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| {{Other uses|Rotation operator (disambiguation)}}
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| {{Quantum mechanics}}
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| This article concerns the '''[[rotation]] [[Operator (physics)|operator]]''', as it appears in [[quantum mechanics]].
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| ==Quantum mechanical rotations==
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| With every physical rotation R, we postulate a quantum mechanical rotation operator D(R) which rotates quantum mechanical states.
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| :<math>| \alpha \rangle_R = D(R) |\alpha \rangle</math>
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| In terms of the generators of rotation,
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| :<math>D (\mathbf{\hat n},\phi) = \exp \left( -i \phi \frac{\mathbf{\hat n} \cdot \mathbf J }{ \hbar} \right)</math>
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| Please see [http://students.washington.edu/tkarin/rotations.pdf "rotations in quantum mechanics"] for details.
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| ==The translation operator==
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| The [[rotation]] [[Operator (physics)|operator]] <math>\,\mbox{R}(z, \theta)</math>, with the first argument <math>\,z</math> indicating the rotation [[Cartesian coordinate system|axis]] and the second <math>\,\theta</math> the rotation angle, can operate through the [[Displacement operator|translation operator]] <math>\,\mbox{T}(a)</math> for infinitesimal rotations as explained below. This is why, it is first shown how the translation operator is acting on a particle at position x (the particle is then in the [[quantum state|state]] <math>|x\rangle</math> according to [[Quantum Mechanics]]).
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| Translation of the particle at position x to position x+a: <math>\mbox{T}(a)|x\rangle = |x + a\rangle</math>
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| Because a translation of 0 does not change the position of the particle, we have (with 1 meaning the [[identity function|identity operator]], which does nothing):
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| :<math>\,\mbox{T}(0) = 1</math>
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| :<math>\,\mbox{T}(a) \mbox{T}(da)|x\rangle = \mbox{T}(a)|x + da\rangle = |x + a + da\rangle = \mbox{T}(a + da)|x\rangle \Rightarrow</math>
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| :<math>\,\mbox{T}(a) \mbox{T}(da) = \mbox{T}(a + da)</math>
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| [[Taylor series|Taylor]] development gives:
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| :<math>\,\mbox{T}(da) = \mbox{T}(0) + \frac{d\mbox{T}(0)}{da} da + ... = 1 - \frac{i}{h}\ p_x\ da</math>
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| with
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| :<math>\,p_x = i h \frac{d\mbox{T}(0)}{da}</math>
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| From that follows:
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| :<math>\,\mbox{T}(a + da) = \mbox{T}(a) \mbox{T}(da) = \mbox{T}(a)\left(1 - \frac{i}{h} p_x da\right) \Rightarrow</math>
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| :<math>\,[\mbox{T}(a + da) - \mbox{T}(a)]/da = \frac{d\mbox{T}}{da} = - \frac{i}{h} p_x \mbox{T}(a)</math>
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| This is a [[differential equation]] with the solution <math>\,\mbox{T}(a) = \mbox{exp}\left(- \frac{i}{h} p_x a\right)</math>.
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| Additionally, suppose a [[Hamilton's equations|Hamiltonian]] <math>\,H</math> is independent of the <math>\,x</math> position. Because the translation operator can be written in terms of <math>\,p_x</math>, and <math>\,[p_x,H]=0</math>, we know that <math>\,[H,\mbox{T}(a)]=0</math>. This result means that linear [[momentum]] for the system is conserved.
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| ==In relation to the orbital angular momentum==
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| Classically we have for the [[angular momentum]] <math>\,l = r \times p</math>. This is the same in [[quantum mechanics]] considering <math>\,r</math> and <math>\,p</math> as operators. Classically, an infinitesimal rotation <math>\,dt</math> of the vector r=(x,y,z) about the z-axis to r'=(x',y',z) leaving z unchanged can be expressed by the following infinitesimal translations (using [[Taylor series|Taylor approximation]]):
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| :<math>\,x' = r \cos(t + dt) = x - y dt + ...</math>
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| :<math>\,y' = r \sin(t + dt) = y + x dt + ...</math>
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| From that follows for states:
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| :<math>\,\mbox{R}(z, dt)|r\rangle</math><math>= \mbox{R}(z, dt)|x, y, z\rangle</math><math>= |x - y dt, y + x dt, z\rangle</math><math>= \mbox{T}_x(-y dt) \mbox{T}_y(x dt)|x, y, z\rangle</math><math>= \mbox{T}_x(-y dt) \mbox{T}_y(x dt)|r\rangle</math>
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| And consequently:
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| :<math>\,\mbox{R}(z, dt) = \mbox{T}_x (-y dt) \mbox{T}_y(x dt)</math> | |
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| Using <math>\,T_k(a) = \exp\left(- \frac{i}{h}\ p_k\ a\right)</math> from above with <math>\,k = x,y</math> and Taylor development we get:
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| :<math>\,\mbox{R}(z, dt) = \exp\left[- \frac{i}{h}\ (x p_y - y p_x) dt\right]</math><math>= \exp\left(- \frac{i}{h}\ l_z dt\right) = 1 - \frac{i}{h} l_z dt + ...</math>
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| with l<sub>z</sub> = x p<sub>y</sub> - y p<sub>x</sub> the z-component of the angular momentum according to the classical [[cross product]].
