Chapter 6

Further Developments of the Theory

In card four we introduced the Schrödinger equation for a charged particle in an electric potential. This equation allows us to determine $\psi(x,y,z,t)$ , that is the distribution of probability amplitudes as a function of time for the triple of measurable quantities $(x,y,z)$ . The central aim of the present card will be to determine the distribution of probability amplitudes for other observable quantities, in particular for momentum, angular momentum and energy.

To achieve this aim it is necessary to review and generalise the principles of Quantum Mechanics in the simplified version summarised below.

Principles of Quantum Mechanics.

For ease of reading we set them out in full, in the statements we shall work on and with the vectors understood to be normalised.

A physical system is characterised by its physical quantities, which can be measured.

When a quantity g is measured, several outcomes are possible, and each outcome has a certain probability of occurring. For every possible result $g=\overline{g}$ there exists a state in which the measurement of g yields the value $g=\overline{g}$ with certainty. Such a state is called an eigenstate of g with eigenvalue $\overline{g}$ .

Any other state α can be considered as a superposition of the eigenstates of a quantity g and can be represented by a vector of complex numbers $|\alpha\rangle=\left(\begin{gathered} \alpha_1 \\ \vdots \\ \alpha_n \end{gathered}\right)$ . Each complex number is associated with one of the possible values that the quantity g can take, ${\overline{g}}_1,{\overline{g}}_2,\dots{\overline{g}}_n.$ The squared modulus of each of these complex numbers, for example α_k, gives the probability that measuring the quantity g yields the particular value ${\overline{g}}_k$ associated with the number considered.

In general, if the system is in a state α and a quantity $g'$ is measured, the probability that $g'={\overline{g}}'$ is obtained is given by the squared modulus of the scalar product between the vector $|\alpha\rangle$ and the vector $|g'={\overline{g}}'\rangle$ associated with the eigenstate $g'={\overline{g}}'$ of $g'$ . $P_{g'={\overline{g}}'}={\left|\left\langle g'={\overline{g}}'|\alpha\right\rangle\right|}^2$

The time evolution of a state αt is continuous and follows a linear law $|\alpha t\rangle=U(t_0\longrightarrow t)|\alpha t_0\rangle$

where $U(t_0\to t)$ is a matrix that depends on the system and on the “ambient conditions”. The matrix $U\left(t_0\to t\right)$ is called the time-evolution matrix. Finally, as long as the system is not observed, the sum of the squared moduli of all the components does not change during the evolution. From this last principle, together with linear superposition, follows the scalar-product conservation theorem: the time-evolution matrix leaves scalar products invariant $\langle\alpha t\mid\beta t\rangle=\langle\alpha t_0\mid\beta t_0\rangle$

Note 02 Why the time-evolution matrix leaves the scalar products unchanged: the proof →

In the statement about measurement the very concept of eigenstate is introduced, that is a state such that if a measurement is performed a given value, called the eigenvalue, is obtained with certainty. In the case of a material particle the eigenstates, for example of the quantity x, are those represented by vectors of the form

\psi(x,y,z)=\delta(x-\overline{x})\psi(y,z)

indeed with this vector one obtains a probability distribution that is zero if $x\neq\overline{x}$ , and is non-zero only if $x=x$ , so when a measurement of x is performed the value $\overline{x}$ will necessarily be obtained. Another example is given by the state represented by the vector

\psi(x,y,z)=\delta(x-\overline{x})\delta(y-\overline{y})\delta(z-\overline{z})

in this case, if a measurement of x, y or z is performed, the values $\overline{x}$ , $\overline{y}$ and $\overline{z}$ will be obtained with certainty; so what we have considered is simultaneously an eigenstate of the quantities x, y and z.

