w_j {\langle {\Psi_j} \rvert} \left( \sum_n {\lvert {a_n} \rangle} {\langle {a_n} \rvert} \right) A {\lvert {\Psi_j} \rangle} \\ &= \sum_j \sum_n w_j \left\langle{{\Psi_j}} \vert {{a_n}}\right\rangle {\langle {a_n} \rvert} A {\lvert {\Psi_j} \rangle} \\ &= \sum_j \sum_n w_j {\langle {a_n} \rvert} A {\lvert {\Psi_j} \rangle} \left\langle{{\Psi_j}} \vert {{a_n}}\right\rangle  \\ &= \sum_n {\langle {a_n} \rvert} A \left( \sum_j w_j {\lvert {\Psi_j} \rangle} {\langle {\Psi_j} \rvert} \right) {\lvert {a_n} \rangle}  \\ \end{aligned}

This inner bit is called the density operator \rho

\begin{aligned}\rho &\equiv \sum_j w_j {\lvert {\Psi_j} \rangle} {\langle {\Psi_j} \rvert}\end{aligned} \hspace{\stretch{1}}(4.77)

Returning to the average we have

\begin{aligned}\left\langle{{A}}\right\rangle_{\text{average}} = \sum_n {\langle {a_n} \rvert} A \rho {\lvert {a_n} \rangle} \equiv \text{Tr}(A \rho)\end{aligned} \hspace{\stretch{1}}(4.78)

The trace of an operator A is

\begin{aligned}\text{Tr}(A) = \sum_j {\langle {a_j} \rvert} A {\lvert {a_j} \rangle} = \sum_j A_{jj}\end{aligned} \hspace{\stretch{1}}(4.79)

Section 5.9, Projection operator.

Returning to the last lecture. From chapter 1, we have

\begin{aligned}P_n = {\lvert {a_n} \rangle} {\langle {a_n} \rvert}\end{aligned} \hspace{\stretch{1}}(4.80)

is called the projection operator. This is physically relavent. This takes a general state and gives you the component of that state associated with that eigenvector. Observe

\begin{aligned}P_n {\lvert {\phi} \rangle} ={\lvert {a_n} \rangle} \left\langle{{a_n}} \vert {{\phi}}\right\rangle =\underbrace{\left\langle{{a_n}} \vert {{\phi}}\right\rangle}_{\text{coefficient}} {\lvert {a_n} \rangle}\end{aligned} \hspace{\stretch{1}}(4.81)

Example: Projection operator for the {\lvert {z+} \rangle} state

\begin{aligned}P_{z+} = {\lvert {z+} \rangle} {\langle {z+} \rvert}\end{aligned} \hspace{\stretch{1}}(4.82)

We see that the density operator

\begin{aligned}\rho &\equiv \sum_j w_j {\lvert {\Psi_j} \rangle} {\langle {\Psi_j} \rvert},\end{aligned} \hspace{\stretch{1}}(4.83)

can be written in terms of the Projection operators

\begin{aligned}{\lvert {\Psi_j} \rangle} {\langle {\Psi_j} \rvert} = \text{Projection operator for state} {\lvert {\Psi_j} \rangle}\end{aligned}

The projection operator is like a dot product, determining the quantity of a state that lines in the direction of another state.

Q: What is the projection operator for spin-up along the z-direction.

\begin{aligned}P_{z+} = {\lvert {z+} \rangle}{\langle {z+} \rvert}\end{aligned} \hspace{\stretch{1}}(4.84)

Or in matrix form with

\begin{aligned}{\langle {z+} \rvert} &=\begin{bmatrix}1 \\ 0\end{bmatrix} \\ {\langle {z-} \rvert} &=\begin{bmatrix}0 \\ 1\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(4.85)


\begin{aligned}P_{z+} = {\lvert {z+} \rangle}{\langle {z+} \rvert} =\begin{bmatrix}1 \\ 0\end{bmatrix}\begin{bmatrix}1 & 0\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(4.87)

