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What are coherence and quantum entanglement? Does it mean that two particles are the same?

I read this in a book called Physics of the Impossible by Michio Kaku. He says that two particles behave in the same way even if they are separated. He also says that this is helpful in teleportation. How can this be possible? Could somebody please explain?

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  • $\begingroup$ Please don't revert any of the edits on this post again. If anyone thinks a rollback is necessary, open a question on Physics Meta to explain why. $\endgroup$
    – David Z
    Nov 12 '13 at 0:57
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Coherent (or pure) state

Consider 2 basic states $\lvert0\rangle$ and $\lvert1\rangle$. (If you never heard about states, treat them as ordinary complex vectors.) Here we suppose that $\lvert0\rangle$ and $\lvert1\rangle$ are orthogonal ($\langle 0\lvert1\rangle =0$).

Now, consider $\lvert c\rangle = \frac{1}{\sqrt{2}} (\lvert0\rangle + e ^{i\phi}\lvert1\rangle)$.

$\lvert c\rangle$ is a (normalized) coherent state, because the phase $\phi$ is constant.

We may look at the density matrix, defined as $\rho = \lvert c\rangle \langle c\lvert$ (that is $\langle i\lvert\rho\lvert j\rangle=\rho_{ij} = c_i^* c_j$, with $c_1 = \frac{1}{\sqrt{2}}, c_2 = \frac{1}{\sqrt{2}}e ^{i\phi}$). We have:

$$\rho =\frac{1}{2} \begin{pmatrix} 1&e ^{i\phi}\\e ^{-i\phi}&1 \end{pmatrix}. \tag{1}$$

This density matrix describes a coherent state. You may verify that $\rho^2 = \rho$, that is $\rho$ is a projector onto the coherent state $\lvert c\rangle$.

Now, suppose the phase $\phi$ is random (that is: the phase difference between $c_1$ and $c_2$ is random), so the mean expectation value of $e ^{i\phi}$ is just zero, and we have a density matrix:

$$\rho' =\frac{1}{2} \begin{pmatrix} 1&0\\0&1 \end{pmatrix}. \tag{2}$$

The density matrix $\rho'$ does not represent no more a coherent state, this is simply a classical statistical probability law. The off-diagonal elements of the matrix $\rho$ have disappeared.

In the 2 cases, we are dealing with only one particle, and the probabilities to find the particle in the $\lvert0\rangle$ state or the $\lvert1\rangle$ state, are the same, and are $\frac{1}{2}$.

$$\langle 0\lvert\rho\lvert0\rangle= \langle 1\lvert\rho\lvert1\rangle= \langle 0\lvert\rho'\lvert0\rangle= \langle 1\lvert\rho'\lvert1\rangle = \frac{1}{2}. \tag{3}$$

Entanglement

Entanglement is about at least $2$ particles, for instance the pure state

$$ \frac{1}{\sqrt{2}}(\lvert0\rangle\lvert0\rangle+\lvert1\rangle\lvert1\rangle )\tag{4}$$is a maximally 2-particles entangled state.

From a given entanged state, you may calculate the correlations for the joint measurement of the 2 particles.

It appears that these entangled quantum correlations are stronger than the classical statistical correlations.

Correlations do not mean that you may exchange instantaneous information; see also this previous answer.

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  • $\begingroup$ But what if you propagate $|c\rangle$? Then you get $|c\rangle_t=\frac{1} {\sqrt{2}}(e^{-i\omega_0 t}|0\rangle+e^{i\Phi}e^{-i\omega_1 t}|1\rangle) $. And there the phase difference $(\omega_0-\omega_1)t-\Phi$ is not constant anymore. $\endgroup$ Jul 28 '17 at 23:05
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What is coherence and quantum entanglement? Does it mean that two particles are the same?

No, coherence means a mathematical relationship that remains invariant . Entanglement is related to coherence, as it is a description of particles that have an invariant relationship in time and space.

An every day description of entanglement would be the following: A set of twins, a boy and a girl, have moved in opposite directions, one lives in California the other in New York. If you meet the boy in California you instantly know that the other twin in New York is a girl. No information has been transmitted over land with any velocity, except the previous knowledge that an invariant relation existed between these two people.

I read this in a book called 'Physics of the impossible' by Michio Kaku. He says two particle behave the same way even if they are separated. He also says this is helpful in teleportation. How can this be possible ?Could somebody please explain?

This I hope is a bad transcription of what he must have said. He must have said that entangled particles allow one to instantly know from detecting one of them the condition of the other. This is trivial, as my twins example shows and carries no utilizable meaning even if teleportation existed, which it does not. Sounds like a science fiction book to me.

Now coherence in quantum mechanics is due to the nature of the wave functions, which describe the underlying stratum of particles and molecules. These are sinusoidal functions which means they not only have an amplitude ( a measure) but also a phase. Coherence means that the phases of the wave function are kept constant between the coherent particles.

Coherence also exists in classical dimensions wherever there are sinusoidal functions describing the situation. Resonances can build up coherently, as in screaming loud speaker or microphone. It is said that soldiers break step crossing old bridges so that the amplitude of their feet hitting the ground does not add up and destroy the bridge.

