Help to understand the state notation

[KFC]

Hi there,
I am reading an article about the evolution of states. It is said that the system is initially in state

$|\Psi\rangle= a|a\rangle + b|b\rangle$

but after evolution, it becomes $|\Psi'\rangle = a|a\rangle|c\rangle + b|b\rangle|d\rangle$

what is that (double ket) notation ($|\rangle|\rangle$) really means?

 

[bhobba]

That means a composite system - |a>|b> denotes system 1 being in state |a> and system 2 in state |b>.

For a second system to be involved there is obviously more than you detailed.

Thanks
Bill

 

[KFC]

That means a composite system - |a>|b> denotes system 1 being in state |a> and system 2 in state |b>.

For a second system to be involved there is obviously more than you detailed.

Thanks
Bill

Thanks a lot. The notation in QM is pretty confusing here. Let's say I have a system H, which is in the initial states$a|a\rangle + b|b\rangle$, the article said it turned on the magnetic field so it becomes $a|a\rangle|c\rangle + b|b\rangle|d\rangle$. But before and after the magnetic field was turned on, I am still dealing with the same system, so why you mean system 1 and system 2 (sorry for the silly question but it is confusing to me ). Also, what's the connection of this double ket notation and the so-called tensor product state?

 

[atyy]

The double ket is the tensor product.

But I don't understand why there is only a single ket before you turn on the magnetic field. Turning on the magnetic field cannot change the dimensionality of the Hilbert space.

 

[KFC]

The double ket is the tensor product.

But I don't understand why there is only a single ket before you turn on the magnetic field. Turning on the magnetic field cannot change the dimensionality of the Hilbert space.

Thanks for clarification. I don't know understand the statement either. It is a personal note and might be wrong. Now I following the reference and reading the other paper about the spin state and measurement in position state. It organizes like this

It said that the system is initially prepared in the spin state $|\Psi_0\rangle = a|x_0\rangle|\text{up}\rangle + b|x_0\rangle|\text{down}\rangle$ where $x_0$ is where the center of the wavepacket locates. Honestly, I don't understand what does it mean. My guessing is the wavefunction is initially prepared around $x_0$ in position space but with the particles in the system has two possible orientation in spin: up and down?

And it applies an unitary operator $U = \exp(-iS_zP\ell/\hbar)$ on $|\Psi_0\rangle$ and it becomes

$|\Psi_1\rangle = a|x_0-\ell\rangle|\text{up}\rangle + b|x_0+\ell\rangle|\text{down}\rangle$

So what is $U = \exp(-iS_zP\ell/\hbar)$ really means in term of experiment or actual physical process?

 

[atyy]

Yes, it means the particle is prepared with a definite initial position, but it is in a superposition of up and down states.

I think the operator $U$ represents the magnetic field in a Stern-Gerlach experiment, but am not sure. Does the book say anything about a Stern-Gerlach experiment?

 

[Simon Bridge]

http://en.wikipedia.org/wiki/Bra–ket_notation#Composite_bras_and_kets
|a>|c> means the same as |a,c>

Per your example post #5
The particle is described in terms of it's position and spin.

The result of the operator on the state is to separate the positions of the two spin components.
The notation is a common abuse of notation for brevity - probably why it is a bit confusing.
"x0-l" is a label for the new state telling you that the initial position wavefunction is shifted in the +x direction by distance l.

Notice how the before and after state vectors have two kets in each term?
Compare with post #1 where the before vector has only one in each term.

 

[KFC]

Yes, it means the particle is prepared with a definite initial position, but it is in a superposition of up and down states.

I think the operator $U$ represents the magnetic field in a Stern-Gerlach experiment, but am not sure. Does the book say anything about a Stern-Gerlach experiment?

Yes, it mentions Stern-Gerlach experiment. But I am still confusing if that's the way how they measure the spin or not.

 

[atyy]

Yes, it mentions Stern-Gerlach experiment. But I am still confusing if that's the way how they measure the spin or not.

The idea of $U |x_0\rangle|\text{up}\rangle = |x_0-\ell\rangle|\text{up}\rangle$ is that the magnetic field acting on an initial down state will cause it to move from $x_{0}$ to $x_0-\ell$ .

Similarly $U |x_0\rangle|\text{down}\rangle = |x_0+\ell\rangle|\text{down}\rangle$ means that the magnetic field acting on an initial down state will cause it to move from $x_{0}$ to $x_0+\ell$.

