 Welcome to the lectures on the principles of quantum mechanics. This is a course offered in the chemistry department of the Indian Institute of Technology, Madras for the beginning masters of science chemistry students. And I would be recording the lectures that I intend to give to these students and also make them available through this particular project. This project through which you are getting the lectures online is known as the National Program on Technology Enhanced Learning. It is abbreviated as NPTEL and my name is Mangalasinder Krishnan. I am a professor of chemistry in the department of chemistry at IIT Madras. And you can see my credentials and my address as well as the site through which these lectures would be made available. So, I request you to take note of this the site where you would find all the supplementary materials is the has the URL NPTEL.IATM.AC.IN and for any email contact regarding the course contents or regarding the assignments and solutions etcetera. Please refer to the website and if you do not find the contents in the website you are most welcome to contact me. And my email addresses are given here the both my institute affiliation and the institute email Mangal at IITM.AC.IN and my personal account which is at gmail.com. This is a very long course in the Indian Institute of Technology this is a four credit course, but what I intend to do is not to have the lectures recorded for with my students, but rather have them recorded for the purpose of not only our students, but also students anywhere else who are at a similar level who have finished their undergraduate chemistry or typically in a North American setting or in an European setting. They are in the final years of the honors program in chemistry where they will have to take a course in quantum mechanics from the chemistry department. So, I would concentrate mostly on the applications from the point of view of chemistry and in this course I shall also record some elementary lectures on mathematics basic mathematics which we need for understanding mechanics. Chemistry students often do not take considerable amount of mathematics in their undergraduate program. Therefore, I would try to supplement the mathematics with some introductory account particularly in three areas namely in vectors vector algebra matrices and on differential equations. The subject that would be covered is in the form of modules. There are approximately 6 to 8 modules and about 50 minutes to 1 hour to 1 hour lectures, but they will be broken up into smaller time pieces smaller segments for the convenience of the reader. It depends on the particular lecture which piece a small or large it also depends on the content that is being considered. The material that I will I will cover in this course I will not write them elaborately the syllabus is there on the website, but basic parts would be vector algebra and an elementary account of vector spaces. It is non mathematical there are other courses in NPTEL which contain formal mathematical account of linear vector spaces where I know the links I will provide or you can see them updated on my website. On the mathematics we will also talk about matrices in reference to some special types like unitary orthogonal Hermitian matrices and the Eigenvalues and Eigenvectors of matrices. These are the various topics the modules will show grouping them in in some specific form. Differential equations first and second order apart from introductory account of the first order differential equation. We will jump to the second order differential equations and in particular to a few special equations which are important in the first course in quantum chemistry or quantum mechanics. The particular equations that I would be looking at from the point of solving them to get functional forms that we can represent using pictures and connect to chemistry etcetera. We will look at the Hermite differential equation. We shall look at the log air differential equation and also we shall look at the Legendre differential equations. Hermite the log air the and Legendre differential equation plus the associated Legendre differential equations. These are important in solving various model problems in chemistry. For example, Hermite's differential equation is fundamentally important in the understanding of molecular vibrations and the simple harmonic oscillators in quantum mechanics. The equations under the name of log air and Legendre differential equations both of them are important in understanding the solutions of the hydrogen atom which are extremely important for chemists. Therefore, this course will contain the full details of the solutions of these equations as part of the lectures. So, the mathematical content will be quite explicit will be taken through the course lectures and supplementary assignments and problems and other things will be provided as part of augmenting whatever that you have learned or supplementing them with other ideas. In the matrix I have already mentioned that I would talk about the matrices, eigenvalues and eigenvectors and we will find the way of bringing in the vectors and matrices together right in the very first lecture. I assume that you are familiar with basic algebra of matrices like addition, subtraction and matrix multiplication and some simple forms of square column vectors, row vectors, square matrices, rectangular matrices. These are all definitions you should be familiar with. We will see where some additional information is required I will provide them then and there. These are things as far as the mathematics is concerned now on the quantum processes and the quantum mechanics that we will look at. We shall look at simple model problems such as particle in a box, harmonic oscillator, the particle on a ring, hydrogen atom and particle on a sphere which is integrated