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Eigen Value Problem Consider some linear operator acting on a ket |V. It will cause a non V V trivial change to the ket. If you remember the Stern-Gerlach Experiment. There a non-uniform magnetic field say in z-direction caused the vector |S to become |Sz+, |Sz- If you remember once we had applied the magnetic field in the zdirection repeated application of SG apparatus with magnetic field in the same direction did not do anything to my vector except for rescaling it. In general if for an operator we have the following relation then we say that |V is an eigen ket of the the operator In Quantum Mechanics one has to determine the eigenkets and eigenvalues. A liitle later we will try to show you where does physics enter here. Or why calculation of eigenket and eigenvalues is made. However, when we start talking of serious quantum mechanics it will be eigen-eigen all the way… राघव वर्ाा S S Z S Z SZ 2 V V Eigenkets of some familiar operators The only eigenvalue of I is 1. All vectors are its eigen vectors with eigen value 1 What about the projection operator IV V Pv Projection operator associated with normalised ket V Any ket V parallel to V is an eigenket w ith eigenvalue 1 PV V V V V V V V V Any ket V perpendicu lar to V is also an eigenket w ith eigen valu e 0 PV V V V V 0 V Eigenvalue s of R iˆ 2 R1 1 Kets that are neither and of the form V V PV ( V V ) V 1 is also an eigenket w ith eigen valu e 1 We will shortly see that eigenkets can only be determined upto a multiplica tive constant. Any other vect or which is not parallel to 1 will get rotated and hence will not be an eigenket of R राघव वर्ाा The characteristic Equation & the solution to the Eigenvalue problem V V I V 0 Multiplyin g by I on both sides we get 1 V I 1 0 Any finite operator acting on a null vector cannot produce a finite vector. The equation cannot be true. The only way it will not be true is if I 1 does not exist. M 1 Cofactor M T / det M If det M 0 M -1 does not exist Coming back to our earlier equation t hen the inverse does not exist if and only if det - I 0 To determine one starts with the eigen valu e equation - I V 0 This is a vector equation. To solve for we need to choose basis. राघव वर्ाा Solving the eigen value equation Choosing the basis and taking projection along the i axis. I V 0 Taking the projection along i axis i I V 0 Having chosen the axis we can now write V also in terms of its components i I v j j 0 ij ij v j 0 Setting the determinan t equal to zero we get c m 0 characteri stic quation m P n ( ) cm m characteri stic polynomial राघव वर्ाा Solve the eigen value problem of R(/2 i) 1 ˆ R i 0 2 0 Characteri stic 0 0 0 1 1 0 Equation 1- 0 0 det (R - I) 0 - - 1 0 1 - The characteri stic equation is (1 )( 2 1) 0 1, i 1 correspond s to 1 x1 x1 1 0 0 x1 x1 0 0 1 x x x x x x 0 3 2 2 3 2 2 0 1 0 x3 x3 x2 x3 One uses the freedom in scale to normalise to write the 1 eigenket as 0 0 Here please note that one can multiply t his eigenket w ith a number of modulus unity with out changing anything. More on this later! !!!! राघव वर्ाा The other two eigen values of R 1 0 0 x1 x1 ix1 0 0 1 x x ix 1 2 3 2 2 0 1 0 x3 x2 ix3 x1 0 1 i i 2 1 Label the kets with thei r eigen valu es x1 0 i 1 The phenomenon of single eigen valu e representi ng more than one eigenvecto r is called degeneracy and correspond s to repeated roots of the characteri stic polynomial The eigenvalue s of a Hermitian Operator are real - - - - (1) Take the adjoint T - - - - (2) But is Hermitian and therefore T Subtract t he second from first 0 ( ) राघव वर्ाा Of the three simultaneous equation the First is not an equation. In general there Will only be n-1 independent equations Hermitian Operators To every Hermitian operator , there exists (ATLEAST) one basis consisting of orthogonal eigenvectors, It is diagonal in this eigen basis & has eigen values as its diagonal entries. Before we prove that lets prove another theorem If |V = 0 implies |V = 0 then -1 exists Let V1 , V2 Vn be linearly independen t basis in V n Then another basis is generated by the action of i.e. V1 , V2 Vn is also a LI basis To see this lets assume to the contrary t hat is there exists a relation of the form i Vi 0 with not all i 0 i Because is linear i Vi 0 i Because is not zero, hence i Vi 0 with not all i i 0 which is not true. If i Vi are then a set of linearly independen t vectors i then any vector V can be written in terms of them V i Vi राघव वर्ाा i Hermitian Operators (Contd…) In terms of V iVi we see that every V in V i may be written as V V where V is unique Every V therefore arises from a unique V under action of Define an operator whose action on any vector V is to take it back to its unique source V If the source of V was not unique say we had two vectors V1 & V2 that are mapped into V we could not define for acting on V it would not know whether to give V1 V2 . The action of is then V V where V V We may identify as inverse of राघव वर्ाा Operator R राघव वर्ाा