Q: A harder problem. What is P_\chi, where

\begin{aligned}{\lvert {\chi} \rangle} =\begin{bmatrix}c_1 \\ c_2\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(4.88)

Note: We want normalized states, with \left\langle{{\chi}} \vert {{\chi}}\right\rangle = {\left\lvert{c_1}\right\rvert}^2 + {\left\lvert{c_2}\right\rvert}^2 = 1.


\begin{aligned}P_{\chi} = {\lvert {\chi} \rangle}{\langle {\chi} \rvert} =\begin{bmatrix}c_1^{*} \\ c_2^{*}\end{bmatrix}\begin{bmatrix}c_1 & c_2\end{bmatrix}=\begin{bmatrix}c_1^{*} c_1 & c_1^{*} c_2 \\ c_2^{*} c_1 & c_2^{*} c_2\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(4.89)

Observe that this has the proper form of a projection operator is that the square is itself

\begin{aligned}({\lvert {\chi} \rangle}{\langle {\chi} \rvert}) ({\lvert {\chi} \rangle}{\langle {\chi} \rvert})&= {\lvert {\chi} \rangle} (\left\langle{{\chi}} \vert {{\chi}}\right\rangle ){\langle {\chi} \rvert} \\ &= {\lvert {\chi} \rangle} {\langle {\chi} \rvert}\end{aligned}

Q: Show that P_{\chi} = a_0 \mathbf{1} + \mathbf{a} \cdot \boldsymbol{\sigma}, where \mathbf{a} = (a_x, a_y, a_z) and \boldsymbol{\sigma} = (\sigma_x, \sigma_y, \sigma_z).

A: See Section 5.9. Note the following about computing (\boldsymbol{\sigma} \cdot \mathbf{a})^2.

\begin{aligned}(\boldsymbol{\sigma} \cdot \mathbf{a})^2&=(a_x \sigma_x+ a_y \sigma_y+ a_z \sigma_z)(a_x \sigma_x+ a_y \sigma_y+ a_z \sigma_z) \\ &=a_x a_x \sigma_x \sigma_x+a_x a_y \sigma_x \sigma_y+a_x a_z \sigma_x \sigma_z+a_y a_x \sigma_y \sigma_x+a_y a_y \sigma_y \sigma_y+a_y a_z \sigma_y \sigma_z+a_z a_x \sigma_z \sigma_x+a_z a_y \sigma_z \sigma_y+a_z a_z \sigma_z \sigma_z \\ &= (a_x^2 + a_y^2 + a_z^2) I+ a_x a_y ( \sigma_x \sigma_y + \sigma_y \sigma_x)+ a_y a_z ( \sigma_y \sigma_z + \sigma_z \sigma_y)+ a_z a_x ( \sigma_z \sigma_x + \sigma_x \sigma_z) \\ &= {\left\lvert{\mathbf{x}}\right\rvert}^2 I\end{aligned}

So we have

\begin{aligned}(\boldsymbol{\sigma} \cdot \mathbf{a})^2 = (\mathbf{a} \cdot \mathbf{a}) \mathbf{1} \equiv \mathbf{a}^2\end{aligned} \hspace{\stretch{1}}(4.90)

Where the matrix representations

\begin{aligned}\sigma_x &\leftrightarrow \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \\ \sigma_y &\leftrightarrow \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \\ \sigma_z &\leftrightarrow \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix},\end{aligned} \hspace{\stretch{1}}(4.91)

would be used to show that

\begin{aligned}\sigma_x^2 = \sigma_y^2 = \sigma_z^2 = I\end{aligned} \hspace{\stretch{1}}(4.94)


\begin{aligned}\sigma_x \sigma_y &= -\sigma_y \sigma_x \\ \sigma_y \sigma_z &= -\sigma_z \sigma_y \\ \sigma_z \sigma_x &= -\sigma_x \sigma_z\end{aligned} \hspace{\stretch{1}}(4.95)


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