with the hydrogen atom concepts. Basically we shall look at spherical harmonics. These are what are known as exactly solvable models. It is a fairly long module elementary. This will be followed by the fundamental aspects of quantum mechanics such as the postulatory basis, the postulates of quantum mechanics namely regarding the wave functions, probabilities, matrix representations, operators and computation of sorry calculation of expectation values etcetera. Then obviously when we do these things we will worry about the eigenvalues and eigenvectors of operators, commuting and non-commuting operators, angular momentum operators, angular momentum and harmonic oscillator operators. That will be more like on the formalism of quantum mechanics. This will be followed by the next topic namely spin angular momentum and the corresponding commutation relationships between components of angular momentum, commutation relations. We shall look at coupling of angular momentum, coupling or vector addition of angular momentum and mention very briefly the specific coefficients known as the Clebsch-Godaian coefficients. This is another module. The in the model problems I mentioned that we will look at hydrogen atom and particle on a sphere, but that will be a full-fledged module by itself namely the solution of the Schrodinger equation, radial functions and angular functions and probabilities. This is to be followed by a brief account of the time dependent Schrodinger equation and different representations of the operators and eigenstates representations particularly the Heisenberg representation or Heisenberg picture, the Schrodinger picture and if time permits a little bit about the interaction picture these are different ways of solving the Schrodinger equation and different ways of representing the problem. The a little bit about continuous and discrete distributions, continuous discrete distributions essentially probabilities and unitary evolution. The last module in this course will contain time independent perturbation theory with some applications to elementary systems. These are the topics that I would cover in this course. So, we shall start the quantum course with a bit of introductory vector algebra. None of what I say will be mathematically rigorous, but I hope it is accurate and that there is no problem of being something wrong if there is any please let me know. We will do the vectors with the help of matrices by representing vectors in the form of two dimensional or three dimensional matrices and follow through the algebra a little bit and I shall start with vectors in two dimensions. Typically, you know that a vector is represented by a magnitude and a direction and if you write an arbitrary vector you can also attach a coordinate system to this in the form of the unit vectors in two perpendicular directions x and y, x unit vector, y unit vector and write down the components of the arbitrary vector a in this coordinate system by the use of these two components A x and A y. So, that you represent the vector a by x times A x which is that plus y times A y which is that and therefore, the sum of the two is the vector that you started with ok. Now, typically A x is the projection of the vector A on the x direction. So, you can write this as x dotted with A scalar product and A y is the component of vector A in the direction of the unit vector y and that is y dotted A. Now, the matrix representation for this is given usually by a 2 by 2 matrix ok. x is usually represented as a column vector 1, 0 and y is represented by the column vector 0, 1 it is 2 rows and 1 column 2 by 1 matrix. When you say that the vector x is unit vector what it means is that the scalar product of x with it itself is unity. In the vector notation this is correct, but in the matrix notation if I have to use the matrices it has to be x transpose dotted with x which is given by 1, 0 the transpose of the matrix 1, 0 with the matrix itself to give you the number 1 ok. In the same way when you say y transpose dotted with y it is 0, 1, 0, 1 and it also gives you 1. The orthogonality or what we call as the perpendicular nature because it is 2 dimensions and probably in 3 dimensions we can write to this by perpendicularityid s. The orthogonality of x and y is given by x t scalar product with y is equal to 0 which is equal to y t scalar product with x is equal to 0 ok. And in the matrix notation it is easy to see that because x of t is 1, 0 and y is 0, 1 the product of the 2 matrices is 0. In a similar way y of t is 0, 1 with the x which is 1, 0 and this is also 0. So, this is orthogonality and this is this is normalization. The vectors are normalized to the unit vectors and the magnitude of the vector is 1 ok. This is a 2 by 2 matrix representation and in this representation if we write the vector a as instead of a x and a y if I follow my lecture notes if we write this as a 1 times x plus a 2 times y then you know that the column is such that it is a 1 times 1, 0 plus a 2 times 0, 1 giving you the column a 1 and a 2. This is the vector a in 2 dimensions. If we have 2 vectors a which is a 1, a 2 and the vector b given by the components b 1 and b 2, then the scalar product between the two or the scalar product of the two vectors is given either by a t dotted with b or with b t dotted with a. A t dotted with b gives you a 1, a 2 the row vector multiplied by the column vector b 1, b 2 to give you a 1, b 1 plus a 2, b 2 and of course, you can see that it is the same thing in this case because we are assuming these to be scalars and odd numbers. Therefore, a b is the same as b a. So, this will also give you the same result named the b 1, b 2 which is b of t and a which is a 1, a 2 will also give the same answer ok. The unit vector associated with a is of course, given by the sorry the unit vector is given by the hat or the