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Photon polarisation

Photon polarisation is the quantum mechanical treatment of light polarisation. Photoelectric effect experiments have shown that the direction of electron emission depends on the polarisation of the incident light. In quantum mechanics, classical polarisation directions are referred to as quantum states, with the states \vert z\rangle and \vert x\rangle representing linear polarisation parallel to the z-axis and the x-axis respectively (see diagram below).

If we consider a unit circle, the state of a photon \vert\psi\rangle polarised at \theta with respect to the z-axis can therefore be written as a linear combination of \vert z\rangle and \vert x\rangle:

\vert\psi\rangle=cos\theta\vert z\rangle+sin\theta\vert x\rangle\; \; \; \; \; \; \; \; 237

where \vert z\rangle and \vert x\rangle become basis state vectors.

Since these basis vectors are unit vectors \vert z\rangle=\begin{pmatrix} 1\\0 \end{pmatrix} and \vert x\rangle=\begin{pmatrix} 0\\1 \end{pmatrix}, we can express the state of the photon as \vert \psi\rangle=\begin{pmatrix} cos\theta\\sin\theta \end{pmatrix}.

 

Question

Can a single photon be unpolarised?

Answer

Each photon in a string of photons (e.g. from the sun) has a definite polarisation, even though one photon’s oscillation direction may be randomly oriented relatively to another photon’s oscillation direction.

 

As per any normalised quantum state of \vert\psi\rangle=\sum_{i=1}^{2}c_i\vert\phi_i\rangle, the square of each coefficient \vert c_i\vert^{2} is interpreted as the probability that a measurement of a system will yield an eigenvalue associated with the corresponding eigenstate \phi_i, i.e. cos^{2}\theta +sin^{2}\theta=1. If we measure the polarisation of a string of photons, whose generic states are given by eq237, using a calcite crystal (see diagram above; the optic axis lies in the plane of the viewer’s screen), photons with oscillations perpendicular to the plane of the screen emerge as part of the o-ray with the state \vert\psi\rangle=\vert x\rangle, while photons with oscillations parallel to the plane emerge as part of the e-ray with the state \vert\psi\rangle=\vert z\rangle. A photon oscillating at \theta=\pm45^{\circ} (\vert\psi\rangle=\frac{1}{\sqrt{2}}[\vert z\rangle\pm\vert x\rangle]) has 50% probability of emerging as part of the o-ray and 50% probability emerging as part of the e-ray.

 

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Pockels cell

A Pockels cell is an electro-optical device that rotates the polarisation of light. It is based on the Pockels Effect, where the birefringence of the crystal in the cell changes linearly with an applied voltage (see diagram below).

An example of a crystal exhibiting the Pockels Effect is ammonium dihydrogen phosphate NH4H2PO4, which has a tetragonal crystal structure. It is aligned with its optic axis parallel to the propagation direction of the light. In the absence of an external electric field, the incoming beam suffers no birefringence. When an external electric field is applied across the crystal, the electron distribution of the crystal is distorted to the extent that the relative values of n_o and n_e changes, with the degree of change being proportional to the applied voltage.

To explain the change in polarisation direction of light, let’s regard the vertically polarised light as a superposition of two circular light components that are rotating in phase with the same angular frequency \omega but in opposite directions (see above diagram). When the external electric field is turned on, one component travels faster than the other in the Pockels cell. For a cell of length l, the difference in time \Delta t travelled by each component is:

\Delta t=\frac{l}{c_o}-\frac{l}{c_e}\; \; \; \; \; \; \; \; 236

where c_o and c_e are the speeds of the two components in the cell.

Substitute the definition of refractive index (n_x=\frac{c}{c_x}, where c is the speed of light in vacuum) and eq235 in eq236

\Delta t=\left ( \frac{n_o}{c}-\frac{n_e}{c} \right )l=\Delta n\frac{l}{c}

Substitute the above equation in the definition of phase difference (\Delta \phi=\omega\Delta t=2\pi\frac{c}{\lambda}\Delta t) between the two components,

\Delta \phi=\Delta n\frac{2\pi l}{\lambda}

As the birefringence of the crystal \Delta n changes proportionally with voltage V, the phase difference \Delta \phi between the two circular components changes with an applied voltage. Since the angle of rotation of polarisation is \Delta\theta=\frac{\Delta \phi}{2} (you can visualise the rotation angle using the diagram above),

\Delta\theta=\kappa V\frac{\pi l}{\lambda}

where \kappa is a proportionality constant, which is specific to the material used.

 

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Birefringence

Birefringence is the optical property of a material in which an incident ray of light is split into two perpendicularly polarised rays by the material. One material exhibiting strong birefringence is calcite (CaCO3), which has a hexagonal unit cell as shown in the diagram below.

When a single crystal of calcite is oriented in a particular direction, we see layers of atoms arranged symmetrically. Each layer consists of carbonate molecules in a plane and calcium atoms directly above the carbon atoms (see above diagram). The layers are stacked in an ABBAABBA… fashion, where the oxygen atoms in B are rotated 60o relative the oxygen atoms in A. Due to the symmetry of the atoms, a ray of unpolarised light passing perpendicular to the layers (oscillation parallel to the layers) emerges as a single ray of unpolarised light, i.e. it suffers no birefringence. The direction in which the ray that suffers no birefringence travels is called the optic axis of the crystal.

Birefringence occurs when a ray of unpolarised light passes through a calcite crystal, which is oriented relative to the ray as shown in the diagram above. For simplicity, only two orthogonal oscillations of the incident ray are depicted, with the up-down arrows representing polarisation along the plane of the viewer’s screen and the blue dots representing polarisation into the plane of the screen. The surface of the calcite crystal facing the reader and the optic axis of the crystal are parallel to the plane of the screen. As mentioned earlier, oscillations that are perpendicular to the optic axis (i.e. blue dots) suffer no birefringence, with its path unaltered. The ray of such oscillations that emerges from the crystal is called an ordinary ray or o-ray.

The polarisation represented by the double arrows are not travelling perpendicular to the optic axis and encounters electron densities that are not symmetrically distributed. The induced oscillations in different parts of those electron densities reradiate light waves that combine to form a refracted wave. The refracted light travels through the crystal with a different speed v and emerges as a separate ray, which is called an extraordinary ray or e-ray. This implies that the crystal has two different refractive indices n_o and n_e. The birefringence \Delta n of a material is defined as:

\Delta n=n_e-n_o\; \; \; \; \; \; \; \; 235

 

Question

Why is the e-ray deflected ‘upwards’ in the crystal?

Answer

The incident ray, with oscillation parallel to the plane of the viewer’s screen, can be further resolved into a component that is perpendicular to the optic axis (oscillation parallel to the optic axis) and one that is parallel to the optic axis (oscillation perpendicular to the optic axis, see diagram below).

We can imagine that the oscillation of the latter component encounters a denser distribution of electrons as it travels along the optic axis in the crystal (refer to the structure of calcite in the earlier diagram) as compared to the other component. Therefore, v_{\parallel}> v_{\perp} and the resultant ray is deflected “upwards”.

 

 

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Light polarisation

Light polarisation is the classical description of oscillation directions of electromagnetic waves. Light from the sun or incandescent lamps is emitted by a large number of atoms, each producing oscillations in random directions at any given time. Since the oscillations at any point fluctuates rapidly and randomly, we call this unpolarised light (see diagram below).

