- #1

- 22,178

- 3,304

There are many apparent paradoxes in quantum mechanics. Luckily, a careful application of math reveals that all is well. But can you figure out why the following ##7## challenges are not paradoxes?

For a particle in one dimension, the operators of momentum ##P## and position ##Q## satisfy Heisenberg’s canonical commutation relation

[tex][P, Q] = \frac{\hbar}{i} \boldsymbol{1}[/tex]

By taking the trace of this relation, one finds a vanishing result for the left-hand side, ##\text{Tr}[P, Q] = 0##, whereas ##\text{Tr}\left(\frac{\hbar}{i}\boldsymbol{1}\right)\neq 0##. What went wrong?

Consider wave functions ##\varphi## and ##\psi## which are square integrable on ##\mathbb{R}## and the momentum ##P = \frac{\hbar}{i}\frac{d}{dx}##. Integration by parts yields

[tex]\int_{-\infty}^{+\infty}\overline{\varphi(x)}(P\psi)(x)dx = \int_{-\infty}^{+\infty} \overline{(P\varphi)(x)}\psi(x)dx + \frac{\hbar}{i} [ (\overline{\varphi}\psi)(x)]_{-\infty}^{+\infty}[/tex]

Since ##\varphi## and ##\psi## are square integrable, one usually concludes that these functions vanish for ##x\rightarrow \pm\infty##. Thus, the last term in the previous equation vanishes, which implies that the operator P is Hermitian.

However, the textbooks of mathematics tell us that square integrable functions do, in general, not admit a limit for ##x\rightarrow \pm \infty## and therefore they do not necessarily vanish at infinity. There are even functions which are continuous and square summable on ##\mathbb{R}## without being bounded at infinity: an example of such a function is given by

[tex]f(x) = x^2 e^{-x^8\sin^2x}[/tex]

Can one conclude that the operator P is Hermitian in spite of these facts and if so, why?

Consider the operators ##P = \frac{\hbar}{i} \frac{d}{dx}## and "##Q = ## multiplication by ##x##" acting acting on wave functions depending on ##x ∈ \mathbb{R}##. Since ##P## and ##Q## are Hermitian operators, the operator

[tex]A = PQ^3 + Q^3P[/tex]

also has this property, because its adjoint is given by

[tex]A^\dagger = (PQ^3 + Q^3P)^\dagger = Q^3P + PQ^3 = A[/tex]

It follows that all eigenvalues of ##A## are real. Nevertheless, one easily verifies that

[tex]Af = \frac{\hbar}{i}f[/tex]

with

[tex]f(x) = \left\{\begin{array}{ll}

\frac{1}{\sqrt{2}} |x|^{-3/2}\text{exp}\left(-\frac{1}{4x^2}\right) & \text{for}~x\neq 0\\

0 & \text{for}~x=0

\end{array}\right.[/tex]

which means that ##A## admits the complex eigenvalue ##\hbar/i##. Note that the function ##f## is infinitely differentiable on ##\mathbb{R}## and that it is square integrable, since

[tex]\int_{-\infty}^{+\infty} |f(x)|^2dx = 2\int_0^{+\infty}|f(x)|^2dx = \int_0^{+\infty} x^{-3} \text{exp}\left(-\frac{1}{2x^2}\right)dx= [\text{exp}(-1/(2x^2))]_0^{+\infty} = 1[/tex]

Where is the error?