carrot on top and it is nothing other than the vector a divided by the magnitude of the vector a and you know the magnitude of the vector a is given by this the square of the magnitude is a dotted with a which in this case is a of t for the matrix and so it will give you a 1, a 2 multiplied by a 1, a 2 row multiplied by a column and the answer is the magnitude of a is the square root of a 1 square plus a 2 square. This product is equal to the square of the magnitude ok. Therefore, the unit vector a is given by the column a 1 by square root of a 1 square plus a 2 square and a 2 by square root of a 1 square plus a 2 square. You can easily now see that the component a 1 can be obtained a 1, 1, 0 which is a 1, 0. This is the vector in the x direction and the quantity a 2, 0, 1 is 0, a 2 vector in the y direction. This is so much for the two dimensions. Let us now look at the vector in three dimensions. Let us now extend to this idea of vectors in two dimensions to three dimensions. The extension is rather straight forward. Let us still do that. The three basis vectors in three dimensions that one has to worry about can be chosen in many different ways, but if you want to restrict to the simplest consequences or the simplest example, we will choose the orthogonal Cartesian coordinate system having the three mutually perpendicular axis x and y and z ok. The unit vector alongone of them is x, another is y and the third is z ok. So, in three dimensions any vector a is represented by three components namely a 1 of x plus a 2 of y times y and a 3 times z. The components a 1, a 2, a 3 are given by the projection of the vector a on to the respective unit directions. So, this is x unit vector dotted with a gives you the component ofthe vector a 1. Suppose the arbitrary vector is in some direction this is better we call this a. The component of this along the x direction is a 1. The component a 2 is thecomponent along the y direction and a 3 is along the z direction of vector a. Now, in three dimensions one writes these unit vectors x, y and z in the form of 3 by 1column matrices namely x as 1, 0, 0, y as 0, 1, 0 and z as 0, 0, 1. Then it is clear that the vector a is given by a 1 times 1, 0, 0 plus etcetera which gives you the vector notationthe matrix notation for this vector as the column a 1, a 2, a 3 and you can see that if you project x t on to vector a or if you project the vector a in the direction of x in the form of matrices it is 1, 0, 0 multiplied by a 1, a 2, a 3 which gives you the component a 1 ok. So, the projection here essentially means matrix multiplication by the transpose of the vector that you have here and likewise for y of t on a which gives you when you multiply the vector a with 0, 1, 0, a 1, a 2, a 3 it gives you the answer a 2 and likewise for a 3. The unit vector a is of course,defined exactly the same way as we did in the 2 dimensions namely the vector a divided by the magnitude of a and in this case the vector a dotted with a in matrix notation is a 1, a 2, a 3, rho multiplied by the column a 1, a 2, a 3 and the answer is a 1 square plus a 2 square plus a 3 square and this is equal to the absolute vector squared and therefore, the absolute vector itself the absolute value itself is the square root leading to a being given by a unit vector being given by a 1 divided by square root of a 1 square plus a 2 square plus a 3 square a 2 divided by square root of a 1 square plus a 2 square plus a 3 square and the third component a 3 divided by square root of a 1 square plus a 2 square plus a 3 square. So, this is the unit vector a in the direction of the vector a. Scalar products are very easy to define obviously because if you have 2 vectors a and b given by 2 columns a 1, a 2, a 3 for a and b 1, b 2, b 3 for b then the scalar product in this case is a t dotted b or it is b t dotted a and the answer is when the components of a and b commute because they are scalars as we have written them down the answer is a 1, b 1 plus a 2, b 2 plus a 3, b 3 the in quantum mechanics we will deal with quantities which are not necessarily commuting if we have the position vector and the momentum vector the position times momentum is not the same as the momentum times the position you will get into the non commutability of vectors and particularly operators the operator form for the position and the operator form for the momentum obviously will be matrices which do not commute with each other. In matrices it is easy to understand that matrices need not have to commute when you take the products right now we have been dealing with numbers so therefore, we do not have any major problems. And now in three dimensions the normalization and orthogonality of the unit vectors also follow the same way namely you have x t dotted with x which is equal to 1 and that is equal to y t dotted with y and that is also equal to z t dotted with z this is normalization. Then you have x t dotted with y is equal to 0 is equal to y t dotted with z is equal to also 0 and this is also z t dotted with x ok x of y x of z y of z. So, we have covered all three of them. So, these this iscalled orthogonality property of the three unit vectors. However, it is interesting that we have been always forming a product of a row with a column and we ended up with numbers what if we do the other way around what if instead of x t dotted with x if we form the products x dotted with x t or x with x t. You can see immediately that x is a column and x t is a row this is a 3 by 1 column this is a 1 by 3 row and simple matrix multiplication gives you the answer that it is a 3 by 3 matrix. We shall now start looking at such quantities in the form of what we call as operators ok. So, we we are looking at what we mean by x x of t or y y of t or x y of t etcetera these are all unit vectors. So, let us start looking at these quantities