 

Question

Do the arrows in the above diagram represent electric field vectors or magnetic field vectors?

Answer

It is more convenient, and in many cases more practical, to define polarisation of light in terms of the oscillation of electric field vectors (a magnetic field does not exert a force on charges at rest).

 

Polarised light can be produced by passing unpolarised light through a polariser. One of the most widely used polarisers is the H-sheet, which is a sheet of polyvinyl alcohol that is heated and stretched to the extent that the polymeric chains are aligned in a particular direction. Long chains of iodine, which are parallel to the alcohol chains, are then formed when the sheet is dipped in iodine solution. As the valence electrons in the iodine chains are only mobile along the chains, light that is polarised parallel to the chains is absorbed, while light that is polarised perpendicular to the chains is transmitted (see diagram below).

 

Question

What happens when light, which is polarised at an angle \theta to the iodine chains where 0^{\circ}< \theta< 90^{\circ}, passes through the H-sheet?

Answer

The polarised light before passing through the H-sheet can be resolved into a component that is parallel to the iodine chains and one that is perpendicular to the iodine chains. Therefore, the transmitted light is polarised perpendicular to the chains.

 

 

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Quantum key distribution

Quantum key distribution is the relay of a cryptographic key, which is generated based on the principles of quantum mechanics, between two parties. In computing and telecommunications, messages are represented by binary codes, e.g. the binary code for the letter ‘m’ is 1101101, where each digit is called a binary digit or a bit. One way for a sender to encrypt a message is to generate a key of the same bit-length as the message, e.g. 0001100, and use it to perform a modular addition on the message (where numbers are not carried or borrowed):

1101101

+ 0001100

1100001

Under the American Standard Code for Information Interchange (ASCII), the encrypted message reads ‘a’ instead of ‘m’. To decode the message, the receiver simply performs a second modular addition on the encrypted message using the same key:

1100001

+ 0001100

1101101

Theoretically, one possible way of encrypting a message is to generate a random key using the principles of quantum entanglement.

Let’s suppose:

    1. The sender and the receiver both have Stern-Gerlach devices, which can only be oriented along the z-axis or the x-axis.
    2. The sender creates seven pairs of entangled spin-\frac{1}{2} For each pair of entangled particles, one particle is measured by the sender using a randomly selected axis and the other by the receiver, also using a randomly selected axis.

We may end up with the following:

 Sender Receiver

Axis

 Bit Axis

Bit

1

 x 0 x

1

2

 z 1 x

1

3

 z 1 z

0

4

 x 0 z

0

5

 z 0 x

1
6 x 0 z

0

7

 z 0 z

1

where 0 and 1 denote an up-spin and a down-spin respectively. Since 0 and 1 are associated with a particle’s spin, which is a quantum mechanical property, we called them quantum bits or qubits.

To complete the protocol, both parties agree to share their axes of measurement (which can be communicated publicly) and agree to construct the key as follows:-

    1. Retain the sender’s bits for measurements made using the same axes
    2. Convert all other qubits to 0

They key in the above example is therefore 0010000 and the letter ‘m’ is encoded as the bracket ‘}’. If a third party intercepts every particle sent to the receiver with another Stern-Gerlach device, we may have the following:

Sender

 Receiver Interceptor

Receiver*

Axis

 State Axis State Axis State Axis

State

1

 x 0 x 1 z 0 x

0

2

 z 1 x 1 x 1 x

1

3

 z 1 z 0 z 0 z

0

4

 x 0 z 0 z 0 z

0

5

 z 0 x 1 z 1 x

1

6

 x 0 z 0 x 1 z

0

7

 z 0 z 1 x 1 z

0

Let’s suppose the interceptor then immediately sends unentangled particles of the same state as the ones he measured to the receiver (i.e. before the axes of the sender and receiver are publicly shared). The receiver’s measurements may be per ‘Receiver*’, where

    1. Receiver measures the same state as the interceptor if both use the same axes (only for 2, 3 and 4).
    2. Receiver may or may or measure the same state as the interceptor otherwise.

When the sender and receiver eventually share their axes of measurement and further share pre-determined parts of their measured states, they will realise, from the discrepancies in 1 and 7, that their channel of communication has been compromised, leading them to discard the key. Of course, the measurements of ‘Receiver*’ for 1 and 7 may be 1 and 1 respectively. On average, the number of times the sender and receiver choosing the same axis for a measurement but the interceptor selecting a different axis is \frac{1}{4}, of which half the time the interceptor will make a different measurement versus the receiver and send the ‘wrong’ state to the receiver. Therefore, the probability of the sender and receiver detecting discrepancies in the retained states when they randomly share parts of their measured states is \frac{1}{8} (assuming the shared states are representative of a randomly selected segment), which is very significant if the qubit-length of the raw key is relatively long. For example, if the raw key has 72 qubits, of which 40 common qubits are eventually selected randomly and shared between the sender and the receiver, 5 retained qubits will be erroneous.

In short, quantum key distribution provides a way to securely encrypt a message, as it allows for the detection of an interceptor. However, it is difficult to construct an infrastructure for quantum key distribution using Stern-Gerlach devices. In practice, entangled photon pairs, photon polarisation states and optical devices are used, as exemplified by the BB84 protocol.

 

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Bell’s inequality

Bell’s inequality, developed by John Bell in 1964, is a non-equal relation (between two expressions) that is based on the ERP paradox. It provides a way to determine whether quantum theory or the EPR paradox is correct. Bell suggested a unique way of measuring the spins of two spin-\frac{1}{2} particles, which are generated from the decay of a spin-0 particle at rest: the two particles are to be passed through two Stern-Gerlach devices, each oriented along one of three non-orthogonal coplanar axes, which are specified by the unit vectors \boldsymbol{\mathit{a}}, \boldsymbol{\mathit{b}} and \boldsymbol{\mathit{c}}.

The average values of the product of the spins in units of \frac{\hbar}{2} (denoted by P(\boldsymbol{\mathit{a}},\boldsymbol{\mathit{b}}), P(\boldsymbol{\mathit{b}},\boldsymbol{\mathit{c}}), and P(\boldsymbol{\mathit{a}},\boldsymbol{\mathit{c}})) are then calculated. As derived in an earlier article (see eq227), quantum mechanics expresses the average values as:

P(\boldsymbol{\mathit{a}},\boldsymbol{\mathit{b}})=P(\boldsymbol{\mathit{b}},\boldsymbol{\mathit{c}})=P(\boldsymbol{\mathit{a}},\boldsymbol{\mathit{c}})=-cos\theta\; \; \; \; \; \; \; \; 228

For the ERP paradox, let’s equate a spin-up measurement to +1 and a spin-down measurement to -1. The average value of the product of the measured spins, for example in the \boldsymbol{\mathit{a}} and \boldsymbol{\mathit{b}} directions, is:

P(\boldsymbol{\mathit{a}},\boldsymbol{\mathit{b}})=\int_{\lambda_i}^{\lambda_f}\rho(\lambda)A(\boldsymbol{\mathit{a}},\lambda)B(\boldsymbol{\mathit{b}},\lambda)d\lambda\; \; \; \; \; \; \; \; 229

where

i) \lambda is a hidden variable.

ii) \rho(\lambda) is the probability density that is a function of \lambda, with \int_{\lambda_i}^{\lambda_f}\rho(\lambda)d\lambda=1 and \rho(\lambda)\geq 0.

iii) A(\boldsymbol{\mathit{a}},\lambda) is a function of the axis of measurement and \lambda. It is associated with the measurement made by the first Stern-Gerlach device, and has output values of \pm1.

iv) B(\boldsymbol{\mathit{b}},\lambda) is a function of the axis of measurement and \lambda. It is associated with the measurement made by the second Stern-Gerlach device, and has output values of \pm1.