The principles we have given are in fact a simplified version, because it is assumed that, given a value $\overline{g}$ , the eigenstate associated with $\overline{g}$ is unique, which in general is not true; for example in the case of a material particle in three-dimensional space the eigenstates associated with an eigenvalue $\overline{x}$ of x are infinite:

\psi(x,y,z)=\delta(x-\overline{x})\psi(y,z)

Therefore we cannot speak of the state $x=x$ , because it is not unique, but we can speak of the state $x=\overline{x},y=\overline{y},z=\overline{z}$ ; indeed this state is uniquely determined and the corresponding probability distribution is

\psi(x,y,z)=\delta(x-\overline{x})\delta(y-\overline{y})\delta(z-\overline{z})

The principle of linear superposition must be generalised by saying that any state can be considered as the superposition of the “simultaneous eigenstates” of a certain set of physical quantities $g_1\cdot\cdot\cdot g_n$ . The coefficients of this superposition will no longer have a single index but as many as there are observable quantities whose eigenstates are taken, so we will not have $\alpha_k$ but $\alpha_{k_1\dots k_n}$ , or in the continuous case we will not have $\psi(x)$ but $\psi(x,y,z)$ .

A set of physical quantities $g_1\cdots g_n$ such that there exists a unique simultaneous eigenstate associated with the eigenvalues $\overline{g}_1\cdots\overline{g}_n$ will be called “complete”. For example the three position variables x, y and z form a complete set, because given a triple of values $\overline{x},\overline{y}$ and $\overline{z}$ , there exists a unique simultaneous eigenstate associated with this triple.

The ket associated with the state $x=\overline{x},y=\overline{y},z=\overline{z}$ may be denoted by the symbol

|x=\overline{x},y=\overline{y},z=\overline{z}\rangle

however we will prefer to commit an abuse of notation and use a more compact notation, denoting the ket by the symbol $|xyz\rangle$ . (There is an abuse of notation because if we wanted to substitute numbers, for example $\overline{x}=1$ , $\overline{y}=2$ and $\overline{z}=3$ , we would have the symbol $|123\rangle$ in which values are indicated but the corresponding quantities are not.)

Now let us consider the probability rule and suppose, for example, that we want to compute the probability amplitude that, measuring x, y and z simultaneously on a particle in a state $|\psi\rangle$ , the values $\overline{x},\overline{y}$ and $\overline{z}$ are obtained; then the sought probability amplitude is given by the scalar product $\langle xyz|\psi\rangle$ and the probability is given by the squared modulus $|\langle\overline{xyz}|\psi\rangle|^2$ .

But if we want the probability that, measuring only x, $\overline{x}$ is obtained? Can we write $P_{x=\overline{x}}={\left|\left\langle\overline{x}|\psi\right\rangle\right|}^2$ ?

No, because the symbol $\langle\overline{x}|$ has no meaning; indeed the eigenvalue $\overline{x}$ does not identify a unique eigenstate.

If we want the probability that, measuring only x, $\overline{x}$ is obtained, then we must first compute the probability

p(\overline{x},\overline{y},\overline{z})={\left|\langle\overline{xyz}\mid\psi\rangle\right|}^2

and then compute the integral

p(\overline{x})=\iint_{\text{All the }\overline{y}\overline{z}\text{ plane}} p(\overline{x},\overline{y},\overline{z})\:d\overline{y}\:d\overline{z}.

At this point we write the generalised principles

A physical system is characterised by its physical quantities, which can be measured.

2.bis In general, given a value $\overline{g}$ , there may exist infinitely many eigenstates associated with this value. However it is always possible to find a complete set of variables $g_1\cdots g_n$ such that, given the n values ${\overline{g}}_1\cdot\cdot\cdot{\overline{g}}_n$ , there exists a unique simultaneous eigenstate associated with these values.

Any other state α can be considered as a superposition of the simultaneous eigenstates of a set of quantities $g_1\cdot\cdot\cdot g_n$ , and can be represented by a vector of complex numbers with n indices $|\alpha\rangle\equiv\left(\alpha_{k_1\dots k_n}\right)$ . Each complex number is associated with one of the possible combinations of values that the n-tuple of quantities $g_1\cdot\cdot\cdot g_n$ can take. The squared modulus of each of these complex numbers, for example $\alpha_{k_1\dots k_n}$ , gives the probability that measuring the quantities $g_1\cdot\cdot\cdot g_n$ simultaneously yields the particular set of values $\overline{g}_{k_1}\cdots\overline{g}_{k_n}$ , those that identify the simultaneous eigenstate $|{\overline{g}}_{k_1}\cdots{\overline{g}}_{k_n}\rangle\leftrightarrow g_1={\overline{g}}_{k_1}\cdots g_n={\overline{g}}_{k_n}$ .