The time evolution operator $U$ is linear, so it will act linearly on an initial state that is a superposition $a|x_0\rangle|\text{up}\rangle + b|x_0\rangle|\text{down}\rangle$.

 

[KFC]

http://en.wikipedia.org/wiki/Bra–ket_notation#Composite_bras_and_kets
|a>|c> means the same as |a,c>

Per your example post #5
The particle is described in terms of it's position and spin.

The result of the operator on the state is to separate the positions of the two spin components.
The notation is a common abuse of notation for brevity - probably why it is a bit confusing.
"x0-l" is a label for the new state telling you that the initial position wavefunction is shifted in the +x direction by distance l.

Notice how the before and after state vectors have two kets in each term?
Compare with post #1 where the before vector has only one in each term.

I appreciate the link, it clarifies some of my doubts though the math is still too vague to me. I didn't receive official training in quantum mechanics but all math I learn from QM is linear algebra in terms of matrices theory, that's why it is hard for me to make a turn to bra-ket notation.

From your first response "The result of the operator on the state is to separate the positions of the two spin components.", does it mean that by applying $U=\exp(-iS_zP\ell/\hbar)$, it actually means the process to measure the wavepacket in position and then measure the spin (e.g. by magnetic field)?

Ok, here is the most confusing part. Why spin states and position states must be defined in different spaces? Because of the physical property of spin and position are very different or other mathematical concern? So if I have a particle, it has spin and it also has momentum and position, to describe the particle in QM, should I write something like this $|\Psi\rangle = |\text{spin}, x, p\rangle$ or just $|\Psi\rangle = |\text{spin}, x\rangle$? I saw few books, they either put position or momentum in the wavefunction but not both, is it because position and momentum are related by Fourier transformation so knowing one of them is knowing the other?

 

[KFC]

The idea of $U |x_0\rangle|\text{up}\rangle = |x_0-\ell\rangle|\text{up}\rangle$ is that the magnetic field acting on an initial down state will cause it to move from $x_{0}$ to $x_0-\ell$ .

Similarly $U |x_0\rangle|\text{down}\rangle = |x_0+\ell\rangle|\text{down}\rangle$ means that the magnetic field acting on an initial down state will cause it to move from $x_{0}$ to $x_0+\ell$.

The time evolution operator $U$ is linear, so it will act linearly on an initial state that is a superposition $a|x_0\rangle|\text{up}\rangle + b|x_0\rangle|\text{down}\rangle$.

wow, that's pretty interesting. So does it mean by measuring the spin, the position will be shifted? I am still stuck in my mind how does this happens in physical point of view. But I learn something here ;)

 

[atyy]

wow, that's pretty interesting. So does it mean by measuring the spin, the position will be shifted? I am still stuck in my mind how does this happens in physical point of view. But I learn something here ;)

The very rough idea is spin is basically a tiny magnet. When you put a magnet in a magnetic field, the magnet will move. So to measure spin, we put the particle in a magnetic field, and depending on its spin, the particle will move one way or the other. From the way it moves, we can infer what spin it has.

 

[Simon Bridge]

The operator has been presented in terms of position coordinate and spin operators, so it is useful to represent the state vector in terms of eigenvectors of position and spin: it makes the maths easier. The state vector can be represented any way you like.

In your example, a measurement of spin leads to a shift in position because that is what the magnetic field in question does: it detects which way the particle magnetic dipole is oriented and then applies a force accordingly. Read "Stern-Gerlach experiment".

It is acceptable to write out a state ket with of any number of independent terms. For instance, the hydrogen state is often represented with 4 quantum numbers |n,l,m,s>, two particles may have two sets i.e. |x1,x2> indicating the positions of particle 1 and particle 2. But I don't think position and momentum can be used together like that.

https://www.physicsforums.com/threads/bras-and-kets-and-tensors.241586/ may help, but I think you should review "vector spaces" a bit more.

 

[KFC]

One more question about spin space. In many reference, the state of a spin defines in this way

$|\Psi_s\rangle = a|\text{up}\rangle \otimes |\psi(\text{up})\rangle + b|\text{down}\rangle \otimes |\psi(\text{down})\rangle$

It is said that $|\text{up}\rangle$ and $|\text{down}\rangle$ are eigenstates of operator $\hat{S}_z$ corresponding to the z component of the spin.