From experiments, we know that

A(\boldsymbol{\mathit{a}},\lambda)=-B(\boldsymbol{\mathit{b}},\lambda)\; \; \; \; \; \; \; \; 230

Substitute eq230 in eq229

P(\boldsymbol{\mathit{a}},\boldsymbol{\mathit{b}})=-\int_{\lambda_i}^{\lambda_f}\rho(\lambda)A(\boldsymbol{\mathit{a}},\lambda)A(\boldsymbol{\mathit{b}},\lambda)d\lambda\; \; \; \; \; \; \; \; 231

Similarly, P(\boldsymbol{\mathit{a}},\boldsymbol{\mathit{c}})=-\int_{\lambda_i}^{\lambda_f}\rho(\lambda)A(\boldsymbol{\mathit{a}},\lambda)A(\boldsymbol{\mathit{c}},\lambda)d\lambda and P(\boldsymbol{\mathit{b}},\boldsymbol{\mathit{c}})=-\int_{\lambda_i}^{\lambda_f}\rho(\lambda)A(\boldsymbol{\mathit{b}},\lambda)B(\boldsymbol{\mathit{c}},\lambda)d\lambda. Therefore,

P(\boldsymbol{\mathit{a}},\boldsymbol{\mathit{b}})-P(\boldsymbol{\mathit{a}},\boldsymbol{\mathit{c}})=-\int_{\lambda_i}^{\lambda_f}\rho(\lambda)A(\boldsymbol{\mathit{a}},\lambda)A(\boldsymbol{\mathit{b}},\lambda)d\lambda+\int_{\lambda_i}^{\lambda_f}\rho(\lambda)A(\boldsymbol{\mathit{a}},\lambda)A(\boldsymbol{\mathit{c}},\lambda)d\lambda

Since [A(\boldsymbol{\mathit{b}},\lambda)]^{2}=1, we can rearrange the above equation to

P(\boldsymbol{\mathit{a}},\boldsymbol{\mathit{b}})-P(\boldsymbol{\mathit{a}},\boldsymbol{\mathit{c}})=-\int_{\lambda_i}^{\lambda_f}\rho(\lambda)[1-A(\boldsymbol{\mathit{b}},\lambda)A(\boldsymbol{\mathit{c}},\lambda)]A(\boldsymbol{\mathit{a}},\lambda)A(\boldsymbol{\mathit{b}},\lambda)d\lambda

Taking the absolute value on both sides of the above equation and using the relation \left | \int f(t)dt\right |\leq \int \left | f(t) \right |dt

\vert P(\boldsymbol{\mathit{a}},\boldsymbol{\mathit{b}})-P(\boldsymbol{\mathit{a}},\boldsymbol{\mathit{c}})\vert\leq \int_{\lambda_i}^{\lambda_f}\left | \rho(\lambda)[1-A(\boldsymbol{\mathit{b}},\lambda)A(\boldsymbol{\mathit{c}},\lambda)]\right | \vert A(\boldsymbol{\mathit{a}},\lambda)A(\boldsymbol{\mathit{b}},\lambda)\vert d\lambda

 

Question

Why is \left | \int_{a}^{b}f(t)dt \right |\leq \int_{a}^{b}\vert f(t)\vert dt?

Answer

For all t\in [a,\cdots,b], we have -\vert f(t)\vert\leq f(t)\leq \vert f(t)\vert. Therefore, -\int_{a}^{b}\vert f(t)\vert dt\leq\int_{a}^{b} f(t) dt\leq \int_{a}^{b}\vert f(t)\vert dt or simply \left | \int_{a}^{b}f(t)dt \right |\leq \int_{a}^{b}\vert f(t)\vert dt, where we have use the identity of \vert x\vert\leq a\; \; \; \; \Leftrightarrow \; \; \; \; -a\leq x\leq a.

 

Since \vert A(\boldsymbol{\mathit{a}},\lambda)A(\boldsymbol{\mathit{b}},\lambda)\vert =1

\vert P(\boldsymbol{\mathit{a}},\boldsymbol{\mathit{b}})-P(\boldsymbol{\mathit{a}},\boldsymbol{\mathit{c}})\vert\leq \int_{\lambda_i}^{\lambda_f}\left | \rho(\lambda)[1-A(\boldsymbol{\mathit{b}},\lambda)A(\boldsymbol{\mathit{c}},\lambda)]\right | d\lambda

Since \rho(\lambda)\geq 0 and 1-A(\boldsymbol{\mathit{b}},\lambda)A(\boldsymbol{\mathit{c}},\lambda)\geq 0, we can ignore the absolute value sign on RHS of the above equation:

\vert P(\boldsymbol{\mathit{a}},\boldsymbol{\mathit{b}})-P(\boldsymbol{\mathit{a}},\boldsymbol{\mathit{c}})\vert\leq \int_{\lambda_i}^{\lambda_f} \rho(\lambda)d\lambda-\int_{\lambda_i}^{\lambda_f}\rho(\lambda)A(\boldsymbol{\mathit{b}},\lambda)A(\boldsymbol{\mathit{c}},\lambda) d\lambda

Substituting \int_{\lambda_i}^{\lambda_f} \rho(\lambda)d\lambda=1 and P(\boldsymbol{\mathit{b}},\boldsymbol{\mathit{c}})=-\int_{\lambda_i}^{\lambda_f}\rho(\lambda)A(\boldsymbol{\mathit{b}},\lambda)A(\boldsymbol{\mathit{c}},\lambda) d\lambda into the above equation gives

\vert P(\boldsymbol{\mathit{a}},\boldsymbol{\mathit{b}})-P(\boldsymbol{\mathit{a}},\boldsymbol{\mathit{c}})\vert\leq 1+P(\boldsymbol{\mathit{b}},\boldsymbol{\mathit{c}})\; \; \; \; \; \; \; \; 232

Eq232 is the Bell’s inequality, which is based on the ERP paradox.

To show that quantum mechanics is incompatible with Bell’s inequality, we let \angle \boldsymbol{\mathit{a}}\boldsymbol{\mathit{c}}=\angle \boldsymbol{\mathit{b}}\boldsymbol{\mathit{c}}=45^{\circ}, i.e. \boldsymbol{\mathit{c}} bisects \angle \boldsymbol{\mathit{a}}\boldsymbol{\mathit{b}}. From eq228,

P(\boldsymbol{\mathit{a}},\boldsymbol{\mathit{b}})=-cos90^{\circ}=0\; \; \; \; \; \; \; \; 233

P(\boldsymbol{\mathit{b}},\boldsymbol{\mathit{c}})=P(\boldsymbol{\mathit{a}},\boldsymbol{\mathit{c}})=-cos45^{\circ}=-\frac{1}{\sqrt{2}}\; \; \; \; \; \; \; \; 234

Substituting eq233 and eq234 in eq232 yields \frac{1}{\sqrt{2}}\leq1-\frac{1}{\sqrt{2}}, which is inconsistent with Bell’s inequality. Therefore, it is possible to experimentally measure the spins of the two particles at non-orthogonal angles to test the predictions of quantum theory versus the ERP paradox. In fact, the results of all experiments conducted at non-orthogonal angles were in agreement with quantum mechanics. This implies that all local hidden-variable hypotheses are invalid.

