Do spin states (for ex: $\langle u| $ & $\langle d| $ and $\langle l| $ & $\langle r| $) along different axes (x-, y-, z- axis) of a quantum object belong to the same Hilbert space (where $\langle u| $ & $\langle d| $ are 2 spin states along z-axis and $\langle l| $ & $\langle r| $ are 2 spin states along y-axis) or spin states along each direction belong to different Hilbert space?

When a spin state, let's say $\langle l| $, is expressed as linear combination of $\langle u| $ & $\langle d| $, is vector $\langle l| $ projected upon the Hilbert space spanned by $\langle u| $ & $\langle d| $? What is the (orthogonality) relationship between $\langle u| $ & $\langle d| $ and $\langle l| $ & $\langle r|$ ?

Thanks a lot

  • $\begingroup$ Note that it's convention to talk about spin states using kets $|s\rangle$, and regard $\langle s |$ as its dual state. $\endgroup$ Commented Mar 21, 2019 at 22:17

3 Answers 3


They all belong to the same Hilbert space, since one is always just a linear combination of the others (i.e. all other vectors can be written as a linear combination of $|u\rangle$ and $|d\rangle$). Linear combinations of elements in a Hilbert space are always also elements of that space.


They all live in the same Hilbert space, and are linear combinations of one another. The important thing to keep in mind is which basis you are working in. If we are in z-spin basis, and use $|+\rangle$ to represent z-spin up, and $|-\rangle$ for z-spin down, then

$$|\pm\rangle_x = \frac{1}{\sqrt{2}}(|+\rangle\pm|-\rangle) \\ |\pm\rangle_y = \frac{1}{\sqrt{2}}(|+\rangle\pm i|-\rangle).$$

One way of deducing these is to find the eigenvectors of the corresponding Pauli spin matrices that have been setup in the spin-z basis. Or, equivalently, with rotations. You can also invert the above expressions:

$$|\pm\rangle = \frac{1}{\sqrt{2}}(|+\rangle_x\pm|-\rangle_x) \\ |+\rangle = \frac{1}{\sqrt{2}}(|+\rangle_y+|-\rangle_y), \:\:\:\: |-\rangle = \frac{1}{i\sqrt{2}}(|+\rangle_y-|-\rangle_y),$$

As far as orthogonality questions go, you can now look at the inner product of any of these states, as they are all represented in the z-basis. For example, let's look at the projection of the spin-x eigenstate $|+\rangle_x$ onto the spin-z eigenstates:

$$\langle +|+\rangle_x = 1/\sqrt{2}, \:\:\: \langle -|+\rangle_x = 1/\sqrt{2},$$

showing that you are just as likely to find this particle with up or down spin when measuring the z-component (take the magnitude squared). In fact, if we let $i,j$ denote any $x,y,z$, all the above expressions show us that

$$|_i\langle\pm|\pm\rangle_j|^2 = 1/2, \:\:\: i\neq j$$

which is just a concise way of stating that you are just as likely to find spin up or down of a definite spin orientation when measured along a different axis. It is when you incorporate multiple particles into the picture that this Hilbert space must be a bit modified. For example, the state of one particle with spin up and the other with spin down would look like

$$|\psi\rangle = |+\rangle\otimes\ |-\rangle\equiv|+\rangle|-\rangle\equiv|\uparrow\downarrow\:\rangle,$$

where I've represented the common ways in which you will encounter this representation.

  • $\begingroup$ Great answer ! But does that mean the eigenvectors along z-axis $$ \psi_{z+}=\left( \begin{array}{l}{1} \\ {0}\end{array}\right), \quad \psi_{z-}=\left( \begin{array}{l}{0} \\ {1}\end{array}\right) $$ are the 2 eigenvectors that span the complex Hilbert state space? And even though $| \pm \rangle_{x}$, $| \pm \rangle_{y}$, and $| \pm \rangle_{z}$ are mutually orthogonal, each state eigenvector set is not orthogonal to the other? $\endgroup$
    – jayjay
    Commented Mar 23, 2019 at 12:29
  • $\begingroup$ Indeed those are two eigenvectors that span the complex Hilbert space, but any of those 6 are linearly independent from the others (not necessarily orthogonal), and will just as well span the Hilbert space. $\endgroup$
    – dsm
    Commented Mar 23, 2019 at 18:07
  • $\begingroup$ Letting $i,j$ denote $x,y,z$, we see that $_i\langle\pm|\pm\rangle_i = 1,\:\:\: _i\langle\mp|\pm\rangle_i = 0,$ and $_i\langle\pm|\pm\rangle_j\neq 0, \:\:\: _i\langle\pm|\mp\rangle_j \neq 0$ for $i\neq j$. It's certainly most convenient to stick with a pair of the same axes, as they are orthonormal, but the relationships stated in the answer give you a fluid way to go between all spin directions. $\endgroup$
    – dsm
    Commented Mar 23, 2019 at 18:14

Let's say you have a particle of spin 1/2. Any spin state in this Hilbert space can be written using an orthonormal basis. The most common basis to use is the $|+z>, |-z>$ basis, the basis of z-component of spin. Every state can then be written as a linear combination of this basis. But this choice of basis is arbitrary. We could also pick $|+x>,|-x>$ as a basis and still represent all the possible states in our Hilbert space.

To answer your question then, if you have a spin state that's aligned on the z-axis and one that's aligned on the y-axis, then yes they do belong to the same Hilbert space, as you can express both state in the same basis. ( The z basis for exemple).

For your second question, $<l|$ is projected onto the Hilbert space spanned by $<u|$ and $<d|$. If those two span the completed space, then we don't really use the word projection. But if you work in a spin-1 space, then 2 eigenstates can't span your entire space, so $<l|$ will always be a vector that's confined to the span set of u and d.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.