The definition for $|\Psi_s\rangle$ is pretty confusing, why we need to $|\text{up}\rangle$ and $|\text{down}\rangle$ as well as $|\psi(\text{up})\rangle$ and $|\psi(\text{down})\rangle$, why don't we write everything onto one space?

$|\psi(\text{up})\rangle$ and $|\psi(\text{down})\rangle$ is the component of wavefunction of particle in the spin-up space and spin-down space separately.

Also, from above statement, it seems that spin-up is one space and spin-down is another one. I always thinks, spin-up and down are defined in one space call spin space.

Honestly, I am learning most of the facts from popular science readings. All in my mind about spin is only up and down (for spin 1/2 system). But here, it defines the spin state in such way and gives me a sense that maybe the so-called up and down is only true while it is lining up with z direction of the magnetic field and in the very general case, it is 3-dimensional but not just up and down? So to write down the most general wavefunction for spin, we use the z component (up and down) as bases and expanded the real wavefunction onto those bases, is my understanding correct?

If it is correct, can I understand the wavefunction $|\Psi_s\rangle = a|\text{up}\rangle \otimes |\psi(\text{up})\rangle + b|\text{down}\rangle \otimes |\psi(\text{down})\rangle$ this way?

We know that each spin (1/2) along z direction has two values: up and down, so the state is $|\text{up}\rangle$ and $|\text{down}\rangle$, for general case, we project the wavefunction onto spin up and down space so to have $|\psi(\text{up})\rangle$ and $|\psi(\text{down})\rangle$. So if everything could be expanded in terms of up and down state, why don't we just write

$|\Psi_s\rangle = a|\text{up} |\psi(\text{up})\rangle + b |\psi(\text{down})\rangle$

why we needs the tensor product?

 

[Simon Bridge]

$| \psi_\uparrow\rangle$ is the position ket associated with the $|\uparrow\rangle$ ket ... there are lots of different ways of writing the labels for kets.

spin up and spin down are labels for single spin states. These are the projection of the particles spin-angular momentum vector onto the z-axis.

classically the particle spin can be in any direction - the z-direction is imposed on space by the magnetic field.
The particle would experience a force in proportion to it's orientation to the z direction. Since this could be anything, the classical prediction is that particles would be displaced in a continuous range between l and - this is your intuition too right?
However - that is not the case in Nature. The SG experiment demonstrated that spin is quantized - turning on the magnetic field sets the spins to one of two possible orientations and so you get one or the other displacement. The equations are how standard quantum mechanics describes this but does not show you why that form of operator is appropriate.

See: http://hyperphysics.phy-astr.gsu.edu/hbase/spin.html
(scroll down to the experiment description)

You need the tensor product because it is easier to work position and spin when they are separated out. Much the same way as it is easier in kinematics to plot the position of the center of mass of an object and it's orientation wrt the center of mass.

Note: You cannot learn physics from popular science readings. Use physics textbooks.

 

[KFC]

$| \psi_\uparrow\rangle$ is the position ket associated with the $|\uparrow\rangle$ ket ... there are lots of different ways of writing the labels for kets.

spin up and spin down are labels for single spin states. These are the projection of the particles spin-angular momentum vector onto the z-axis.

classically the particle spin can be in any direction - the z-direction is imposed on space by the magnetic field.
The particle would experience a force in proportion to it's orientation to the z direction. Since this could be anything, the classical prediction is that particles would be displaced in a continuous range between l and - this is your intuition too right?
However - that is not the case in Nature. The SG experiment demonstrated that spin is quantized - turning on the magnetic field sets the spins to one of two possible orientations and so you get one or the other displacement. The equations are how standard quantum mechanics describes this but does not show you why that form of operator is appropriate.

See: http://hyperphysics.phy-astr.gsu.edu/hbase/spin.html
(scroll down to the experiment description)

You need the tensor product because it is easier to work position and spin when they are separated out. Much the same way as it is easier in kinematics to plot the position of the center of mass of an object and it's orientation wrt the center of mass.

Note: You cannot learn physics from popular science readings. Use physics textbooks.

Thanks a lot for all those comments and explanation. I will try to read more information from formal books to learn deeper.

 

[dextercioby]

Hi there,
I am reading an article about the evolution of states. It is said that the system is initially in state

$|\Psi\rangle= a|a\rangle + b|b\rangle$

but after evolution, it becomes $|\Psi'\rangle = a|a\rangle|c\rangle + b|b\rangle|d\rangle$

what is that (double ket) notation ($|\rangle|\rangle$) really means?

Maybe there's some poor notation involved:

$$|\Psi\rangle= a|a\rangle + b|b\rangle$$

should have been written

$$\Psi\rangle= a|a\rangle|1\rangle + b|1\rangle|b\rangle$$

 

[KFC]

Maybe there's some poor notation involved:

$$|\Psi\rangle= a|a\rangle + b|b\rangle$$

should have been written

$$\Psi\rangle= a|a\rangle|1\rangle + b|1\rangle|b\rangle$$

Thanks. I still have hard time to understand the tensor product. I found some introduction material but they use pretty vague mathematics. Is that any materials explaining the tensor product in terms of matrices?

 

[bhobba]

Thanks. I still have hard time to understand the tensor product. I found some introduction material but they use pretty vague mathematics. Is that any materials explaining the tensor product in terms of matrices?

Its not hard.

Given a vector space with basis |bi> and another vector space with basis |dj> one can consider a new space with basis |bi,dj> called its tensor product. Its easier to see for finite dimensional spaces. If i has values 1 to n and j has values 1 to m you can consider |b1>, |b2>, ... |bn>,
|d1>, |d2>, ... |dm> as the basis of another vector space of dimension n+m.

Thanks
Bill

 

[KFC]

Hi all,
I borrow a text from the library to learn some basic idea in QM and the related math. I read the first two chapters about the linear space, ket, bra vector representation, etc. I also read a thin book for QM very beginner. There it mentions one statement, a state vector in Hilbert space in ket notation is written as $|\psi\rangle$, if we want to find the wavepacket in position space, we could find it by dot product $\langle x|\psi\rangle$. For the state vector $|\psi\rangle$ from which we find the wavefunction in position space, any specific requirement? I mean if I use the state vector in momentum space what does this wavefunction $\langle x|\psi_p\rangle$ tell me?

 

[bhobba]

I mean if I use the state vector in momentum space what does this wavefunction $\langle x|\psi_p\rangle$ tell me?

In $\langle x|\psi_p\rangle$ $|\psi_p\rangle$ is a state of definite momentum. $\langle x|\psi_p\rangle$ is it's representation in terms of the position basis.

If that doesn't make sense you need to have a look at a good textbook.

I recommend the first two chapters of Ballentine:
https://www.amazon.com/dp/9810241054/?tag=pfamazon01-20

Thanks
Bill

 

[wle]

I mean if I use the state vector in momentum space what does this wavefunction $\langle x|\psi_p\rangle$ tell me?

There isn't a distinction between a state vector in momentum space and a state vector in position space. They're just different ways of writing the same state. You can expand any state vector $\lvert \psi \rangle$ in the position basis in terms of its wavefunction: $$\lvert \psi \rangle = \int \mathrm{d}x \, \psi(x) \lvert x \rangle \,.$$ In units where $\hbar = 1$, momentum states are related to position states by $$\lvert p \rangle = \frac{1}{\sqrt{2 \pi}} \int \mathrm{d}p \, e^{i p x} \lvert x \rangle$$ (which is just saying that a momentum state is a de Broglie wave $\psi_{p}(x) \propto e^{ipx}$ in the position basis) and the inverse relation $$\lvert x \rangle = \frac{1}{\sqrt{2 \pi}} \int \mathrm{d}p \, e^{-ipx} \lvert p \rangle \,.$$ Insert the inverse relation into the state above and you get a Fourier transform of the wavefunction: $$\begin{eqnarray}
\lvert \psi \rangle &=& \int \mathrm{d}x \, \psi(x) \frac{1}{\sqrt{2\pi}} \int \mathrm{d}p \, e^{-ipx} \lvert p \rangle \\
&=& \int \mathrm{d}p \biggl[ \frac{1}{\sqrt{2 \pi}} \int \mathrm{d} x \, \psi(x) e^{-ipx} \biggr] \lvert p \rangle \\
&=& \int \mathrm{d}p \, \tilde{\psi}(p) \lvert p \rangle \,.
\end{eqnarray}$$ This is still the same state vector as before, just written in terms of the momentum states instead of the position states.

From the expression for $\lvert p \rangle$ above, you can work out what the dot product $\langle x \vert p \rangle$ is: $$\begin{eqnarray}
\langle x \vert p \rangle &=& \langle x \rvert \frac{1}{\sqrt{2 \pi}} \int \mathrm{d}x' \, e^{ipx'} \lvert x' \rangle \\
&=& \frac{1}{\sqrt{2 \pi}} \int \mathrm{d}x' \, e^{ipx'} \langle x \vert x' \rangle \\
&=&\frac{1}{\sqrt{2 \pi}} \int \mathrm{d}x' \, e^{ipx'} \delta(x - x') \\
&=& \frac{1}{\sqrt{2 \pi}} e^{ipx} \,.
\end{eqnarray}$$ So if you have a state expressed in the momentum basis, there's no problem working out its scalar product with a position state: $$\begin{eqnarray}
\langle x \vert \psi \rangle &=& \langle x \rvert \int \mathrm{d}p \, \tilde{\psi}(p) \lvert p \rangle \\
&=& \int \mathrm{d}p \, \tilde{\psi}(p) \langle x \vert p \rangle \\
&=& \frac{1}{\sqrt{2 \pi}} \int \mathrm{d}p \, \tilde{\psi}(p) e^{ipx} \,,
\end{eqnarray}$$ which is just the inverse Fourier transform of the wavefunction $\tilde{\psi}(p)$ in momentum space.

 

[KFC]

There isn't a distinction between a state vector in momentum space and a state vector in position space. They're just different ways of writing the same state. You can expand any state vector $\lvert \psi \rangle$ in the position basis in terms of its wavefunction: $$\lvert \psi \rangle = \int \mathrm{d}x \, \psi(x) \lvert x \rangle \,.$$ In units where $\hbar = 1$, momentum states are related to position states by $$\lvert p \rangle = \frac{1}{\sqrt{2 \pi}} \int \mathrm{d}p \, e^{i p x} \lvert x \rangle$$ (which is just saying that a momentum state is a de Broglie wave $\psi_{p}(x) \propto e^{ipx}$ in the position basis) and the inverse relation $$\lvert x \rangle = \frac{1}{\sqrt{2 \pi}} \int \mathrm{d}p \, e^{-ipx} \lvert p \rangle \,.$$ Insert the inverse relation into the state above and you get a Fourier transform of the wavefunction: $$\begin{eqnarray}
\lvert \psi \rangle &=& \int \mathrm{d}x \, \psi(x) \frac{1}{\sqrt{2\pi}} \int \mathrm{d}p \, e^{-ipx} \lvert p \rangle \\
&=& \int \mathrm{d}p \biggl[ \frac{1}{\sqrt{2 \pi}} \int \mathrm{d} x \, \psi(x) e^{-ipx} \biggr] \lvert p \rangle \\
&=& \int \mathrm{d}p \, \tilde{\psi}(p) \lvert p \rangle \,.
\end{eqnarray}$$ This is still the same state vector as before, just written in terms of the momentum states instead of the position states.

From the expression for $\lvert p \rangle$ above, you can work out what the dot product $\langle x \vert p \rangle$ is: $$\begin{eqnarray}
\langle x \vert p \rangle &=& \langle x \rvert \frac{1}{\sqrt{2 \pi}} \int \mathrm{d}x' \, e^{ipx'} \lvert x' \rangle \\
&=& \frac{1}{\sqrt{2 \pi}} \int \mathrm{d}x' \, e^{ipx'} \langle x \vert x' \rangle \\
&=&\frac{1}{\sqrt{2 \pi}} \int \mathrm{d}x' \, e^{ipx'} \delta(x - x') \\
&=& \frac{1}{\sqrt{2 \pi}} e^{ipx} \,.
\end{eqnarray}$$ So if you have a state expressed in the momentum basis, there's no problem working out its scalar product with a position state: $$\begin{eqnarray}
\langle x \vert \psi \rangle &=& \langle x \rvert \int \mathrm{d}p \, \tilde{\psi}(p) \lvert p \rangle \\
&=& \int \mathrm{d}p \, \tilde{\psi}(p) \langle x \vert p \rangle \\
&=& \frac{1}{\sqrt{2 \pi}} \int \mathrm{d}p \, \tilde{\psi}(p) e^{ipx} \,,
\end{eqnarray}$$ which is just the inverse Fourier transform of the wavefunction $\tilde{\psi}(p)$ in momentum space.

Thanks a lot. I read and go over all your maths, I learn something.

In the book I read, it gives me an example about the relation between spatial wave function and the counterpart in momentum space for free particle. Just like what you presented, they are related by Fourier transformation. But is this always true for any particle in any environment? Because it use free particle as example, I wonder if it is still true to derive a momentum wave function by Fourier transformation on the given spatial wave function even it is in a potential well?

One more question on your statement " They're just different ways of writing the same state. You can expand any state vector $\lvert \psi \rangle$ in the position basis in terms of its wavefunction ...". The state vectors in Hilbert space is still confusing me. I saw the statement often in the text that we write the position state vector in the spin-1/2 space and something like that. In my mind, I always use the Cartesian coordinates as example to understand this. In 3D Cartesian, we could use any 3 vectors such that any other vectors could be written in terms of those 3 vectors. So it is not hard to understand your statement, we could expand a state vector in some basis ... For momentum and position, they are physically associated by Fourier transformation, that's easy to understand. But for spin and position, what's the physics to connect them so I can write one in the other basis? Also, I have a doubt that for spin1/2 along z direction, it only has two orientation so the state vector has 2 components, but position state should be 3D and has 3 components, how can we write a 3D state vector in terms of 2D one?

 

[wle]

In the book I read, it gives me an example about the relation between spatial wave function and the counterpart in momentum space for free particle. Just like what you presented, they are related by Fourier transformation. But is this always true for any particle in any environment? Because it use free particle as example, I wonder if it is still true to derive a momentum wave function by Fourier transformation on the given spatial wave function even it is in a potential well?

Yes. Their relation as Fourier transformations of each other just follows from the expression $\lvert p \rangle = \int \mathrm{d} x \, e^{ipx} \lvert x \rangle$. This is just a statement that a momentum state corresponds to a de Broglie wave in position space. It's practically a postulate of quantum mechanics. The operator commutation relation $[X, P] = i \hbar$ amounts to saying the same thing.

One more question on your statement " They're just different ways of writing the same state. You can expand any state vector $\lvert \psi \rangle$ in the position basis in terms of its wavefunction ...". The state vectors in Hilbert space is still confusing me. I saw the statement often in the text that we write the position state vector in the spin-1/2 space and something like that. In my mind, I always use the Cartesian coordinates as example to understand this. In 3D Cartesian, we could use any 3 vectors such that any other vectors could be written in terms of those 3 vectors. So it is not hard to understand your statement, we could expand a state vector in some basis ... For momentum and position, they are physically associated by Fourier transformation, that's easy to understand. But for spin and position, what's the physics to connect them so I can write one in the other basis? Also, I have a doubt that for spin1/2 along z direction, it only has two orientation so the state vector has 2 components, but position state should be 3D and has 3 components, how can we write a 3D state vector in terms of 2D one?

Spin and position/momentum are just different degrees of freedom and aren't directly related to each other in quantum mechanics the way position and momentum are to each other. A spin 1/2 particle simply has a position/momentum state and a spin state, so its full state vector in 3D space might be written as something like $$\lvert \Psi \rangle = \int \mathrm{d}x \int \mathrm{d}y \int \mathrm{d}z \sum_{s \in \{\uparrow,\, \downarrow\}} \Psi(x, y, z, s) \lvert x \rangle \lvert y \rangle \lvert z \rangle \lvert s \rangle \,,$$
with a wavefunction $\Psi(x, y, z, s)$ that depends on the $x$, $y$, and $z$ coordinates and on the spin state $s$ which can be $s = \uparrow$ or $s = \downarrow$.

If you just concentrate on the spin 1/2 states, by the way, different spin projections are also related to each other similar to how position and momentum states are related to each other. For example, if you call the $z$-projection spin states $\lvert \uparrow \rangle$ and $\lvert \downarrow \rangle$ ("spin up" and "spin down"), and you call the $x$-projection spin states $\lvert \rightarrow \rangle$ and $\lvert \leftarrow \rangle$ ("spin right" and "spin left"), then they're related by $$\begin{eqnarray}
\lvert \rightarrow \rangle &=& \frac{1}{\sqrt{2}} \bigl( \lvert \uparrow \rangle + \lvert \downarrow \rangle \bigr) \,, \\
\lvert \leftarrow \rangle &=& \frac{1}{\sqrt{2}} \bigl( \lvert \uparrow \rangle - \lvert \downarrow \rangle \bigr)
\end{eqnarray}$$
(where I'm imagining that "up" is the $+z$ direction and "right" is the $+x$ direction).