Let us consider a particle confined to the interval ##[0, 1]## and described by a wave function ##\psi## satisfying the boundary conditions ##\psi(0) = 0 = \psi(1)##. Then the momentum operator

[tex]P = \frac{\hbar}{i}\frac{d}{dx}[/tex]

is Hermitian, since the surface term appearing upon integration by parts vanishes:

[tex]\int_0^1 \left(\overline{\varphi}(P\psi) - \overline{(P\varphi)}\psi\right)(x)dx = \frac{\hbar}{i}[(\overline{\varphi}\psi)(x)]_0^1 = 0[/tex]

Since P is Hermitian, its eigenvalues are real. In order to determine the latter, we note that the eigenvalue equation,

[tex](P\psi_p)(x) = p\psi_p(x)[/tex]

is solved by ##\psi_p(x) = c_p \text{exp}\left(\frac{i}{\hbar}px\right)## with ##p\in \mathbb{R}## and ##c_\in \mathbb{C}\setminus\{0\}##. The boundary condition ##\psi_p(0) = 0## now implies ##\psi_p = 0##, therefore ##P## does not admit any eigenvalues. Nevertheless, the spectrum of ##P## is the entire complex plane and ##P## does not represent an observable. How can one understand these results which seem astonishing?

If one introduces polar coordinates in the plane or spherical coordinates in space, then the polar angle ##\varphi## and the component ##L_z## of angular momentum are canonically conjugate variables in classical mechanics. In quantum theory, the variable ##\varphi## becomes the operator of "multiplication of the wave function ##\psi(\varphi)## by ##\varphi## " and ##L_z = \frac{\hbar}{i} \frac{\partial}{\partial \varphi}##, which implies the commutation relation

[tex][L_z,\varphi] = \frac{\hbar}{i}\boldsymbol{1}[/tex]

These operators acting on periodic wave functions (i.e. ##\psi(0) = \psi(2\pi)##) are Hermitian. Furthermore, ##L_z## admits a complete system of orthonormal eigenfunctions ##\psi_m## for ##m\in \mathbb{Z}## and

[tex]L_z \psi_m = m\hbar\psi_m~\text{with}~\psi_m(\varphi) = \frac{1}{\sqrt{2\pi}}\text{exp}(im\varphi).[/tex]

By evaluating the average value of the operator ##[L_z,\varphi]## in the state ##\psi_m## and by taking

into account the fact that ##L_z## is Hermitian, one finds that

[tex]\begin{align*}

\frac{\hbar}{i}

& = \left\langle \psi_m, \frac{\hbar}{i}\boldsymbol{1} \psi_m\right\rangle\\

& = \left\langle \psi_m, L_z\varphi \psi_m\right\rangle - \left\langle \psi_m, \varphi L_z\psi_m\right\rangle\\

& = \left\langle L_z^\dagger\psi_m, \varphi \psi_m\right\rangle - m\hbar\left\langle \psi_m, \varphi \psi_m\right\rangle\\

& = (m\hbar - m\hbar)\left\langle \psi_m, \varphi \psi_m\right\rangle\\

& = 0

\end{align*}[/tex]

Where is the problem?

Let us add a bit to the confusion of the previous example! In 1927, Pauli noted that the

canonical commutation relation implies Heisenberg’s uncertainty relation ##\Delta P \Delta Q\geq \frac{\hbar}{2}## by virtue of the Cauchy-Schwarz inequality. One can derive, in the same way, the uncertainty relation

[tex] \Delta L_z \Delta \varphi\geq \frac{\hbar}{2}[/tex]

The following physical reasoning shows that this inequality cannot be correct. One can always find a state for which ##\Delta L_z < \frac{\hbar}{4\pi}## and then the uncertainty for the angle ##\varphi## has to be larger than ##2\pi##, which does not have any physical sense, since ##\varphi## takes values in the interval ##[0, 2\pi)##. How is this possible?

By the way, this example shows that the uncertainty relation ##\Delta A \Delta B\geq \frac{1}{2}|\left\langle [A,B]\right\rangle|## for any two observables ##A## and ##B## (whose derivation can be found in most quantum mechanics texts) is not valid in such a generality.

