- #1

dyn

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If I know the probability of it being spin up and spin down how do I know if it exists as a pure state of a superposition of spin up and spin down or a mixed state of spin up and spin down ?

Thanks

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- #1

dyn

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If I know the probability of it being spin up and spin down how do I know if it exists as a pure state of a superposition of spin up and spin down or a mixed state of spin up and spin down ?

Thanks

- #2

ChrisVer

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yes because there is nothing else to alter this.If I have a spin 1/2 particle eg electron in an isolated box can I state for definite that there is a 50% chance of it being spin up and 50% spin down ?

what do you mean by mixed state vs superposition state?If I know the probability of it being spin up and spin down how do I know if it exists as a pure state of a superposition of spin up and spin down or a mixed state of spin up and spin down ?

it will exist (spin) in :[itex] \frac{1}{\sqrt{2}} \Big( |\uparrow> + |\downarrow> \Big)[/itex]

this is a superposition with no other choice.

- #3

Truecrimson

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If I have a spin 1/2 particle eg electron in an isolated box can I state for definite that there is a 50% chance of it being spin up and 50% spin down ?

Without any measurement, how can you know? The principle of indifference?

If I know the probability of it being spin up and spin down how do I know if it exists as a pure state of a superposition of spin up and spin down or a mixed state of spin up and spin down ?

Knowing the probabilities of getting the spin up and spin down state along the z axis will restrict possible states to be on a plane cutting the Bloch sphere perpendicular to the z axis. Only subsequent measurements of the spin in two more linearly independent directions (each being done on a separate but identically prepared spin system) can tell you the state of the spin.

Or if you allow joint measurements on many identical copies of the spin system, there are other ways.

- #4

dyn

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The pure state would be the superposition state but could it also exist as a mixed state of a spin up state and a spin down state ?yes because there is nothing else to alter this.

what do you mean by mixed state vs superposition state?

it will exist (spin) in :[itex] \frac{1}{\sqrt{2}} \Big( |\uparrow> + |\downarrow> \Big)[/itex]

this is a superposition with no other choice.

- #5

Truecrimson

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yes because there is nothing else to alter this.

The OP doesn't state the state the spin is in.

- #6

blue_leaf77

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You can't, because you don't know the state of the electron in the box.can I state for definite that there is a 50% chance of it being spin up and 50% spin down ?

For atoms, you can determine whether it's in pure or mixed state by observing its emission. An atom in pure and superposition states of its eigenstates will have time-dependent dipole moment, which means it will emit EM radiation at certain frequencies. Conversely, if this atom is in a mixed state of its eigenstates, it will not emit anything. As for hypothetical individual electrons without spatial wavefunction, I don't know a way to distinguish.If I know the probability of it being spin up and spin down how do I know if it exists as a pure state of a superposition of spin up and spin down or a mixed state of spin up and spin down ?

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dyn

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- #8

blue_leaf77

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- #9

Truecrimson

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https://en.wikipedia.org/wiki/Quantum_tomography

- #10

ChrisVer

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what do you mean by state of the electron?The OP doesn't state the state the spin is in.

The electron's spin eigenfunction will be like the one I posted... there is no reason however [if the electron is isolated] to say that the probability for its spin to be up and down are different.

- #11

Truecrimson

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what do you mean by state of the electron?

The electron's spin eigenfunction will be like the one I posted... there is no reason however [if the electron is isolated] to say that the probability for its spin to be up and down are different.

Why must it be in the eigenfunction of the Pauli ## \sigma_x ## if I don't know anything about the the spin e.g. has it been subject to a magnetic field in some direction?

Moreover, there are infinitely many pure states that have the same probabilities of being up or down. $$ \frac{1}{\sqrt{2}}(|\uparrow \rangle + e^{i \theta} |\downarrow \rangle) $$

- #12

ChrisVer

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Why must it be in the eigenfunction of the Pauli σx \sigma_x if I don't know anything about the the spin e.g. has it been subject to a magnetic field in some direction?

You can always write in as a sum of eigenfunctions of [itex]\sigma_i[/itex] since any state can be written then in that basis. In my case I wrote it in [itex]\sigma_3[/itex] eigenfunctions [itex]|\uparrow>[/itex] and [itex]|\downarrow>[/itex].

That means I wrote down something like:

[itex] |ele_spin> = a |\uparrow> + b |\downarrow>[/itex]

Where [itex]|a|^2 + |b|^2 =1[/itex]... a simple choice (and physical one) is [itex]|a|=|b|= \frac{1}{\sqrt{2}}[/itex] and chose then that.

The magnetic field in some direction would indicate that the electron is not isolated within the box (something that I took as given by the OP). As I noted "there is nothing else to alter this":

because there is nothing else to alter this.

the one you posted however preserves the point which I made; the probabilities for being up or down remain 50%... the phase you added does not alter this result as long as the states don't "communicate" (something that a magnetic field could do).Moreover, there are infinitely many pure states that have the same probabilities of being up or down.

[itex]P(\uparrow) =\Big|\Big| <\uparrow| \frac{1}{\sqrt{2}} \Big(|\uparrow>+ e^{i\theta} |\downarrow> \Big) \Big| \Big|^2[/itex]

[itex]P(\uparrow) = \frac{1}{2}[/itex]

In fact what you can do with that extra phase is eg vary the basis you exist in (I guess). Which shouldn't alter the result.

- #13

Truecrimson

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I'm not sure how to respond to this. If you are given a task of predicting the state an isolated spin is in, you will do poorly by estimate it to be in a pure state. Because if the true spin state is the orthogonal pure state, no Bayesian updating will be able to bring your estimate to the pure state, even if you collect an infinite amount of data.

The guess that the probabilities of getting the spin up or down are the same is good by the principle of indifference. But asserting that the state is pure is unjustified.

Edit: typo

The guess that the probabilities of getting the spin up or down are the same is good by the principle of indifference. But asserting that the state is pure is unjustified.

Edit: typo

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- #14

ChrisVer

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I don't understand what you mean by "pure" state. Could you elaborate?

- #15

ChrisVer

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In fact I would say it's much more than just indifference...The guess that the probabilities of getting the spin up or down are the same is good by the principle of indifference.

As long as up and down are up to the observer to decide [in other words they are connected via a symmetry], if don't break that symmetry (by eg setting a preferred orientation), you shouldn't really expect something else... the opposite would need a very good justification...

If you found that

- #16

Truecrimson

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Ah! That must be where our disagreement came from.

The general statistical description of a quantum system is given by a density operator: a trace-1, positive semidefinite matrix ## \rho ##. This comes from the fact that if there is a set of states ## |\psi_j \rangle ## that the system can be in, and a probability distribution ##p_j## over the set, I can describe the system by the operator $$ \rho = \sum_j p_j |\psi_j \rangle \langle \psi_j| $$. This is enough to calculate any expectation value of any observable ## A ## by the Born rule in the form of ## \text{Tr}(A \rho)##. Then the properties of the density operator that I gave in the beginning is the requirement that the probabilities I can calculate are positive and sum to 1.

A state is pure iff the density operator is rank 1. That is, ## \rho = |\psi \rangle \langle \psi |## for some ## |\psi \rangle ##. Otherwise it is a mixture or mixed state (this terminology is standard even though a mixture does not have a definite state).

For a spin-1/2, all density operators are depicted on the Bloch ball. The interior is where the mixed states live.

So there are infinitely many mixed and pure states consistent with the statement that the probabilities of getting a spin up or down are the same; they are all the states on the plane perpendicular to the z axis cutting the ball through the origin.

Edit: Typo in an equation

The general statistical description of a quantum system is given by a density operator: a trace-1, positive semidefinite matrix ## \rho ##. This comes from the fact that if there is a set of states ## |\psi_j \rangle ## that the system can be in, and a probability distribution ##p_j## over the set, I can describe the system by the operator $$ \rho = \sum_j p_j |\psi_j \rangle \langle \psi_j| $$. This is enough to calculate any expectation value of any observable ## A ## by the Born rule in the form of ## \text{Tr}(A \rho)##. Then the properties of the density operator that I gave in the beginning is the requirement that the probabilities I can calculate are positive and sum to 1.

A state is pure iff the density operator is rank 1. That is, ## \rho = |\psi \rangle \langle \psi |## for some ## |\psi \rangle ##. Otherwise it is a mixture or mixed state (this terminology is standard even though a mixture does not have a definite state).

For a spin-1/2, all density operators are depicted on the Bloch ball. The interior is where the mixed states live.

So there are infinitely many mixed and pure states consistent with the statement that the probabilities of getting a spin up or down are the same; they are all the states on the plane perpendicular to the z axis cutting the ball through the origin.

Edit: Typo in an equation

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- #17

Truecrimson

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