In general, if the system is in a state α and a complete set of quantities $g_1\cdots g_n$ is measured simultaneously, the probability of obtaining $g_1'={\overline{g}}_1'\dots g_n'={\overline{g}}_n'$ is given by the squared modulus of the scalar product between the vector $|\alpha\rangle$ and the vector $|g_1'\cdots g_n'\rangle$ associated with the simultaneous eigenstate $g_1'={\overline{g}}_1'\dots g_n'={\overline{g}}_n'$ . That is $P_{g_1'={\overline{g}}_1'\dots g_n'={\overline{g}}_n'}={\left|\left\langle{\overline{g}}_1'\dots{\overline{g}}_n'|\alpha\right\rangle\right|}^2$ .

The time evolution of a state αt is continuous and follows a linear law $|\alpha t\rangle=U(t_0\longrightarrow t)|\alpha t_0\rangle$

The generalisation of these principles to the continuous case is immediate: it suffices to replace the index representation $\alpha_{k_1\dots k_n}$ with the functional representation $\psi(x_1\dots x_n)$ .

With the functional representation the state of a particle is described by a complex function of position: in this form we recover the premises stated in the introduction, to which is added agreement with Newton’s mechanics — the requirement that in card four allowed us to derive the Schrödinger equation.

We observe that two kets $|g_1\cdots g_n\rangle$ and $|g_1'\cdots g_n'\rangle$ are orthogonal, unless they have all values equal ${\overline{g}}_1={\overline{g}}_1'\dots{\overline{g}}_n={\overline{g}}_n'$ . Indeed, if we suppose that one of the values, for example the first, is different $g_1\neq g_1'$ , then the probability of obtaining the values ${\overline{g}}_1\cdots{\overline{g}}_n$ in the state $|g_1'\cdots g_n'\rangle$ is zero, hence:

\begin{aligned} |\langle{\overline{g}}_1\cdots{\overline{g}}_n|{\overline{g}}_1'\cdots{\overline{g}}_n'\rangle|^2 & =0\Leftrightarrow \\ \langle{\overline{g}}_1\cdots{\overline{g}}_n|{\overline{g}}_1'\cdots{\overline{g}}_n'\rangle & =0 \end{aligned}

⇔ are orthogonal.

If the kets $|g_1\cdots g_n\rangle$ , besides being orthogonal, are also normalised, then the components of a generic ket $|\alpha\rangle$ in the basis $\{|\overline{g}_1\cdots\overline{g}_n\rangle\}$ can be written with the scalar product $\alpha_{k_1\dots k_n}=\langle{\overline{g}}_{k_1}\dots{\overline{g}}_{k_n}|\alpha\rangle$ , so we will have

\begin{aligned} |\alpha\rangle & =\sum_{k_1\dots k_n}\alpha_{k_1\dots k_n}|{\overline{g}}_{k_1}\dots{\overline{g}}_{k_n}\rangle \\ & =\sum_{k_1\dots k_n}|{\overline{g}}_{k_1}\dots{\overline{g}}_{k_n}\rangle\langle{\overline{g}}_{k_1}\dots{\overline{g}}_{k_n}|\alpha\rangle \\ & =\left[\sum_{k_1\dots k_n}|{\overline{g}}_{k_1}\dots{\overline{g}}_{k_n}\rangle\langle{\overline{g}}_{k_1}\dots{\overline{g}}_{k_n}|\right]|\alpha\rangle \end{aligned}

These identities must hold for every choice of the ket $|\alpha\rangle$ , so we can deduce the formula

\sum_{k_1\dots k_n}|{\overline{g}}_{k_1}\cdots{\overline{g}}_{k_n}\rangle\langle{\overline{g}}_{k_1}\cdots{\overline{g}}_{k_n}|=I

where by I we have denoted the identity matrix. This formula will be extremely useful on many future occasions.

