Eigenstates of particle with 1/2 spin (qbit)

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• cianfa72
cianfa72
TL;DR Summary
About the eigenstates of particles with 1/2 spin (qbit)
A very basic doubt about a QM system (particle) with spin 1/2 (qbit).

From the Bloch sphere representation we know that a qbit's pure state is represented by a point on the surface of the sphere. Picking a base, for instance the pair of vector/states ##\ket{\uparrow}## and ##\ket{\downarrow}##, any qbit's pure state can be given as (normalized) complex linear combination of the above states $$\ket{\psi} = \alpha \ket{\uparrow} + \beta \ket{\downarrow}$$
Does that mean that for any qbit's pure state there always exist an Hermitian/self-adjoint operator (an observable) such that that state is eigenstate of ? Thanks.

cianfa72 said:
TL;DR Summary: About the eigenstates of particles with 1/2 spin (qbit)

A very basic doubt about a QM system (particle) with spin 1/2 (qbit).

From the Bloch sphere representation we know that a qbit's pure state is represented by a point on the surface of the sphere. Picking a base, for instance the pair of vector/states ##\ket{\uparrow}## and ##\ket{\downarrow}##, any qbit's pure state can be given as (normalized) complex linear combination of the above states.

Does that mean for any qbit's pure state there always exist an Hermitian/self-adjoint operator (an observable) such that that state is eigenstate of ? Thanks.
Yes.

Edit: the recipe is the following, a generic state can be written as

$$|\psi\rangle=\cos(\theta/2)|\uparrow\rangle+e^{i\varphi}\sin(\theta/2)|\downarrow\rangle$$

with ##|\alpha|^2+|\beta|^2=1## (pure state). Then this state is an eigenstate of the spin in the direction of the unit vector ##\hat{\mathbf{n}}=(\sin \theta \cos\varphi, \sin\varphi \sin\theta, \cos\theta)##:
$$\mathbf S_{\hat{\mathbf n}}=\hat{\mathbf n}\cdot \mathbf S=\frac12 \hat{\mathbf n}\cdot \boldsymbol \sigma= \frac{1}{2}\begin{pmatrix} \cos \theta & \sin\theta e^{-i\varphi} \\ \sin\theta e^{i\varphi}&-\cos\theta \end{pmatrix},$$
where ##\boldsymbol \sigma =(\sigma_x,\sigma_y,\sigma_z)## are the Pauli matrices.

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DrClaude and cianfa72
pines-demon said:
...a generic state can be written as$$|\psi\rangle=\cos(\theta/2)|\uparrow\rangle+e^{i\varphi}\sin(\theta/2)|\downarrow\rangle$$with ##|\alpha|^2+|\beta|^2=1## (pure state).
Did you mean to write ##|\psi\rangle=\alpha\cos(\theta/2)|\uparrow\rangle+\beta e^{i\varphi}\sin(\theta/2)|\downarrow\rangle##?

anuttarasammyak
renormalize said:
Did you mean to write ##|\psi\rangle=\alpha\cos(\theta/2)|\uparrow\rangle+\beta e^{i\varphi}\sin(\theta/2)|\downarrow\rangle##?
I think he meant to include that ##\alpha = \cos (\theta / 2)## and ##\beta = e^{i \varphi} \sin (\theta / 2)##.

renormalize
renormalize said:
Did you mean to write ##|\psi\rangle=\alpha\cos(\theta/2)|\uparrow\rangle+\beta e^{i\varphi}\sin(\theta/2)|\downarrow\rangle##?
PeterDonis said:
I think he meant to include that ##\alpha = \cos (\theta / 2)## and ##\beta = e^{i \varphi} \sin (\theta / 2)##.
Indeed my bad, I was trying to write a general state using ##\alpha## and ##\beta## but I just dropped it, forget about ##\alpha,\beta## in #2.

pines-demon said:
Then this state is an eigenstate of the spin in the direction of the unit vector ##\hat{\mathbf{n}}=(\sin \theta \cos\varphi, \sin\varphi \sin\theta, \cos\theta)##:
$$\mathbf S_{\hat{\mathbf n}}=\hat{\mathbf n}\cdot \mathbf S=\frac12 \hat{\mathbf n}\cdot \boldsymbol \sigma= \frac{1}{2}\begin{pmatrix} \cos \theta & \sin\theta e^{-i\varphi} \\ \sin\theta e^{i\varphi}&-\cos\theta \end{pmatrix},$$
where ##\boldsymbol \sigma =(\sigma_x,\sigma_y,\sigma_z)## are the Pauli matrices.
Ok, the above means that if one performs a spin measurement along the direction $$\hat{\mathbf{n}}=(\sin \theta \cos\varphi, \sin\varphi \sin\theta, \cos\theta)$$ on an ensemble of particles all in that eigenstate, one get spin-up with a percentage of occurrences of 100%.