Let us consider a particle of mass ##m## in the infinite potential well (for ##a>0##)

[tex]V(x) = \left\{\begin{array}{ll}

0 & \text{if}~|x|\leq a\\

\infty & \text{otherwise}

\end{array}\right.[/tex]

The Hamiltonian for the particle confined to the inside of the well is simply

[tex]H = - \frac{\hbar^2}{2m} \frac{d^2}{dx^2}[/tex]

Let

[tex]\psi(x) = \left\{\begin{array}{ll}

\frac{\sqrt{15}}{4a^{5/2}}(a^2 - x^2) & \text{if}~|x|\leq a\\

0 & \text{otherwise}

\end{array}\right.[/tex]

be the normalized wave function of the particle at a given time. Since

[tex]H^2\psi = \frac{\hbar^4}{4m^2}\frac{d^4}{dx^4}\psi = 0,[/tex]

the average value of the operator ##H^2## in the state ##ψ## vanishes:

[tex]\left\langle H^2\right\rangle_\psi = \left\langle\psi,H^2\psi\right\rangle = \int_{-a}^a\overline{\psi(x)}(H^2\psi)(x)dx = 0[/tex]

This average value can also be determined from the eigenvalues and eigenfunctions of H,

[tex]H\varphi_n = E_n \varphi_n[/tex]

with

[tex]E_n = \frac{\pi^2\hbar^2}{8ma^2} n^2~\text{for}~n=1,2,3,...[/tex]

by applying the formula

[tex]\left\langle H^2\right\rangle_\psi = \sum_{n=1}^{+\infty} E_n^2 p_n[/tex]

with ##p_n = |\left\langle\varphi_n,\psi\right\rangle|^2##.

Proceeding in this way, one definitely does not find a vanishing result, because ##E_n^2>0## and ##0\leq p_n\leq 1##, ##\sum_{n=1}^{\infty} p_n = 1##. In fact, the calculation yields ##\left\langle H^2\right\rangle_\psi = \frac{15\hbar^4}{8m^2a^4}##. Which one of these two results is correct and where does the inconsistency come from?

[1] Mathematical surprises and Dirac's formalism in quantum mechanics by François Gieres.

https://arxiv.org/pdf/quant-ph/9907069

**Rules:**- Do not look at paper [1] before answering. It contains all the answers in detail. Any other use of outside sources is allowed, but make sure to quote them.
- Give a careful explanation why you think the speciic situation is not a paradox. Use any math/physics you wish.

**Challenge 1**For a particle in one dimension, the operators of momentum ##P## and position ##Q## satisfy Heisenberg’s canonical commutation relation

[tex][P, Q] = \frac{\hbar}{i} \boldsymbol{1}[/tex]

By taking the trace of this relation, one finds a vanishing result for the left-hand side, ##\text{Tr}[P, Q] = 0##, whereas ##\text{Tr}\left(\frac{\hbar}{i}\boldsymbol{1}\right)\neq 0##. What went wrong?

**Challenge 2**Consider wave functions ##\varphi## and ##\psi## which are square integrable on ##\mathbb{R}## and the momentum ##P = \frac{\hbar}{i}\frac{d}{dx}##. Integration by parts yields

[tex]\int_{-\infty}^{+\infty}\overline{\varphi(x)}(P\psi)(x)dx = \int_{-\infty}^{+\infty} \overline{(P\varphi)(x)}\psi(x)dx + \frac{\hbar}{i} [ (\overline{\varphi}\psi)(x)]_{-\infty}^{+\infty}[/tex]

Since ##\varphi## and ##\psi## are square integrable, one usually concludes that these functions vanish for ##x\rightarrow \pm\infty##. Thus, the last term in the previous equation vanishes, which implies that the operator P is Hermitian.

However, the textbooks of mathematics tell us that square integrable functions do, in general, not admit a limit for ##x\rightarrow \pm \infty## and therefore they do not necessarily vanish at infinity. There are even functions which are continuous and square summable on ##\mathbb{R}## without being bounded at infinity: an example of such a function is given by

[tex]f(x) = x^2 e^{-x^8\sin^2x}[/tex]

Can one conclude that the operator P is Hermitian in spite of these facts and if so, why?