Now suppose we want to compute the distribution of probability amplitudes associated with physical quantities other than position, for example momentum $\overline{p}$ . (From now on, symbols with an overbar will denote vector quantities.)

Probability amplitudes for momentum.

If we know the state of a material particle, and we know the distribution of probability amplitudes for the position variable $\psi(x,y,z)$ , how can we determine the distribution of probability amplitudes for the quantity momentum?

If we knew the eigenstates of momentum then we could apply the fourth postulate, but we do not know these eigenstates and we must determine them.

To find the solution to our problem we must take the other end of the thread, that is we suppose we know what the eigenstates of momentum are, and from these we try to compute a value that we already know by another route, so that we can have a link between the things we know and those we have not yet determined.

Suppose for example that we want to compute the mean value of the component along x of the momentum $\langle p_x\rangle$ , assuming we know the state of the system $|\psi\rangle$ and the eigenstates of momentum.

To keep generality we cannot assume that the quantity $p_x$ by itself constitutes a complete set, so to “complete the set” we introduce a set of auxiliary quantities that we will denote briefly by a single contracted symbol $\xi$ . To represent an eigenstate associated with an eigenvalue $p_x'$ we will use the ket $|p_x',\xi'\rangle$ , where $\xi'$ is a multi-index relative to the set of auxiliary quantities $\xi$ .

(Note that it would be wrong to write the ket $|p_x'\rangle$ , because it is not possible to identify a state by giving only the value $p_x'$ taken by the quantity $p_x$ ).

On the basis of these notations we can write the probability:

\begin{aligned} \rho(p_x,\xi) & ={\left|\langle p_x,\xi|\psi\rangle\right|}^2 \\ & =\langle\psi|p_x,\xi\rangle\langle p_x,\xi|\psi\rangle \end{aligned}

where we have denoted the probability by the symbol $\rho$ so as not to create confusion with the p of momentum.

On the basis of this formula we can compute the mean value:

\begin{aligned} \langle p_x\rangle & =\int_\xi d\xi'\int_{-\infty}^{+\infty}\rho(p_x',\xi')p_x'\:dp_x' \\ & =\int_\xi d\xi'\int_{-\infty}^{+\infty}\langle\psi|p_x',\xi'\rangle p_x'\langle p_x',\xi'|\psi\rangle\:dp_x' \\ & =\langle\psi|\left[\int_\xi d\xi'\int_{-\infty}^{+\infty}|p_x',\xi'\rangle p_x'\langle p_x',\xi'|\:dp_x'\right]|\psi\rangle \end{aligned}

but we also know that

\langle p_x\rangle=\langle\psi|P_x|\psi\rangle

where $P_x=-i\hbar D_x$ is the momentum operator that we defined in the fourth card. On the basis of these two equalities we can write

\langle\psi|P_x|\psi\rangle=\langle\psi|\left[\int_\xi d\xi'\int_{-\infty}^{+\infty}|p_x',\xi'\rangle\:p_x'\langle p_x',\xi'|\:dp_x'\right]|\psi\rangle

since this equation is true for every $\psi$ , we can deduce that

P_x=\int_\xi d\xi'\int_{-\infty}^{+\infty}|p_x',\xi'\rangle p_x'\langle p_x',\xi'|dp_x'

This formula shows us that the operator $P_x$ “contains within itself” the eigenstates we are looking for $|p_x',\xi'\rangle$ , so we must find a way to extract them.