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  • $\begingroup$ i except the first part. 'Physics of the impossible' is no science fiction book. It's author is one of the top 10 physicists in the world. $\endgroup$
    – hegde98
    Sep 16 '13 at 9:48
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    $\begingroup$ Your twin example does not violate Bell inequalities whereas experiments on entangled particles do. Entanglement is not trivial. I accept there may be a new model to explain entanglement in a non 'Bell local' way but at the moment you can't claim that quantum teleportation is not real until you give a better theory that explains the experiments - your twin example is not sufficient. $\endgroup$
    – Matta
    Jun 2 '15 at 10:59
  • $\begingroup$ @centralcharge, wait, really? $\endgroup$
    – Len
    Mar 13 '18 at 22:48
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I think a more basic answer would be helpful.

When we think of particles and wavefunctions we think of each particle having its own wavefunction $\psi(x)$ - the tendency for a particle to be found at some position $x$. When we have two particles, it is natural to think that we have two wavefunctions, $\psi_1(x)$ and $\psi_2(x)$, each describing its own particle. But entanglement is the statement that we should describe the two particles with a single wavefunction $\psi(x_1, x_2)$ instead. In other words, a measure of the tendency for particle 1 to be in position $x_1$ WHEN particle 2 is in position $x_2$.

There are many caveats to this, and I hope the technically inclined will forgive their omission.

Now, because particles can have more to them than just position, we can do the same thing with spin to the mix. Let $\psi(s_1, s_2)$ give the tendency for particle 1 to have spin $s_1$ (up or down) and particle 2 to have spin $s_2$ etc. The wavefunction still depends on position but I've ignored it momentarily. Then suppose we measure the spin of particle 2 and find it to be "up". Because the two particles are described by the same function, we know that particle 1's tendency to be in a certain state $s_1$ is $\psi(s_1, up)$. Therefore, knowledge of the second particle's state has affected the behaviour of the first particle. This is the essence of entanglement.

We can do the same thing with one particle, and say that it's spin and it's position are entangled. Similarly, we can describe as many particles as we wish. It is better to think of there being one and only one wavefunction that describes all the particles in the universe. The reason we can sometimes get away with single particle wavefunctions is because this "grand" wavefunction often factors neatly into a product of wavefunctions. For instance, if two particles are not entangled, then we can write $\psi(x_1, x_2) = \psi_1(x_1) \times \psi_2(x_2)$. This is obviously not always the case.

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An entangled state is well defined mathematically. Looking at it through this lens should demystify the matter. Assume you have a two-part system described by the state spaces $H_{1}\otimes H_{2}$. Any vector from the space has the form

$ \left | \psi \right \rangle = \sum_{i=1}^{n} \sum_{j=1}^{m} c_{i,j} ( \left | \phi_{i} \right \rangle \left | \varphi _{j} \right \rangle)$, (1)

where $\left | \phi_{i} \right \rangle$ is the basis of $H_{1}$ and $\left | \varphi _{j} \right \rangle$ is the basis of $H_{2}$. A separable (not entangled) state is defined as a state for which you can find vectors $ \left | \phi \right \rangle \in H_{1}$, $\left | \varphi \right \rangle \in H_{2}$ such that

$\left | \psi \right \rangle = \left | \phi \right \rangle \left | \varphi \right \rangle$.

Since $ \left | \phi \right \rangle \in H_{1}$ and $\left | \varphi \right \rangle \in H_{2}$ we have the following basis representations:

$\left | \phi \right \rangle = \sum_{i=1}^{n} a_{i,j} \left | \phi_{i} \right \rangle $,

$\left | \varphi \right \rangle = \sum_{j=1}^{m} c_{i,j} \left | \varphi _{j} \right \rangle$.

if you apply the definition of the tensor product and "multiply these out" you get:

$ \left | \psi \right \rangle = \sum_{i=1}^{n} \sum_{j=1}^{m} a_{i}b_{j} ( \left | \phi_{i} \right \rangle \left | \varphi _{j} \right \rangle)$. (2)

If you compare (1) and (2) you can see that for a state to be separable you need the coefficients to be related in the following way:

$c_{i,j} = a_{i}b_{j}$ for all $i, j$. (3)

The following example will hopefully provide ultimate clarity. Suppose you are to decide whether the following state is separable:

$\left | \psi \right \rangle = \frac{1}{\sqrt{2}}(\left | 00 \right \rangle + \left | 11 \right \rangle)$.

If you suppose that this state arose as a product of two "elementary" states from $H_{1}$, $H_{2}$ you would get a contradiction (i.e. no such set of coefficients that (3) is satisfied exist). This shows us that the state is not-separable (i.e. entangled). The main takeaways are:

1) There are "more" vectors in $H_{1}\otimes H_{2}$ then just the ones of the form $\left | \psi \right \rangle = \left | \phi \right \rangle \left | \varphi \right \rangle$.

2) The entangled states have a "special structure". Suppose you would perform a von Neumann measurement on the first qbit of state $\left | \psi \right \rangle$. Regardless of what result you would get you would collapse the second part of the system to either state $\left | 0 \right \rangle$ or $\left | 1 \right \rangle$. This interdependency is where these types of states get their name from.

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