cianfa72 said:
Ok, the above means that if one performs a spin measurement along the direction $$\hat{\mathbf{n}}=(\sin \theta \cos\varphi, \sin\varphi \sin\theta, \cos\theta)$$ on an ensemble of particles all in that eigenstate, one get spin-up with a percentage of occurrences of 100%.
Yes, for an ensemble of spins aligned along ##\hat{\mathbf n}##, the measurement of ##\mathbf S_{\hat{\mathbf n}}## is always ##+\tfrac12## (earlier I used ##\hbar=1## but if you include it, you get ##+\hbar/2##).

cianfa72
Consider the position/momentum Hilbert space for a particle (i.e. the space ##L^2(\mathbb R##) in one dimension) and a vector/state ##\ket{\phi} \in L^2(\mathbb R)##.

Does the statement in post#2 hold true even for it (namely for any state ##\ket{\phi}## always exists a corresponding Hermitian/self-adjoint operator ##A## such that that state is eigenstate of) ?

cianfa72 said:
Consider the position/momentum Hilbert space for a particle (i.e. the space ##L^2(\mathbb R##) in one dimension) and a vector/state ##\ket{\phi} \in L^2(\mathbb R)##.

Does the statement in post#2 hold true even for it (namely for any state ##\ket{\phi}## always exists a corresponding Hermitian/self-adjoint operator ##A## such that that state is eigenstate of) ?
It holds true for any pure state, since for any pure state ##\ket{\phi}## you can define the projection operator ##\ket{\phi} \bra{\phi}##, for which ##\ket{\phi}## is an eigenstate, and which is easily shown to be self-adjoint.

What you can't assume is that there will be a self-adjoint operator for which ##\ket{\phi}## is an eigenstate that has any reasonably understandable physical meaning.

cianfa72 said:
Consider the position/momentum Hilbert space for a particle (i.e. the space ##L^2(\mathbb R##) in one dimension) and a vector/state ##\ket{\phi} \in L^2(\mathbb R)##.

Does the statement in post#2 hold true even for it (namely for any state ##\ket{\phi}## always exists a corresponding Hermitian/self-adjoint operator ##A## such that that state is eigenstate of) ?
There is one way to carry out that measurement if you know how to prepare the state. You just run the preparation backwards (Hermitian conjugate of the gates you used to build the state). This is a common technique to check the fidelity of your state. Start from a given state (let's say ##|\uparrow\rangle##), prepare a given state (let's say ##\frac{1}{\sqrt2}[|\uparrow\rangle+i|\downarrow\rangle)##] and then un-prepare the state back to the initial state. If everything is done correctly you will measure the initial state (up spin in this example).

PeterDonis said:
What you can't assume is that there will be a self-adjoint operator for which ##\ket{\phi}## is an eigenstate that has any reasonably understandable physical meaning.
Ok, so the difference between spin 1/2 two-state system (qubit) and particle position/momentum is that for the former any pure state is eigenstate of a physical observable (namely the spin measured along a physical axis) while for the latter one can not always assume that.

In a sense a qubit has a physical "element of reality" according to EPR's view (if we know the state the qubit is in then one can predict with 100% certain the counterfactual value of the spin along that axis).

cianfa72 said:
Ok, so the difference between spin 1/2 two-state system (qubit) and particle position/momentum is that for the former any pure state is eigenstate of a physical observable (namely the spin measured along a physical axis) while for the latter one can not always assume that.
Other observables have a clear picture. Positions relates to positions in space, momentum to momentum measurements and so on. The main distinction is that the Hilbert space for a single 1/2-spin particle is 2 dimensional.
cianfa72 said:
In a sense a qubit has a physical "element of reality" according to EPR's view (if we know the state the qubit is in then one can predict with 100% certain the counterfactual value of the spin along that axis).
Please clarify, up to now we have not discussed EPR in any way, this is coming out of nowhere. Note that EPR is about entangled particles, and for entangled particles it is not possible to make a measurement on a single particle that will results 100% predicted values (without measuring the second particle).