Let us try to multiply the operator $P_x$ by an eigenstate $|p_x,\xi\rangle$ :

\begin{aligned} P_x|p_x,\xi\rangle & =\int_\xi d\xi'\int_{-\infty}^{+\infty}|p_x',\xi'\rangle p_x'\langle p_x',\xi'|dp_x'|p_x,\xi\rangle= \end{aligned}

\begin{aligned} & \\ & =\int_\xi d\xi'\int_{-\infty}^{+\infty}|p_x',\xi'\rangle p_x'\langle p_x',\xi'|p_x,\xi\rangle\:dp_x'= \end{aligned}

Observing that $\langle p_x',\xi'|p_x,\xi\rangle$ is zero if $p_x'\neq p_x$ , within the integral we can replace the “central” term $p_x'$ with $p_x$ .

\begin{aligned} & \\ & =\int_\xi d\xi'\int_{-\infty}^{+\infty}|p_x',\xi'\rangle p_x\langle p_x',\xi'|p_x,\xi\rangle dp_x' \\ & =p_x\int_\xi d\xi'\int_{-\infty}^{+\infty}|p_x',\xi'\rangle\langle p_x',\xi'|p_x,\xi\rangle dp_x'= \end{aligned}

since $\langle p_x',\xi'|p_x,\xi\rangle$ are the components of the vector $|p_x,\xi\rangle$ in the basis $\{|p_x',\xi'\rangle\}$ , the integral is nothing but the combination that expresses the vector $|p_x,\xi\rangle$ in the basis $\{|p_x',\xi'\rangle\}$ , hence we have found

P_x\:|p_x,\xi\rangle=p_x\:|p_x,\xi\rangle

that is, the state vectors $|p_x,\xi\rangle$ are “eigenvectors” of the operator $P_x$ . This is the reason why the eigenstates are so called.

At this point we have determined a very clear picture: with the observable quantity $p_x$ is associated an operator $P_x$ , the eigenstates of the quantity $p_x$ are associated with the eigenvectors of the operator $P_x$ , while the values that the quantity $p_x$ can take are the eigenvalues of the operator $P_x$ .

If we solve the eigenvalue problem

P_x\:|p_x,\xi\rangle=p_x\:|p_x,\xi\rangle

we determine the eigenstates of the operator $|p_x,\xi\rangle$ and, applying postulate four, we manage to solve our problem, that is we manage to determine the probability distribution for momentum.

Now we generalise the results obtained for momentum with the following very important theorem:

With every physical quantity $g$ is associated a Hermitian operator $G$ such that:

The mean value of the quantity $g$ , when the system is in a state represented by the ket $|\psi\rangle$ , is given by the formula $\langle\psi|G|\psi\rangle$ ;

The eigenvalues of the operator $G$ are all the possible values that the quantity $g$ can take, and the eigenvectors associated with an eigenvalue $g'$ represent the eigenstates of the quantity $g$ in which the value $g'$ is obtained with certainty.

Proof:

Let us consider a quantity $g$ , let $g'$ be the possible values of the quantity $g$ , and let $|g',\xi\rangle$ be the kets representing the eigenstates of the quantity $g$ and of the set of auxiliary quantities $\xi$ ; then the operator

G=\iint_{g'\:\xi}|g',\xi\rangle g'\langle g',\xi|\:d\xi\:dg'

is Hermitian and satisfies the two points of the statement. To verify it, one need only follow exactly the same steps that we followed for the operator $P_x$ .

Hermiticity is immediate: conjugating and transposing the integral swaps the bra and the ket and conjugates the eigenvalue

G^+=\iint_{g'\:\xi}|g',\xi\rangle{g'}^*\langle g',\xi|\:d\xi\:dg'=G

and the last step holds because the eigenvalues $g'$ are real numbers: they are the results of a measurement.

Compatible quantities and incompatible quantities.

Suppose we want to perform a simultaneous measurement of $x$ and of $p_x$ . After such a measurement the system should settle into a simultaneous eigenstate of the quantities $x$ and $p_x$ , and the resulting state ket should be an eigenket of the operators $X$ and $P_x$ , but this is impossible. Indeed, denoting by $|x,p_x,\xi\rangle$ a hypothetical eigenstate of the quantities $x$ , $p_x$ and of a suitable set of auxiliary quantities $\xi$ , we have

\begin{aligned} XP_x|x,p_x,\xi\rangle & =xp_x|x,p_x,\xi\rangle \\ & =P_xX|x,p_x,\xi\rangle \end{aligned}

from which

\left(XP_x-P_xX\right)|x,p_x,\xi\rangle=0

but we know that $(XP_x-P_xX)=i\hbar I$ , hence we have

\begin{aligned} i\hbar I|x,p_x,\xi\rangle & =0\Leftrightarrow \\ |x,p_x,\xi\rangle & =0 \end{aligned}

We conclude that the simultaneous eigenket of the operators $X$ and $P_x$ cannot exist because it would necessarily be zero. Therefore it is not possible to perform a precise and simultaneous measurement of the quantities $x$ and $p_x$ , whereas it is perfectly possible in other cases, for example with the quantities $x$ , $p_y$ .