 

 

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Einstein-Podolsky-Rosen (EPR) paradox

The Einstein-Podolsky-Rosen paradox (EPR paradox), conceived in 1935, proposes that a particle (or a system) has definite attributes (i.e., physical properties) prior to any measurements. It also suggests that each particle is influenced only by its immediate surroundings (locally), rather than by another particle at a distance. This contrasts with the statistical interpretation of a particle’s state as theorised by quantum mechanics, where the certainty of an attribute exists only after a measurement is made, and particles are considered entangled.

Let’s examine the case of the decay of a spin-0 particle at rest into two spin-\frac{1}{2} particles with zero relative orbital angular momentum, and the subsequent passing of the two particles through two Stern-Gerlach devices. Setting aside the EPR paradox for a moment, experimental results show that if the spin of the first particle, which is measured in the +z-direction, is an up spin, the spin of the second particle measured in the same direction is always a down spin. This is of course due to the principle of conservation of spin angular momentum. If we instead measure the spin of the second particle in the +x-direction (see diagram below), we may record either an up spin or a down spin.

By repeating the experiment, say a hundred times, the results (assuming a random distribution of spins after decay) are as follows:-

Particle 1 (measured in +z) Particle 2 (inferred in +z) Particle 2 (measured in +x)
EV_1 Population   EV_2 Population
+z 50 -z +x 25
-x 25
-z 50 +z +x 25
-x 25

where +z and +x under EV_1 and EV_2 refer to up spins, while -z and -x refer to down spins.

The data revealed that amongst all the fifty runs of particle 1 returning results of up spins in the +z-direction, twenty five corresponding particle 2 have up spins in the +x-direction, while the remaining twenty five have down spins. Similarly, for all fifty runs of particle 1 returning results of down spins in the +z-direction, twenty five corresponding particle 2 have up spins in the +x-direction, while the remaining twenty five have down spins.

Quantum theory, which states that we can only be certain that a system possesses an attribute after a measurement is made, accurately predicted the above results via eq226. The EPR paradox, on the other hand, proposes that each particle must have the definite spin angular momentum information necessary for the Stern-Gerlach device to reveal before the particle passes through it. This suggests that the population of particle 2 can be categorised as follows:

Particle 2
(-z,+x)
(-z,-x)
(+z,+x)
(+z,-x)

which implies that the corresponding particle 1 can be grouped along with particle 2 as:

Groups, \lambda_j Particle 1 Particle 2
\lambda_1 (+z,-x) (-z,+x)
\lambda_2 (+z,+x) (-z,-x)
\lambda_3 (-z,-x) (+z,+x)
\lambda_4 (-z,+x) (+z,-x)

 

Question

Does such a categorisation of spin-\frac{1}{2} particle pairs violate the uncertainty principle, which states that we cannot precisely determine the z-component of the spin angular momentum and the x-component of the spin angular momentum of a particle simultaneously?

Answer

No, because we are only measuring the spin angular momentum of each particle either in the +z-direction or the +x-direction but not in both directions simultaneously.

 

Therefore, it appears that the EPR paradox is consistent with the results obtain from the double Stern-Gerlach device measurements for orthogonal axes. If \lambda_j is a variable that is measurable, we would be able to distinguish the particles. The EPR paradox suggests that such a variable is hidden from us (a hidden variable) and that quantum mechanics is incomplete. This challenge to quantum mechanics remained unresolved until John Bell, in 1964, conceptualised a method to determine whether quantum mechanics or the EPR paradox is correct.

 

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Quantum entanglement

Quantum entanglement is a phenomenon where each particle constituting a composite system cannot be described by its own characteristic state. Instead, the particles must be expressed as a single composite state. One such entangled composite system is the decay of a spin-0 particle at rest into two spin-\frac{1}{2} particles with zero relative orbital angular momentum. According to the conservation of linear momentum, the particles move in opposite directions after decay. If we measure the spin of the particles by passing them through two Stern-Gerlach devices, we would get correlated results due to the conservation of spin momentum. For example, if one particle is measured to be spin up with respect to the z-axis (see diagram below), the other particle is always measured to be spin down with respect to the same axis, and vice versa.

The quantum mechanical explanation of the results is that the singlet state \vert\psi\rangle=\frac{1}{\sqrt{2}}[\vert+z\rangle_1\otimes\vert-z\rangle_2-\vert-z\rangle_1\otimes\vert+z\rangle_2] collapses into \vert\phi_1\rangle=\vert+z\rangle_1\otimes\vert-z\rangle_2 if the first measurement is spin up. If the first measurement is spin down, the composite state collapses into \vert\phi_2\rangle=\vert-z\rangle_1\otimes\vert+z\rangle_2. Whether the first measurement is spin up or spin down is purely random, as evident from the coefficient of the singlet state. In other words, quantum mechanics provides, via \vert\psi\rangle, a statistical distribution of possible outcomes of a measurement, in a way that the certainty that a system possesses a physical property is only determined after a measurement is made.

 

Question

How does the coefficient of the singlet state show that the collapse of the state is random?

Answer

The singlet state is normalised and is a linear combination of the states \vert\phi_1\rangle=\vert+z\rangle_1\otimes\vert-z\rangle_2 and \vert\phi_2\rangle=\vert-z\rangle_1\otimes\vert+z\rangle_2; i.e. \vert\psi\rangle=\sum_{i=1}^{2}c_i\vert\phi_i\rangle, where c_1=c_2=\frac{1}{\sqrt{2}}. Since \vert c_i\vert^{2} is interpreted as the probability that a measurement of a system will yield an eigenvalue associated with the eigenfunction \phi_i, the probability of obtaining an up-spin (or a down-spin) for the first measurement is 50%.

 

If we conduct the experiment a hundred times, with the first Stern-Gerlach device in the +z direction and the second device in the +x direction, we have (assuming a random distribution of spins after decay) the following results:

 

Particle 1 (measured in +z) Particle 2 (inferred in +z) Particle 2 (measured in +x)
EV_1 Population EV_{2,z} EV_2 Population EV_1EV_2 Average EV_1EV_2
+\frac{\hbar}{2} 50 -\frac{\hbar}{2} +\frac{\hbar}{2} 25 +\frac{\hbar^{2}}{4} \frac{(25\times\frac{\hbar^{2}}{4})+[25\times(-\frac{\hbar^{2}}{4})]}{50}=0
-\frac{\hbar}{2} 25 -\frac{\hbar^{2}}{4}
-\frac{\hbar}{2} 50 +\frac{\hbar}{2} +\frac{\hbar}{2} 25 -\frac{\hbar^{2}}{4} \frac{[25\times(-\frac{\hbar^{2}}{4})]+(25\times\frac{\hbar^{2}}{4})}{50}=0
-\frac{\hbar}{2} 25 +\frac{\hbar^{2}}{4}

 