Edit: note that the original EPR article used position and space, no need of spin.

pines-demon said:
Please clarify, up to now we have not discussed EPR in any way, this is coming out of nowhere. Note that EPR is about entangled particles, and for entangled particles it is not possible to make a measurement on a single particle that will results 100% predicted values (without measuring the second particle).
Yes, there is PF thread I started about EPR view of quantum physics. EPR paper assumes the following definition of "element of reality":

"if, without in any way disturbing a system, we can predict with certainty (i.e. with probability equal to unity) the value of a physical quantity, then there exists an element of physical reality corresponding to this physical quantity."

Hence, if we know that the qubit is in an eigenstate of observable ##\mathbf S_{\hat{\mathbf n}}## , then that physical quantity (the spin value along that axis) is an "element of physical reality" (i.e. one can predict with certainty the outcome that a spin measurement along that axis would have without disturbing it).

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cianfa72 said:
the difference between spin 1/2 two-state system (qubit) and particle position/momentum is that for the former any pure state is eigenstate of a physical observable (namely the spin measured along a physical axis) while for the latter one can not always assume that.
A qubit is actually highly unusual in that every self-adjoint operator on its Hilbert space has an easily understood physical meaning. I believe even going to just the spin-1 Hilbert space makes that no longer true.

cianfa72 said:
In a sense a qubit has a physical "element of reality" according to EPR's view (if we know the state the qubit is in then one can predict with 100% certain the counterfactual value of the spin along that axis).
EPR considered position and momentum to be "elements of reality" in the same way, so they did not see any difference between position/momentum and spin in that respect.

cianfa72
PeterDonis said:
EPR considered position and momentum to be "elements of reality" in the same way, so they did not see any difference between position/momentum and spin in that respect.
You mean any difference between particle's position/momentum and 1/2 spin along two different axes.

cianfa72 said:
You mean any difference between particle's position/momentum and 1/2 spin along two different axes.
Yes.

cianfa72 said:
Yes, there is PF thread I started about EPR view of quantum physics. EPR paper assumes the following definition of "element of reality":

"if, without in any way disturbing a system, we can predict with certainty (i.e. with probability equal to unity) the value of a physical quantity, then there exists an element of physical reality corresponding to this physical quantity."

Hence, if we know that the qubit is in an eigenstate of observable ##\mathbf S_{\hat{\mathbf n}}## , then that physical quantity (the spin value along that axis) is an "element of physical reality" (i.e. one can predict with certainty the outcome that a spin measurement along that axis would have without disturbing it).
For entangled particles the state in the ##z##-axis is something like this
$$|\Psi\rangle=\frac1{\sqrt2}(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle)$$
It is not separable. Thus there is no ##\hat{\mathbf n}## and no single particle operator ##\mathbf S_{\hat{\mathbf n}}## that will give you a 100% assured result. You have to construct a more complicate two particle operator that has no single direction in space. Note that in every axis it is also entangled, in the ##x##-axis the same state looks like:
$$|\Psi\rangle=\frac1{\sqrt2}(|\leftarrow\rightarrow\rangle-|\rightarrow\leftarrow\rangle)$$

cianfa72
We know the quantum system's wave function evolves in time according Schrodinger equation.

In the case of qubit the wave function in ##\{ \ket{\uparrow}, \ket{\downarrow}\}## basis is given by the complex coefficients ##\alpha## and ##\beta## such that for the generic qubit state ##\ket{\psi}## $$\ket{\psi} = \alpha \ket{\uparrow} + \beta \ket{\downarrow}$$ Which is the form of the Schrodinger equation applied to coefficients ##\alpha## and ##\beta## for this specific case ?

cianfa72 said:
Which is the form of the Schrodinger equation applied to coefficients ##\alpha## and ##\beta## for this specific case ?
To formulate the Schrodinger equation, you need to know the Hamiltonian. But the Hamiltonian depends on the specific physical situation. So your question has no single answer.