Before proceeding with the study, let us generalise to the continuous case the formula we proved at the end of the first paragraph:

if $\xi$ is a complete set of observable quantities, and $|\xi\rangle$ is the orthonormal set of the “eigenkets” associated with its eigenstates, then we have

\int_\xi|\xi\rangle\langle\xi|\:d\xi=I

For clarity we repeat the proof: if we multiply the integral on the left by a generic ket $|\psi\rangle$ we have

\left[\int_\xi|\xi\rangle\langle\xi|d\xi\right]|\psi\rangle=\int_\xi|\xi\rangle\left\langle\xi|\psi\right\rangle d\xi

since $\langle\xi|\psi\rangle$ are the components of the vector $|\psi\rangle$ in the basis $\{|\xi\rangle\}$ , we have

\left[\int_\xi|\xi\rangle\langle\xi|\:d\xi\right]|\psi\rangle=|\psi\rangle

which, having to be true for every $|\psi\rangle$ , proves the statement.

Angular momentum.

In Classical Physics the angular momentum with respect to a pole O is defined by the formula

L_o={\overline{r}}_o\wedge\overline{p}

In the scheme of Quantum Mechanics the angular-momentum quantity must be consistent with the classical definition.

Hence for the mean values we must have:

\langle L_o\rangle=\langle{\overline{r}}_o\wedge\overline{p}\rangle

To simplify the work it is convenient to consider the individual components; moreover, to avoid burdening the notation, we will omit the subscript O.

\overline{L}=\overline{r}\wedge\overline{p}=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k} \\ x & y & z \\ p_x & p_y & p_z\end{vmatrix}=\hat{\imath}(yp_z-zp_y)+\hat{\jmath}(zp_x-xp_z)+\hat{k}(xp_y-yp_x)

For the components we have

\left\{\begin{gathered} L_x=\left(yp_z-zp_y\right) \\ L_y=\left(zp_x-xp_z\right) \\ L_z=\left(xp_y-yp_x\right) \end{gathered}\right.

Let us consider the component $L_z$ . To have agreement between Classical Mechanics and Quantum Mechanics we must have:

\langle L_z\rangle=\langle xp_y\rangle-\langle yp_x\rangle

Let us focus attention on the term $\langle xp_y\rangle$

\langle xp_y\rangle=\int_{x,p_y}\rho(x,p_y)xp_y\:dxdp_y

We denote by $|x,p_y,\xi\rangle$ an eigenket representing a simultaneous eigenstate of the quantities $x$ , $p_y$ and of the set of auxiliary quantities $\xi$ .

This eigenket exists because x and p_y are compatible quantities: it is the very example with which the previous paragraph closed. From the same commutation a second fact follows, one we shall need: since XP_y = P_yX, the product does not depend on the order of the factors and is hermitian, as the operator of a physical quantity must be. For xp_x neither of the two would have held, and the construction would have called for a further device.

Using these kets we have

\begin{aligned} \rho(x,p_y,\xi) & =|\langle x,p_y,\xi|\psi\rangle|^2 \\ & =\langle\psi|x,p_y,\xi\rangle\langle x,p_y,\xi|\psi\rangle \end{aligned}

integrating with respect to the variables $\xi$ we have

\rho(x,p_y)=\int_\xi\langle\psi|x,p_y,\xi\rangle\langle x,p_y,\xi|\psi\rangle d\xi