In general, if the second Stern-Gerlach device is rotated at an angle \theta relative to the first device, the composite spin angular momentum operator associated with EV_1EV_2 (the product of EV_1 and EV_2) is:

\hat{S}_z^{\; (1)}\otimes\hat{S}_r^{\; (2)}=\left (\hat{S}_z^{\; (1)}\otimes I\right )\left (I\otimes\hat{S}_r^{\; (2)}\right )

Substitute eq174 and eq184 (where we let \phi=0^{\circ}) into the above equation,

\hat{S}_z^{\; (1)}\otimes\hat{S}_r^{\; (2)}=\frac{\hbar^{2}}{4}\begin{pmatrix} cos\theta &sin\theta &0 &0 \\ sin\theta &-cos\theta &0 &0 \\ 0 &0 &-cos\theta &-sin\theta \\ 0 &0 &-sin\theta &cos\theta \end{pmatrix}

The average value of EV_!EV_2 over multiple experiments is:

\small \langle\psi\vert\hat{S}_z^{\; (1)}\otimes\hat{S}_r^{\; (2)}\vert\psi\rangle=\frac{1}{\sqrt{2}}\begin{pmatrix} 0 &1 &-1 &0 \end{pmatrix}\frac{\hbar^{2}}{4}\begin{pmatrix} cos\theta &sin\theta &0 &0 \\ sin\theta &-cos\theta &0 &0 \\ 0 &0 &-cos\theta &-sin\theta \\ 0 &0 &-sin\theta &cos\theta \end{pmatrix} \frac{1}{\sqrt{2}}\begin{pmatrix} 0\\1 \\ -1 \\ 0 \end{pmatrix}

\langle\psi\vert\hat{S}_z^{\; (1)}\otimes\hat{S}_r^{\; (2)}\vert\psi\rangle=-\frac{\hbar^{2}}{4}cos\theta\; \; \; \; \; \; \; \; 226

or in spin units of \frac{\hbar}{2}

\langle\psi\vert\hat{S}_z^{\; (1)}\otimes\hat{S}_r^{\; (2)}\vert\psi\rangle=-cos\theta\; \; \; \; \; \; \; \; 227

Finally, an interesting behaviour of an entangled system is that the particles constituting the system remain entangled if they are separated by a large distance, even at a distance greater than light can travel within the time between measurements. If so, and if the outcome of the first measurement is random, one may perceive that the two particles are able to influence each other instantaneously. This property of a composite system seems to be at odds with special relativity, which implies that nothing (including communication of any form) can travel faster than the speed of light. In opposition to the quantum mechanical interpretation of the results, Albert Einstein, along with Boris Podolsky and Nathan Rosen, proposed an alternate explanation called the Einstein-Podolsky-Rosen paradox.

 

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The singlet and triplet states

The singlet state describes a composite system with a coupled spin eigenstate of \small \vert j=0,m_j=0\rangle, while a triplet state refers to a composite system with coupled spin eigenstates of \small \vert 1,+1\rangle, \small \vert 1,0\rangle and \small \vert 1,-1\rangle.

Consider a system of two spin-\small \frac{1}{2} particles ( \small s_1=\frac{1}{2} and \small s_2=\frac{1}{2}). According to the Clebsch-Gordan series, the allowed angular momenta of the system are \small j=1,0, which implies that the system is characterised by the spin eigenstates \small \vert 1,+1\rangle, \small \vert 1,0\rangle, \small \vert 1,-1\rangle and \small \vert 0,0\rangle. The basis states forming these four spin eigenstates are given by eq193, with the general spin eigenstate of the system being a linear combination of these basis states (see eq195).

Let’s begin with the determination of the values of the coefficients for the spin eigenstates \small \vert 1,0\rangle and \small \vert 0,0\rangle, which must adhere to the condition: \small \hat{S}_z^{\; (T)}\vert\psi\rangle=0\hbar\vert\psi\rangle. Letting the operator \small \hat{S}_z^{\; (T)} act on eq195 and using eq193 and 197,

\hat{S}_z^{\; (T)}\vert\psi\rangle=c_1\hbar\begin{pmatrix} 1\\0 \\ 0 \\ 0 \end{pmatrix}-c_4\hbar\begin{pmatrix} 0\\0 \\ 0 \\ 1 \end{pmatrix}+0\hbar\left [c_2\begin{pmatrix} 0\\1 \\ 0 \\ 0 \end{pmatrix}+c_3\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix}\right ]

For the eigenvalue equation \hat{S}_z^{\; (T)}\vert\psi\rangle=0\hbar\vert\psi\rangle to be valid, c_1=c_4=0, i.e. one possible spin eigenstate when m_j=0 is \vert\psi\rangle=c_2\vert+z\rangle_1\otimes\vert-z\rangle_2+c_3\vert-z\rangle_1\otimes\vert+z\rangle_2, while the other possible spin eigenstate is \vert\psi\rangle=c_2\vert+z\rangle_1\otimes\vert-z\rangle_2-c_3\vert-z\rangle_1\otimes\vert+z\rangle_2. These two spin eigenstates must correspond to either \vert 1,0\rangle or \vert0,0\rangle. To distinguish them, we use eq203 and let the operator act on the two spin eigenstates to give:

\hat{{S}^{2}}^{(T)}\left [ c_2\begin{pmatrix} 0\\1 \\0 \\0 \end{pmatrix}\pm c_3\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix}\right ] =\hbar^{2}\left [ c_2\begin{pmatrix} 0\\1 \\1 \\0 \end{pmatrix}\pm c_3\begin{pmatrix} 0\\1 \\ 1 \\ 0 \end{pmatrix}\right ]

Since the two eigenstates are characterised by j=1 and j=0, the expected results are \hat{{S}^{2}}^{(T)}\vert\psi\rangle=j(j+1)\hbar^{2}\vert\psi\rangle=2\hbar^{2}\vert\psi\rangle and \hat{{S}^{2}}^{(T)}\vert\psi\rangle=j(j+1)\hbar^{2}\vert\psi\rangle=0\hbar^{2}\vert\psi\rangle respectively. Therefore, the above equation is valid only if c_2=c_3=c, such that:

\small \hat{{S}^{2}}^{(T)}\left [ c\begin{pmatrix} 0\\1 \\0 \\0 \end{pmatrix}+ c\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix}\right ] =\hbar^{2}\left [ c\begin{pmatrix} 0\\1 \\1 \\0 \end{pmatrix}+ c\begin{pmatrix} 0\\1 \\ 1 \\ 0 \end{pmatrix}\right ]=2\hbar^{2}c\begin{pmatrix} 0\\1 \\ 1 \\ 0 \end{pmatrix}=2\hbar^{2}\left [ c\begin{pmatrix} 0\\1 \\0 \\0 \end{pmatrix}+ c\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix}\right ]

and

\small \hat{{S}^{2}}^{(T)}\left [ c\begin{pmatrix} 0\\1 \\0 \\0 \end{pmatrix}- c\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix}\right ] =\hbar^{2}\left [ c\begin{pmatrix} 0\\1 \\1 \\0 \end{pmatrix}- c\begin{pmatrix} 0\\1 \\ 1 \\ 0 \end{pmatrix}\right ]=\hbar^{2}c\begin{pmatrix} 0\\0 \\ 0 \\ 0 \end{pmatrix}=0\hbar^{2}\left [ c\begin{pmatrix} 0\\1 \\0 \\0 \end{pmatrix}- c\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix}\right ]

Therefore, \small \vert1,0\rangle=c(\vert+z\rangle_1\otimes\vert-z\rangle_2+\vert-z\rangle_1\otimes\vert+z\rangle_2) and \small \vert0,0\rangle=c(\vert+z\rangle_1\otimes\vert-z\rangle_2-\vert-z\rangle_1\otimes\vert+z\rangle_2). The spin state \small \vert0,0\rangle as mentioned earlier is the singlet state, which after normalisation becomes

\small \vert0,0\rangle=\frac{1}{\sqrt{2}}(\vert+z\rangle_1\otimes\vert-z\rangle_2-\vert-z\rangle_1\otimes\vert+z\rangle_2)\; \; \; \; \; \; \; \; 220

 

Question

Show that the normalisation constant is \small \frac{1}{\sqrt{2}}.