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| To get a rotation for the angle <math>\,t</math>, we construct the following differential equation using the condition <math>\mbox{R}(z, 0) = 1</math>:
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| :<math>\,\mbox{R}(z, t + dt) = \mbox{R}(z, t) \mbox{R}(z, dt) \Rightarrow</math>
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| :<math>\,[\mbox{R}(z, t + dt) - \mbox{R}(z, t)]/dt = d\mbox{R}/dt</math><math>\,= \mbox{R}(z, t) [\mbox{R}(z, dt) - 1]/dt</math><math>\,= - \frac{i}{h} l_z \mbox{R}(z, t) \Rightarrow</math>
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| :<math>\,\mbox{R}(z, t) = \exp\left(- \frac{i}{h}\ t\ l_z\right)</math>
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| Similar to the translation operator, if we are given a Hamiltonian <math>\,H</math> which rotationally symmetric about the z axis, <math>\,[l_z,H]=0</math> implies <math>\,[\mbox{R}(z,t),H]=0</math>. This result means that angular momentum is conserved.
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| For the spin angular momentum about the y-axis we just replace <math>\,l_z</math> with <math>\,S_y = \frac{h}{2} \sigma_y</math> and we get the [[spin (physics)|spin]] rotation operator <math>\,\mbox{D}(y, t) = \exp\left(- i \frac{t}{2} \sigma_y\right)</math>.
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| ==Effect on the spin operator and quantum states==
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| {{Main|Spin (physics)#Spin and rotations}}
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| Operators can be represented by [[Matrix (mathematics)|matrices]]. From [[linear algebra]] one knows that a certain matrix <math>\,A</math> can be represented in another [[basis (linear algebra)|basis]] through the transformation
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| :<math>\,A' = P A P^{-1}</math>
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| where <math>\,P</math> is the basis transformation matrix. If the vectors <math>\,b</math> respectively <math>\,c</math> are the z-axis in one basis respectively another, they are perpendicular to the y-axis with a certain angle <math>\,t</math> between them. The spin operator <math>\,S_b</math> in the first basis can then be transformed into the spin operator <math>\,S_c</math> of the other basis through the following transformation:
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| :<math>\,S_c = \mbox{D}(y, t) S_b \mbox{D}^{-1}(y, t)</math>
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| From standard quantum mechanics we have the known results <math>\,S_b |b+\rangle = \frac{\hbar}{2} |b+\rangle</math> and <math>\,S_c |c+\rangle = \frac{\hbar}{2} |c+\rangle</math> where <math>\,|b+\rangle</math> and <math>\,|c+\rangle</math> are the top spins in their corresponding bases. So we have:
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| :<math>\,\frac{\hbar}{2} |c+\rangle = S_c |c+\rangle = \mbox{D}(y, t) S_b \mbox{D}^{-1}(y, t) |c+\rangle \Rightarrow</math>
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| :<math>\,S_b \mbox{D}^{-1}(y, t) |c+\rangle = \frac{\hbar}{2} \mbox{D}^{-1}(y, t) |c+\rangle</math> | |
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| Comparison with <math>\,S_b |b+\rangle = \frac{\hbar}{2} |b+\rangle</math> yields <math>\,|b+\rangle = D^{-1}(y, t) |c+\rangle</math>.
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| This means that if the state <math>\,|c+\rangle</math> is rotated about the y-axis by an angle <math>\,t</math>, it becomes the state <math>\,|b+\rangle</math>, a result that can be generalized to arbitrary axes. It is important, for instance, in [[Sakurai's Bell inequality]].
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| ==See also==
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| *[[Symmetry in quantum mechanics]]
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| *[[Spherical basis]]
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| ==References==
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| *L.D. Landau and E.M. Lifshitz: ''Quantum Mechanics: Non-Relativistic Theory'', Pergamon Press, 1985
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| *P.A.M. Dirac: ''The Principles of Quantum Mechanics'', Oxford University Press, 1958
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| *R.P. Feynman, R.B. Leighton and M. Sands: ''The Feynman Lectures on Physics'', Addison-Wesley, 1965
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| *Karin, Todd. [http://students.washington.edu/tkarin/rotations.pdf ''Rotations in Quantum Mechanics.''] Unpublished Work.
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| ==See also==
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| *[[Optical phase space]]
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| {{Physics operator}}
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| {{DEFAULTSORT:Rotation Operator (Quantum Mechanics)}}
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| [[Category:Rotational symmetry]]
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| [[Category:Quantum mechanics]]
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