Hence for the mean value we have

\langle xp_y\rangle=\int_{x,p_y}\left[\int_\xi\langle\psi|x,p_y,\xi\rangle\langle x,p_y,\xi|\psi\rangle\:d\xi\right]xp_y\:dx\:dp_y

recalling that $XP_y\:|x,p_y,\xi\rangle=xp_y\:|x,p_y,\xi\rangle$ we can write

\begin{aligned} & \langle xp_y\rangle \\ & \qquad=\int_{x,p_y}\int_\xi\langle\psi|XP_y|x,p_y,\xi\rangle\langle x,p_y,\xi|\psi\rangle\:d\xi\:dx\:dp_y \\ & \qquad=\langle\psi|XP_y\left[\int_{x,p_y}\int_\xi|x,p_y,\xi\rangle\langle x,p_y,\xi|\:d\xi\:dx\:dp_y\right]|\psi\rangle \end{aligned}

since

\iint|x,p_y,\xi\rangle\langle x,p_y,\xi|\:d\xi\:dx\:dp_y=I

we have the final formula

\langle xp_y\rangle=\langle\psi|XP_y|\psi\rangle

The steps are analogous for all the other terms, so we obtain:

\left\{\begin{gathered} \langle L_x\rangle=\langle\psi|(YP_z-ZP_y)|\psi\rangle \\ \langle L_y\rangle=\langle\psi|(ZP_x-XP_z)|\psi\rangle \\ \langle L_z\rangle=\langle\psi|(XP_y-YP_x)|\psi\rangle \end{gathered}\right.

Thus with the quantity $L_z$ , for example, we associate the operator $L_z=XP_y-YP_x$ , and analogously for the others.

At this point we have found an operator that gives us a formula to compute the mean value of angular momentum in the formalism of Quantum Mechanics and that agrees with the formula of Classical Mechanics.

This same operator can give us all the possible values that angular momentum can take and the corresponding eigenstates; to obtain this information it suffices to solve the eigenvalue problem

L_z|\psi\rangle=l_z|\psi\rangle

Energy.

As the last quantity of study we consider energy.

In the formalism of Classical Mechanics, for a charged particle in an electric potential, we have the energy

E=qV+\frac{1}{2m}{\overline{p}}^2

so we must look for an operator that gives us a mean value consistent with this classical formula

\begin{aligned} \langle E\rangle & =\left\langle qV+\frac{1}{2m}{\overline{p}}^2\right\rangle \\ & =q\langle V\rangle+\frac{1}{2m}\left\langle{\overline{p}}^2\right\rangle \end{aligned}

But we already have such an operator, namely H, indeed

\begin{aligned} \langle\psi|H|\psi\rangle & =\langle\psi|\left(qV(X)+\frac{1}{2m}{\overline{P}}^2\right)|\psi\rangle \\ & =q\langle\psi|V(X)|\psi\rangle+\frac{1}{2m}\langle\psi|{\overline{P}}^2|\psi\rangle \\ & =q\langle V\rangle+\frac{1}{2m}\langle{\overline{P}}^2\rangle \end{aligned}

Therefore, if we want to find the values that energy can take, and the eigenstates into which the particle settles after a measurement of energy, then we must solve the following eigenvalue problem:

\begin{aligned} H|\psi\rangle & =E|\psi\rangle\Leftrightarrow \\ \left(qV(X)+\frac{1}{2m}{\overline{P}}^2\right)|\psi\rangle & =E|\psi\rangle \end{aligned}

Energy levels for the hydrogen atom.

The hydrogen atom consists of a nucleus with charge equal and opposite to that of the electron, and a single orbiting electron.

To determine the motion of the electron around the nucleus one must solve the Schrödinger equation, whereas if one wants to determine the values that the energy of the atom can take, it suffices to solve the eigenvalue problem we posed in the previous paragraph in the case where we have a Coulomb potential $V(r)=e/(4\pi\varepsilon_0 r)$ , where e is the charge of the nucleus, equal and opposite to the charge q of the electron appearing in H.

Energy can take only discrete values given by the formula

E_n=-\frac{me^4}{8\varepsilon_0^2h^2n^2}

Note 07 The full calculation leading to this formula: the derivation →