Answer

\langle\psi\vert\psi\rangle=1\; \; \; \; \Rightarrow \; \; \; \; c^{*}\left [ \begin{pmatrix} 0\\1 \\ 0 \\ 0 \end{pmatrix}-\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix} \right ] ^{\dagger}c\left [ \begin{pmatrix} 0\\1 \\ 0 \\ 0 \end{pmatrix}-\begin{pmatrix} 0\\0 \\ 1 \\ 0 \end{pmatrix} \right ]=1

\vert c\vert^{2}\begin{pmatrix} 0 &1 &-1 &0 \end{pmatrix}\begin{pmatrix} 0\\1 \\ -1 \\ 0 \end{pmatrix}=1\; \; \; \; \Rightarrow \; \; \; \; c=\frac{1}{\sqrt{2}}

 

Likewise, the state \vert1,0\rangle after normalisation is

\vert1,0\rangle=\frac{1}{\sqrt{2}}(\vert+z\rangle_1\otimes\vert-z\rangle_2+\vert-z\rangle_1\otimes\vert+z\rangle_2)\; \; \; \; \; \; \; \; 221

We have two more spin eigenstates of the system to consider: \vert1,+1\rangle and \vert1,-1\rangle. Since the allowed values of the total magnetic quantum number m_j are the sum of the allowed values of the two contributing magnetic quantum numbers (see eq207), \vert1,+1\rangle=\vert+z\rangle_1\otimes\vert+z\rangle_2 and \vert1,-1\rangle=\vert-z\rangle_1\otimes\vert-z\rangle_2., both of which are already normalised because they are basis states.

 

Question

Verify that \hat{{S}^{^{2}}}^{{T}}\vert1,\pm1\rangle=2\hbar^{2}\vert1,\pm1\rangle and \hat{S}_z^{\; {T}}\vert1,\pm1\rangle=\pm\hbar\vert1,\pm1\rangle.

Answer

Using eq203

\hat{{S}^{2}}^{(T)}\vert1,+1\rangle=\hbar^{2}\begin{pmatrix} 2 &0 &0 &0 \\ 0 &1 &1 &0 \\ 0& 1 & 1 & 0\\ 0& 0& 0& 2 \end{pmatrix}\begin{pmatrix} 1\\0 \\ 0 \\ 0 \end{pmatrix}=2\hbar^{2}\begin{pmatrix} 1\\0 \\ 0 \\ 0 \end{pmatrix}

Similarly, \hat{{S}^{2}}^{(T)}\vert1,-1\rangle=2\hbar^{2}\vert1,-1\rangle. Using eq197,

\hat{S}_z^{\; {T}}\vert1,+1\rangle=\hbar\begin{pmatrix} 1 &0 &0 &0 \\ 0 &0 &0 &0 \\ 0& 0 & 0 & 0\\ 0& 0& 0& -1 \end{pmatrix}\begin{pmatrix} 1\\0 \\ 0 \\ 0 \end{pmatrix}=\hbar\begin{pmatrix} 1\\0 \\ 0 \\ 0 \end{pmatrix}

Similarly, \hat{S}_z^{\; (T)}\vert1,-1\rangle=-\hbar\vert1,-1\rangle.

 

Therefore, we have:

Triplet\: state=\left\{\begin{matrix} \vert1,+1\rangle&=\vert+z\rangle_1\otimes\vert+z\rangle_2 \\ \vert1,0\rangle &=\frac{1}{\sqrt{2}}(\vert+z\rangle_1\otimes\vert-z\rangle_2+\vert-z\rangle_1\otimes\vert+z\rangle_2)\\ \vert1,-1\rangle&=\vert-z\rangle_1\otimes\vert-z\rangle_2 \end{matrix}\right.

Singlet\: state=\vert0,0\rangle=\frac{1}{\sqrt{2}}[\vert+z\rangle_1\otimes\vert-z\rangle_2-\vert-z\rangle_1\otimes\vert+z\rangle_2]

or in the wave function notation:

Triplet\: state=\left\{\begin{matrix} \sigma_a&=\alpha(1)\alpha(2)\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; 222 \\ \sigma_b &=\frac{1}{\sqrt{2}}[\alpha(1)\beta(2)+\beta(1)\alpha(2)]\; \; \; \; \; \; 223\\ \sigma_c&=\beta(1)\beta(2)\; \; \; \; \; \; \; \;\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; 224 \end{matrix}\right.

Singlet\: state=\sigma_d=\frac{1}{\sqrt{2}}[\alpha(1)\beta(2)-\beta(1)\alpha(2)]\; \; \; \; \; \; \; \; 225

The helium atom with the excited state configuration of 1s12s1 is commonly used to illustrate the singlet and triplet states. If the electrons have anti-parallel spins, the excited atom is characterised by the singlet state. If the electrons have parallel spins, the excited helium is in the triplet state.

 

 

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Composite systems – beyond spin operators

In the previous article, we showed that the total z-component spin angular momentum operator\hat{S}_z^{\; (T)}  is \hat{S}_z^{\; (T)}=\hat{S}_z^{\; (1)}\otimes I+I\otimes\hat{S}_z^{\; (2)}, which is a special case of the general form:

\hat{J}_i=\hat{M}_i^{\; (1)}\otimes I+I\otimes\hat{M}_i^{\; (2)}\; \; \; \; \; \; \; \; 205

\hat{J}_i (where i=x,y,z) is the total angular momentum component operator. \hat{M}_i^{\; (1)} and \hat{M}_i^{\; (2)} are component operators of \hat{M}^{ (1)} and \hat{M}^{ (2)} respectively. \hat{M}^{ (1)} and \hat{M}^{ (2)} are operators of two sources of angular momentum, where they may be: 1) the orbital angular momentum operator of particle 1 and orbital angular momentum operator of particle 2 respectively; 2) the spin angular momentum operator of particle 1 and spin angular momentum operator of particle 2 respectively; 3) the orbital angular momentum operator and spin angular momentum operator respectively of a particle.

Also mentioned in the previous article is that L_z=\sum_{i=1}^{n}l_{z,i} and S_z=\sum_{i=1}^{n}s_{z,i}. It is therefore easy to accept the validity of points 1) and 2). For point 3, the proposal that j_z=l_z+s_z may seem untenable. However, spin angular momentum, like orbital angular momentum, is a form of angular momentum. In fact, the total angular momentum \boldsymbol{\mathit{J}} of a system is defined as the vector sum \boldsymbol{\mathit{J}}=\boldsymbol{\mathit{L}}+\boldsymbol{\mathit{S}}. If point 3) is valid, \hat{J}_i must satisfy the same commutation relations as described by eq99, eq100 and eq101.

 

Question

Show that \hat{J}_i satisfies the same commutation relations as described by eq99, eq100 and eq101.

Answer

\left [ \hat{J}_x,\hat{J}_y \right ]=\left [\hat{M}_x^{\; (1)}\otimes I+I\otimes\hat{M}_x^{\; (2)},\hat{M}_y^{\; (1)}\otimes I+I\otimes\hat{M}_y^{\; (2)}\right ]

Expanding the RHS of the above equation and noting that \hat{M}_i^{\; (1)} and \hat{M}_i^{\; (2)} commute because they act on different vector spaces, we have

\left [ \hat{J}_x,\hat{J}_y \right ]=\left [\hat{M}_x^{\; (1)}\otimes I,\hat{M}_y^{\; (1)}\otimes I\right ] +\left [I\otimes\hat{M}_x^{\; (2)},I\otimes\hat{M}_y^{\; (2)}\right ]

\left [ \hat{J}_x,\hat{J}_y \right ]=\left [\hat{M}_x^{\; (1)} ,\hat{M}_y^{\; (1)}\right ]\otimes I +I\otimes\left [\hat{M}_x^{\; (2)},\hat{M}_y^{\; (2)}\right ]

With reference to eq99, eq100, eq101, eq165, eq166 and eq167, \left [\hat{M}_i^{\; (l)} ,\hat{M}_j^{\; (l)}\right ]=i\hbar\epsilon_{ijk}\hat{M}_k^{\; l}, where l=1,2 and \epsilon_{ijk} is the Levi-Civita symbol. So,

\left [ \hat{J}_x,\hat{J}_y \right ]=i\hbar\hat{M}_z^{\; (1)} \otimes I +i\hbar I\otimes\hat{M}_z^{\; (2)}=i\hbar\hat{J}_z

Similarly, we have \left [ \hat{J}_y,\hat{J}_z \right ]=i\hbar\hat{J}_x and \left [ \hat{J}_z,\hat{J}_x \right ]=i\hbar\hat{J}_y.

 

Since the total angular momentum component operators satisfy the form of commutation relations as described by eq99, eq100 and eq101, the raising and lowering operators also apply to the total angular momentum operator \hat{J}. We would therefore expect

\hat{J}^2\psi=j(j+1)\hbar^2\psi\; \; \; \; and\; \; \; \; \; \hat{J}_z\psi=m_j\hbar\psi\; \; \; \; \; \; \; \; \; 205a

You’ll realise from the workings of the above Q&A that we can simplify the notation \hat{J}_z of eq205 as

\hat{J}_z=\hat{M}_{z,1}+\hat{M}_{z,2}\; \; \; \; \; \; \; \; 206

To show that \hat{J}_z commutes with \hat{M}_{z,i}, where \hat{M}_{z,i} can be either \hat{l}_{z,i} or \hat{s}_{z,i}, we have \left [ \hat{J}_z,\hat{l}_z \right ]=\left [ \hat{l}_z,\hat{l}_z \right ]+\left [ \hat{s}_z,\hat{l}_z \right ]=0 and \left [ \hat{J}_z,\hat{s}_z \right ]=\left [ \hat{l}_z,\hat{s}_z \right ]+\left [ \hat{s}_z,\hat{s}_z \right ]=0. Therefore, the eigenstate of \hat{J}_z is simultaneously the eigenstates of \hat{M}_{z,1} and \hat{M}_{z,2}. This implies that the eigenvalues of \hat{J}_z are the sum of the eigenvalues of \hat{M}_{z,1} and \hat{M}_{z,2}, i.e. m_j\hbar=m_1\hbar+m_2\hbar, or

m_j=m_1+m_2\; \; \; \; \; \; \; \; 207

In other words, the allowed values of the total magnetic quantum number m_j are the sum of the allowed values of the two contributing magnetic quantum numbers. As for the allowed values of the total angular momentum quantum number j, let’s further define the eigenvalues of \hat{M}_1^{\; 2} and \hat{M}_2^{\; 2} as M_1(M_1+1)\hbar^{2} and M_2(M_2+1)\hbar^{2} respectively. This allows us to work with the quantum numbers M_1 and M_2.

Now, the maximum value of m_j in eq207 is m_{j,max}=m_{1,max}+m_{2,max}. Since the maximum value of a magnetic quantum number is the angular momentum quantum number (i.e. m_{j,max}=j_{max} and m_{i,max}=M_{i}, where i=1,2), the highest value of j is

j_{max}=M_1+M_2\; \; \; \; \; \; \; \; 208

Furthermore, for a particular value of \boldsymbol{\mathit{j}} in the coupled representation, there are 2j+1 values of m_j and therefore 2j+1 states. So j_{max} has 2j_{max}+1=2M_1+2M_2+1 states. These states are \vert j_{max},m_{j,max}\rangle,\vert j_{max},m_{j,max}-1\rangle,\vert j_{max},m_{j,max}-2\rangle,\cdots. The states for the next lower value of j (denoted by j') are \vert j',m'_{j,max}\rangle,\vert j',m'_{j,max}-1\rangle,\vert j',m'_{j,max}-2\rangle,\cdots. The same logic applies for states all the way to the lowest value of j.

 

Question

Show that the total number of states in the uncoupled representation is (2M_1+1)(2M_2+1).

Answer

In eq193, the total number of states in the uncoupled representation \vert M_1,m_1;M_2,m_2\rangle is the number of ways to form Kronecker products of basis vectors from each vector space. Since there are 2M_1+1 basis vectors in the 1st vector space and 2M_2+1 basis vectors in the 2nd vector space,

total\: states\: in\: uncoupled\: representation=(2M_1+1)(2M_2+1)\; \; \; \; \; \; \; \; 209

 

To determine the lower values of j, we consider the lower values of m_{j,max}, the first being m_{j,max}-1=m'_{j,max}. There are two possible ways to obtain this value, with m_{1,max}-1 and m_{2,max}, or m_{1,max} and m_{2,max}-1.  Since each state is characterised by a unique value of m_j for a particular value of j, one of the two possibilities is accounted for by the state \vert j_{max},m_{j,max}-1\rangle. The remaining possibility must be due to \vert j',m'_{j,max}\rangle. Since m_{j,max}=j_{max}, we must have m'_{j,max}=j'. Furthermore, because m_{j,max}-1=m'_{j,max}, we have j'=m_{j,max}-1=j_{max}-1. The state \vert j',m'_{j,max}\rangle is therefore \vert j_{max}-1,m_{j,max}-1\rangle.

For m_{j,max}-2, there are three possible ways to obtain it. Again, one of the possible ways is accounted for by \vert j_{max},m_{j,max}-2\rangle and the second way by \vert j',m'_{j,max}-1\rangle=\vert j_{max}-1,m_{j,max}-2\rangle. The remaining possibility must be due to the state \vert j'',m''_{j,max}\rangle=\vert j'-1,m'_{j,max}-1\rangle=\vert j_{max}-2,m_{j,max}-2\rangle.