**Challenge 3**Consider the operators ##P = \frac{\hbar}{i} \frac{d}{dx}## and "##Q = ## multiplication by ##x##" acting acting on wave functions depending on ##x ∈ \mathbb{R}##. Since ##P## and ##Q## are Hermitian operators, the operator

[tex]A = PQ^3 + Q^3P[/tex]

also has this property, because its adjoint is given by

[tex]A^\dagger = (PQ^3 + Q^3P)^\dagger = Q^3P + PQ^3 = A[/tex]

It follows that all eigenvalues of ##A## are real. Nevertheless, one easily verifies that

[tex]Af = \frac{\hbar}{i}f[/tex]

with

[tex]f(x) = \left\{\begin{array}{ll}

\frac{1}{\sqrt{2}} |x|^{-3/2}\text{exp}\left(-\frac{1}{4x^2}\right) & \text{for}~x\neq 0\\

0 & \text{for}~x=0

\end{array}\right.[/tex]

which means that ##A## admits the complex eigenvalue ##\hbar/i##. Note that the function ##f## is infinitely differentiable on ##\mathbb{R}## and that it is square integrable, since

[tex]\int_{-\infty}^{+\infty} |f(x)|^2dx = 2\int_0^{+\infty}|f(x)|^2dx = \int_0^{+\infty} x^{-3} \text{exp}\left(-\frac{1}{2x^2}\right)dx= [\text{exp}(-1/(2x^2))]_0^{+\infty} = 1[/tex]

Where is the error?

**Challenge 4**Let us consider a particle confined to the interval ##[0, 1]## and described by a wave function ##\psi## satisfying the boundary conditions ##\psi(0) = 0 = \psi(1)##. Then the momentum operator

[tex]P = \frac{\hbar}{i}\frac{d}{dx}[/tex]

is Hermitian, since the surface term appearing upon integration by parts vanishes:

[tex]\int_0^1 \left(\overline{\varphi}(P\psi) - \overline{(P\varphi)}\psi\right)(x)dx = \frac{\hbar}{i}[(\overline{\varphi}\psi)(x)]_0^1 = 0[/tex]

Since P is Hermitian, its eigenvalues are real. In order to determine the latter, we note that the eigenvalue equation,

[tex](P\psi_p)(x) = p\psi_p(x)[/tex]

is solved by ##\psi_p(x) = c_p \text{exp}\left(\frac{i}{\hbar}px\right)## with ##p\in \mathbb{R}## and ##c_\in \mathbb{C}\setminus\{0\}##. The boundary condition ##\psi_p(0) = 0## now implies ##\psi_p = 0##, therefore ##P## does not admit any eigenvalues. Nevertheless, the spectrum of ##P## is the entire complex plane and ##P## does not represent an observable. How can one understand these results which seem astonishing?

**Challenge 5**If one introduces polar coordinates in the plane or spherical coordinates in space, then the polar angle ##\varphi## and the component ##L_z## of angular momentum are canonically conjugate variables in classical mechanics. In quantum theory, the variable ##\varphi## becomes the operator of "multiplication of the wave function ##\psi(\varphi)## by ##\varphi## " and ##L_z = \frac{\hbar}{i} \frac{\partial}{\partial \varphi}##, which implies the commutation relation

[tex][L_z,\varphi] = \frac{\hbar}{i}\boldsymbol{1}[/tex]

These operators acting on periodic wave functions (i.e. ##\psi(0) = \psi(2\pi)##) are Hermitian. Furthermore, ##L_z## admits a complete system of orthonormal eigenfunctions ##\psi_m## for ##m\in \mathbb{Z}## and

[tex]L_z \psi_m = m\hbar\psi_m~\text{with}~\psi_m(\varphi) = \frac{1}{\sqrt{2\pi}}\text{exp}(im\varphi).[/tex]