PeterDonis said:
To formulate the Schrodinger equation, you need to know the Hamiltonian. But the Hamiltonian depends on the specific physical situation.
Could you give an example of Hamiltonian with the corresponding Schrodinger equation for the qubit system? Thanks.

cianfa72 said:
Could you give an example of Hamiltonian with the corresponding Schrodinger equation for the qubit system? Thanks.
Let us write
$$|\uparrow\rangle=\begin{pmatrix}1\\0\end{pmatrix};|\downarrow\rangle=\begin{pmatrix}0\\1\end{pmatrix}$$ then your initial state is generally written as
$$|\psi(0)\rangle=\begin{pmatrix}\alpha\\\beta\end{pmatrix}$$
thus you need to solve Schrödinger's equation
$$i\hbar \frac{\partial }{\partial t}|\psi(t)\rangle=H|\psi(t)\rangle$$.

If your Hamiltonian does not depend on time, this is easy, for example
$$H=\hbar\omega_0\begin{pmatrix}0&1\\1 & 0\end{pmatrix}$$
where ##\omega_0## is a constant. ##H## has eigenvalues ##\pm E_0## and eigenstates ##|\leftrightarrow\rangle=\frac{1}{\sqrt2}(|\uparrow\rangle\pm|\downarrow\rangle)##
then
$$|\psi(t)\rangle=\mathcal{N}\begin{pmatrix}\alpha\cos(\omega_0t)+i\beta \sin (\omega_0 t)\\ i\alpha\sin(\omega_0t)+\beta \cos(\omega_0 t) \end{pmatrix}$$
where ##\mathcal{N}## is a normalization factor that I did not bother to calculate.

Ah ok, so from the eigenvalue equation (time-independent Schrodinger equation) $$H\ket{\psi} = \lambda \ket{\psi}$$ one gets the two eigenvalues ##E_0=\pm \hbar \omega_0## associated to ##\ket{\rightarrow} =\frac{1}{\sqrt2}(\ket{\uparrow} + \ket{\downarrow})## and ##\ket{\leftarrow} =\frac{1}{\sqrt2}(\ket{\uparrow} - \ket{\downarrow})## respectively.

What about last step ? How does one get that expression for the time evolution of ##\ket{\psi(t)}## ?

ps. MathJax supports \ket{ } and \bra{ } for ket and bra respectively.

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cianfa72 said:
Could you give an example of Hamiltonian with the corresponding Schrodinger equation for the qubit system?
There are plenty of examples of this in textbooks. Have you tried looking them up?

pines-demon said:
$$H=\hbar\omega_0\begin{pmatrix}0&1\\1 & 0\end{pmatrix}$$
This operator is the spin-x measurement operator, i.e., the Pauli matrix ##\sigma_x##. You might want to explain to the OP how, and under what physical circumstances, it can be interpreted as a Hamiltonian, since the Hamiltonian operator measures energy, not spin.

cianfa72 said:
Ah ok, so from the eigenvalue equation (time-independent Schrodinger equation) $$H\ket{\psi} = \lambda \ket{\psi}$$ one gets the two eigenvalues ##E_0=\pm \hbar \omega_0## associated to ##\ket{\rightarrow} =\frac{1}{\sqrt2}(\ket{\uparrow} + \ket{\downarrow})## and ##\ket{\leftarrow} =\frac{1}{\sqrt2}(\ket{\uparrow} - \ket{\downarrow})## respectively.

What about last step ? How does one get that expression for the time evolution of ##\ket{\psi(t)}## ?
If you start with an eigenstate of ##H##, ##H|\phi_i\rangle=E_i|\phi_i\rangle## then you can show that the state evolves as
$$|\psi(t)\rangle=e^{iE_i t/\hbar}|\phi_i\rangle$$
So now if you start with a sum of eigenstates ##\sum_j \alpha_j |\phi_j\rangle##, then
$$|\psi(t)\rangle=\sum_j e^{iE_j t/\hbar }\alpha_j |\phi_j\rangle$$
where the ##\alpha_j## are the amplitudes associated to the initial state.

So how to proceed:
1. Write your initial state in the eigenbasis of the Hamiltonian
2. Evolve the state, each eigenstate in the sum evolves with a factor of ##e^{i E_j t/\hbar}##
3. Transform back to the basis that you prefer

Please try to find some resources on introductory quantum mechanics before asking more questions, this is covered in the first chapters.
cianfa72 said:
ps. MathJax supports \ket{ } and \bra{ } for ket and bra respectively.
\ket is not usually available in most of mathjax or latex-like implementation, I have gotten used to write states without it to avoid this problem.

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cianfa72

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