Therefore, the allowed values of  are

j=j_{max},j_{max}-1,j_{max}-2,\cdots,j_{min}

or

j=M_1+M_2,M_1+M_2-1,M_1+M_2-2,\cdots,j_{min}

To determine j_{min}, we note that the total number of states for the system can be written as \sum_{j=j_{min}}^{j_{max}}(2j+1) because there are 2j+1 states associated with each value of j. Since j_{min}\geq 0, we can further split the sum as:

total\: states=\sum_{j=0}^{j_{max}}(2j+1)-\sum_{j=0}^{j_{min}-1}(2j+1)\; \; \; \; \; \; \; \; 210

 

Question

Show that \sum_{j=0}^{n-1}(2j+1)=n^{2} and hence \sum_{j=0}^{x}(2j+1)=(x+1)^{2}.

Answer

\sum_{j=0}^{n-1}(2j+1)=1+\sum_{j=1}^{n-1}(2j+1)=1+2\sum_{j=1}^{n-1}j+\sum_{j=1}^{n-1}1\; \; \; \; \; \; \; \; 211

For the 2nd term on RHS of 2nd equality of eq211, \sum_{j=1}^{n-1}j=1+2+\cdots+(n-2)+(n-1), which if written in the reverse order becomes \sum_{j=1}^{n-1}j=(n-1)+(n-2)+\cdots+2+1. Adding the two sums, we have

2\sum_{j=1}^{n-1}j=n(n-1)\; \; \; \; \; \; \; \; 214

For the 3rd term on RHS of 2nd equality of in eq211

\sum_{j=1}^{n-1}1=\sum_{j=1}^{n-1}j^{0}=1+1+\cdots+1=n-1\; \; \; \; \; \; \; \; 215

Substitute eq214 and eq215 back in eq211, we have \sum_{j=0}^{n-1}(2j+1)=n^{2}. Let x=n-1, we have

\sum_{j=0}^{n-1}(2j+1)=(x+1)^{2}\; \; \; \; \; \; \; \; 216

 

Using eq216, where x=j_{max} for \sum_{j=0}^{j_{max}}(2j+1), and x=j_{min}-1 for \sum_{j=0}^{j_{min}-1}(2j+1), eq210 becomes

total\: states=(j_{max}+1)^{2}-j_{min}^{\; \;\; \; \; \; 2}=j_{max}^{\; \;\; \; \; \; 2}+2j_{max}+1-j_{min}^{\; \;\; \; \; \; 2}\; \; \; \; \; \; \; \; 217

The total number of states (energy levels) of a system must be independent of the chosen representation. Substituting eq209 in LHS of eq217 and eq208 in RHS of eq217 and simplifying,

j_{min}=\pm(M_1-M_2)\; \; \; \; \; \; \; \; 218

Eq218 is equivalent to j_{min}=\vert M_1-M_2\vert because j_{min}\geq 0, M_1\geq 0, M_2\geq 0, and M_2 may be a larger value than M_1. Therefore, for a given value of M_1 and a given value of M_2, the allowed values of the total angular momentum quantum number j are:

j=M_1+M_2,M_1+M_2-1,\cdots,\vert M_1-M_2\vert\; \; \; \; \; \; \; \; 219

which is called the Clebsch-Gordan series.

 

Question

Write all the eigenstates (in the form of \vert j,m_j\rangle) and basis states (in the form of \vert m_1,m_2\rangle) of a system with two sources of angular momentum, M_1=2 and M_2=1.

Answer

There are a total of 15 eigenstates and also 15 basis states. The allowed values of j are 3, 2 and 1. The eigenstates are \vert 3,3\rangle, \vert 3,2\rangle, \vert 3,1\rangle, \vert 3,0\rangle, \vert 3,-1\rangle, \vert 3,-2\rangle, \vert 3,-3\rangle, \vert 2,2\rangle, \vert 2,1\rangle, \vert 2,0\rangle, \vert 2,-1\rangle, \vert 2,-2\rangle, \vert 1,1\rangle, \vert 1,0\rangle and \vert 1,-1\rangle. The basis states are \vert 2,1\rangle, \vert 1,1\rangle, \vert 0,1\rangle, \vert -1,1\rangle, \vert -2,1\rangle, \vert 2,0\rangle, \vert 2,-1\rangle, \vert 1,0\rangle, \vert 1,-1\rangle, \vert 0,0\rangle, \vert 0,-1\rangle, \vert -1,0\rangle, \vert -1,-1\rangle, \vert -2,0\rangle and \vert -2,-1\rangle. Each spin eigenstate of the system is a linear combination of the 15 basis states.

 

What we have described so far pertains to a system with two sources of angular momentum. If the system has more than two sources of angular momentum, the Clebsch-Gordan series is applied repeatedly, i.e. a first series is written with M_1 and M_2, and then the Clebsch-Gordan procedure is again applied to each value of this series with M_3 to form a second resultant series, and the procedure is repeated until a final resultant series is developed with M_{final}. For example, a system with three sources of angular momentum, s_1=\frac{1}{2}, s_2=\frac{1}{2} and s_3=\frac{1}{2}, has the following allowed values of S:

1st series using s_1 and s_2, S=1,0

2nd and final series using 1,0 and s_3, S=\frac{3}{2},\frac{1}{2},\frac{1}{2}

For this system, there are 8 basis states, whose explicit forms can be expressed as follows:

\small \begin{pmatrix} 1\\0 \end{pmatrix}\otimes\begin{pmatrix} 1\\0 \end{pmatrix} \otimes\begin{pmatrix} 1\\0 \end{pmatrix} \; \; \; \;\begin{pmatrix} 0\\1 \end{pmatrix}\otimes\begin{pmatrix} 1\\0 \end{pmatrix} \otimes\begin{pmatrix} 1\\0 \end{pmatrix} \; \; \; \;\begin{pmatrix} 1\\0 \end{pmatrix}\otimes\begin{pmatrix} 1\\0 \end{pmatrix} \otimes\begin{pmatrix} 0\\1 \end{pmatrix} \; \; \; \;\begin{pmatrix} 0\\1 \end{pmatrix}\otimes\begin{pmatrix} 1\\0 \end{pmatrix} \otimes\begin{pmatrix} 0\\1 \end{pmatrix}

\small \begin{pmatrix} 1\\0 \end{pmatrix}\otimes\begin{pmatrix} 0\\1 \end{pmatrix} \otimes\begin{pmatrix} 1\\0 \end{pmatrix} \; \; \; \;\begin{pmatrix} 0\\1 \end{pmatrix}\otimes\begin{pmatrix} 0\\1 \end{pmatrix} \otimes\begin{pmatrix} 1\\0 \end{pmatrix} \; \; \; \;\begin{pmatrix} 1\\0 \end{pmatrix}\otimes\begin{pmatrix} 0\\1 \end{pmatrix} \otimes\begin{pmatrix} 0\\1 \end{pmatrix} \; \; \; \;\begin{pmatrix} 0\\1 \end{pmatrix}\otimes\begin{pmatrix} 0\\1 \end{pmatrix} \otimes\begin{pmatrix} 0\\1 \end{pmatrix}

 

Question

What are the allowed angular momenta of a system with three sources of angular momentum, \small l_1=1, \small l_2=1 and \small l_3=1, and how many basis states are there in total?

Answer

\small j=3,2,2,1,1,1,0. The total number of basis states is 27.

 

 

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