By evaluating the average value of the operator ##[L_z,\varphi]## in the state ##\psi_m## and by taking

into account the fact that ##L_z## is Hermitian, one finds that

[tex]\begin{align*}

\frac{\hbar}{i}

& = \left\langle \psi_m, \frac{\hbar}{i}\boldsymbol{1} \psi_m\right\rangle\\

& = \left\langle \psi_m, L_z\varphi \psi_m\right\rangle - \left\langle \psi_m, \varphi L_z\psi_m\right\rangle\\

& = \left\langle L_z^\dagger\psi_m, \varphi \psi_m\right\rangle - m\hbar\left\langle \psi_m, \varphi \psi_m\right\rangle\\

& = (m\hbar - m\hbar)\left\langle \psi_m, \varphi \psi_m\right\rangle\\

& = 0

\end{align*}[/tex]

Where is the problem?

**Challenge 6**Let us add a bit to the confusion of the previous example! In 1927, Pauli noted that the

canonical commutation relation implies Heisenberg’s uncertainty relation ##\Delta P \Delta Q\geq \frac{\hbar}{2}## by virtue of the Cauchy-Schwarz inequality. One can derive, in the same way, the uncertainty relation

[tex] \Delta L_z \Delta \varphi\geq \frac{\hbar}{2}[/tex]

The following physical reasoning shows that this inequality cannot be correct. One can always find a state for which ##\Delta L_z < \frac{\hbar}{4\pi}## and then the uncertainty for the angle ##\varphi## has to be larger than ##2\pi##, which does not have any physical sense, since ##\varphi## takes values in the interval ##[0, 2\pi)##. How is this possible?

By the way, this example shows that the uncertainty relation ##\Delta A \Delta B\geq \frac{1}{2}|\left\langle [A,B]\right\rangle|## for any two observables ##A## and ##B## (whose derivation can be found in most quantum mechanics texts) is not valid in such a generality.

**Challenge 7**Let us consider a particle of mass ##m## in the infinite potential well (for ##a>0##)

[tex]V(x) = \left\{\begin{array}{ll}

0 & \text{if}~|x|\leq a\\

\infty & \text{otherwise}

\end{array}\right.[/tex]

The Hamiltonian for the particle confined to the inside of the well is simply

[tex]H = - \frac{\hbar^2}{2m} \frac{d^2}{dx^2}[/tex]

Let

[tex]\psi(x) = \left\{\begin{array}{ll}

\frac{\sqrt{15}}{4a^{5/2}}(a^2 - x^2) & \text{if}~|x|\leq a\\

0 & \text{otherwise}

\end{array}\right.[/tex]

be the normalized wave function of the particle at a given time. Since

[tex]H^2\psi = \frac{\hbar^4}{4m^2}\frac{d^4}{dx^4}\psi = 0,[/tex]

the average value of the operator ##H^2## in the state ##ψ## vanishes:

[tex]\left\langle H^2\right\rangle_\psi = \left\langle\psi,H^2\psi\right\rangle = \int_{-a}^a\overline{\psi(x)}(H^2\psi)(x)dx = 0[/tex]

This average value can also be determined from the eigenvalues and eigenfunctions of H,

[tex]H\varphi_n = E_n \varphi_n[/tex]

with

[tex]E_n = \frac{\pi^2\hbar^2}{8ma^2} n^2~\text{for}~n=1,2,3,...[/tex]

by applying the formula

[tex]\left\langle H^2\right\rangle_\psi = \sum_{n=1}^{+\infty} E_n^2 p_n[/tex]

with ##p_n = |\left\langle\varphi_n,\psi\right\rangle|^2##.

Proceeding in this way, one definitely does not find a vanishing result, because ##E_n^2>0## and ##0\leq p_n\leq 1##, ##\sum_{n=1}^{\infty} p_n = 1##. In fact, the calculation yields ##\left\langle H^2\right\rangle_\psi = \frac{15\hbar^4}{8m^2a^4}##. Which one of these two results is correct and where does the inconsistency come from?

**References**[1] Mathematical surprises and Dirac's formalism in quantum mechanics by François Gieres.

https://arxiv.org/pdf/quant-ph/9907069

Last edited: