I found this explanation.

"The Hadamard gate can also be expressed as a 90º rotation around the Y-axis, followed by a 180º rotation around the X-axis. So $H=XY^{1/2}H = X Y^{1/2}H=XY^{1/2}$."

Can everything in QM be explained with degrees instead of matrices and vectors?


3 Answers 3


In principle, yes, you can always do it. The Bloch representation can be generalised to arbitrary dimensions, and you can always parametrise states in it by their "angle coordinates".

For example, you can write an arbitrary 3-modes pure state as $$|\psi\rangle=\cos\alpha|0\rangle + e^{i\theta}\sin\alpha\cos\beta|1\rangle+e^{i\phi}\sin\alpha\sin\beta|2\rangle,$$ for $\alpha,\beta,\theta,\phi\in\mathbb R$.

It should be noted, however, that things get much with spaces of dimension larger than $2$. For example, it's harder to interpret arbitrary unitary gates as rotations in this larger space. By this, I mean that even if it is always true that for any given unitary $U$ there is some (and in general an infinity of) Hermitians $H$ such that $e^{-iHt}=U$ at some time $t$, whether these should be considered "rotations" is arguable.

On the one hand, if you represent $H$ as a point in the Bloch representation (you can always do this because the Bloch representation associates a point to any Hermitian matrix, even though this point will in general fall outside of the region representing the set of physical states), then you can think of $H$ as the "axis" of the rotation, as the direction associated with this point will be fixed by the rotation.

On the other hand, in general $t\mapsto e^{-iHt}$ is not a rotation, in the sense that it doesn't "loop back" as rotations do in the Bloch sphere. By this I mean that in general there is no $t>0$ such that $e^{-iHt}=I$, which is what you would expect from a rotation. An easy example of this is: $$H\equiv \begin{pmatrix}\alpha&0&0\\0&\beta&0\\0&0&\gamma\end{pmatrix},\quad\alpha,\beta,\gamma\in\mathbb R.$$ Then, $e^{-iHt}=\mathrm{diag}(e^{-i\alpha t},e^{-i\beta t},e^{-i\gamma t})$, and $e^{-iH t}=I$ if and only if $e^{-i\alpha t}=e^{-i\beta t}=e^{-i\gamma t}=1$, which aren't simultaneously satisfiable for incommensurable $\alpha,\beta,\gamma$.


The state of single qubit can be described as a point on the Bloch sphere. All the allowed transformations of a single qubit can then be described as rotations on the Bloch sphere. Unfortunately, bigger quantum systems can no longer be described as fitting on a sphere like geometry. As a result, this idea of considering transformations as rotations does not hold for all quantum systems.

  • 1
    $\begingroup$ The set of pure states, in arbitrary (finite) dimensions, lives in a projective Hilbert space. As such, isn't it arguably more "natural" to describe everything in terms of generalised angles and rotations (at least as long as one considers only pure states)? $\endgroup$
    – glS
    Commented Oct 3, 2019 at 14:57
  • $\begingroup$ I think it depends on what you want to do. If you want to calculate how close together two pure states are, sure, use the Hilbert space inner product which essentially calculates the angle between them. But when I'm dealing with quantum circuits I just want to calculate the linear maps of the gates involved, and not deal with some complex highly dimentional geometry. $\endgroup$
    – John
    Commented Oct 4, 2019 at 11:33
  • $\begingroup$ See quantumcomputing.stackexchange.com/questions/4699/… $\endgroup$ Commented Oct 4, 2019 at 12:05
  • $\begingroup$ "Hilbert space inner product which essentially calculates the angle between them". Can you explain in simple term what is outer product? $\endgroup$
    – guest
    Commented Oct 5, 2019 at 2:59

Short answer: Yes, except for measurement.

There two postulates in play here:
1- the evolution of a closed QM system can always be described by a unitary matrix.
2- measurement operators (observable operator) are always hermitian

Hermitian operators can't (always) be described as a rotation in any space.

Unitary operators can always be considered as rotation around some axis since they always preserve the inner products. However the dimensions of the space where the rotation happens increases as the dimension of the system increase:

  • operation on 1 qubit is a rotation on the 2D surface of a 3D sphere (Bloch sphere).
  • operation on m qubits is a rotation on the $2^{m+1}$ − 2 dimensions surface of some $2^{m+1}$ − 1 sphere.

Which makes this visualization (degrees of rotation) not very helpful when applied to a multi-qubit system.

  • $\begingroup$ I don't understand this "multi-dimensions". If we have 10 REAL ELECTRONS, how they can make multidimension surface of some sphere? They are 10 real electrons, and we should represent them with 10 bloch spheres. As you can see I'm engineer and think about real objects. $\endgroup$
    – guest
    Commented Oct 4, 2019 at 4:52
  • $\begingroup$ you can represent the state of 10 independent qubits on 10 bloch spheres. However take bell state |00>+|11> There is no two points on 2 bloch spheres that represent this state. If you picked the points |0>+|1> on both of the spheres it would be the the state :|00>+|01>+|10>+|11> which is not the bell state. That is because 2 bloch spheres gives you 4 degrees of freedom, while 2 qubit system has 6 degrees of freedom. You wouldn't be able to represent the "entangled" part of the two qubits system. Back to the main question a unitary U: |00> -> |00>+|11> must be rotation on sphere with 6D surface $\endgroup$ Commented Oct 4, 2019 at 6:27
  • $\begingroup$ Maybe, better way to describe 10 qbits system/circuit is to use one matrix? What are Bra and Ket then? Ket represent columns of that matrix, and Bra represent rows of that 10 qbit matrix? Or I get everything wrong about qbits matrices, Bra, Ket, etc!!! $\endgroup$
    – guest
    Commented Oct 5, 2019 at 3:01
  • $\begingroup$ I'm afraid you've got everything wrong. kets represent the state of a system of qubits. Matrices represent an operation on a given state. You can look at a matrix and see what it does to a state by looking at it's columns. Column i is what happens to the ith bases. so an operation [c1 c2 c3..] will map the states: [1 0 0 .. ]* to c1, [0 1 0 .. ]* to c2.. etc. A state that is in a superposition of states [a b c ..]* will be mapped to ac1+bc2+... etc. For a matrix to be a valid QM operation it needs to be unitary. $\endgroup$ Commented Oct 5, 2019 at 13:33
  • $\begingroup$ Thanks for this explanation! "kets represent the state of a system of qubits". Ket is column vector written in different form, right? That is same? Second question, what represent "bra" (row vector?) and what represent "outer product"? $\endgroup$
    – guest
    Commented Oct 6, 2019 at 0:50

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