first by doing a two dimensional calculation and in two dimensions of course, you know that x is 1 0 and x of t is 1 0. Therefore, x x of t is called an operator and the form is that it is 1 0 multiplied by 1 0 and the answer is the two dimensional matrix 2 by 2 matrix 1 0 0 0 and likewise y y of t in two dimensions is given by 0 1 multiplied by 0 1 to give you the answer 0 0 0 1 ok. And therefore, what you have is that x x of t plus y y of t is equal to the sum of the two matrices to give you the identity matrix in two dimensions identity matrix in 2 D. This is an operator identity operator what about x y of t that is going to be given by x is 1 0 and y is 0 1. So, the matrix that comes out of it is 0 1 0 0 and y x of t will be 0 1 1 0 to give you 0 0 1 0. So, you have four matrices which are called basis operators these are x x of t y y of t x y of t and y x of t. The any 2 by 2 matrix which is given by simply a 1 a 2 a 3 a 4 or more standard notation that very often one uses is a 1 1 a 1 2 a 2 1 a 2 2 does not matter which one do you use. Any such 2 by 2 matrix is expressed in the form of these four matrices along with the coefficients and the breakup of this is given by a 1 times 1 0 0 0 plus a 2 times 0 1 0 0 plus a 3 times 0 0 1 0 plus a 4 times 0 0 0 1. So, this is the expansion of an operator of a general 2 by 2 matrix of course, we are using real numbers we have not yet gone into complex space within that sense you must know general here refers only to real vector spaces. So, general 2 by 2 matrix in terms of four basis operators. Therefore, if we want to know individual component like a 1 it is easy to see that a 1 is obtained from the product of these three quantities 1 0 the matrix a 1 a 2 a 3 a 4 followed by the column vector 1 0. If we take the product of these three matrices you will get a 1. So, what does that mean? In terms of our notation it means that if we call this matrix by a let us write 1 0 by x of t and we call this matrix not by the arrow, but by a curly a tilde or whatever curly notation a and if you write x then that is given by a 1 or it is called a x x a 2 will be x t a y which is given by a x y this is like 1 1 this is like 1 2 in the matrix this is the row and column this is the row and the column if you are associating these with rows and columns and a 3 will be y of t a x and this is like a y x or it is a 2 1 and a 4 is y t a y these are all unit vectors. So, let us make sure that we have them with the tilde with the the carrots on top or with the what do you call the hats on top. So, this is a y y it is also known as a 2 2. So, the elements themselves or projections of the the operator a on both sides on the left and the right you pick the right matrix element by projecting on the corresponding quantities corresponding basis vectors. Therefore, a notation may be introduced at this point of time which will be very useful later on known as the bracket notation and the column vector x will be x the row vector the transpose will be written by the bra state x this is called the ket this is called the bra state and the scalar product will be written by this bracket x x the scalar product this is x t dotted with x and the scalar product of course, x x all are hats x y will be 0 and likewise for yeah this is in two dimensions we do not need to worry about z we are still in two dimensions with respect to the operators and then you can see that what you have done with respect to the quantity a tilde which has been written as a 1 a 2 a 3 a 4 is nothing but x t which is x a x let us go back and look up to this one a 1 is x of t a x. So, x of t is given by the bra state x and then this is the operator a and then the ket state x ok. So, the quantity a 1 is given by that the quantity a 2 is given by x a y the quantity a 3 is given by y a x and the last one is y a y and these are for obvious reasons called matrix elements of a matrix elements of. So, if we denote the vector x by 1 the symbol 1 and if we denote the vector y by the symbol 2 then you can write now the matrix a in terms of a 1 1 a 1 2 a 2 1 a 2 2 which simply means that in the ordering of the vectors x and y as bra in this order x and y and as ket in this order for the column in the ordering of this you see that what you get is x a x the row operator a column the row operator a column and the row operator a this is y row operator a column and the row operator a column ok. So, now you can see that this is the matrix element 2 1 in the basis set 2 and in the basis set 1 a being projected on both sides and so on. So, this is the way of representing a 2 dimensional linear operator using the basis vectors x and y in 3 dimensions what do we do we extend to this idea by adding the z. Therefore, the vector a in 3 dimension so not the vector sorry the operator a in 3 dimension is given by the the operator a in 3 dimension is given by 9 quantities because it is a general a 3 by 3 matrix let us follow the notation a 1 1 a 1 2 a 1 3 a 2 1 a 2 2 a 2 3 a 3 1 a 3 2 a 3 3 and that is also follow the notation that 1 essentially is the vector x 2 means if you want to put the kets here as well 2 means the vector y and 3 means the vector z and now you can see that the coefficients a 1 1 a 2 2 are given by the matrix element 1 and the operator a and the matrix element 1 this is a vector 1 this is vector 1 is given by 1 1 and likewise for any a i j matrix element of i th row and j th column is given by the vector i projecting on to the matrix a on the left and the vector j projecting on to the matrix a on the right. So, this is the matrix element of one lasting in 3 dimensions if we write x x of t plus y y of t plus z z of t if we do that you can see immediately that in 3 dimensions the x is given by 1 0 0 multiplied by 1 0 0 plus the y is given by 0 1 0 multiplied by 0 1 0 and the z is given by 0 0 1 multiplied by 0 0 1 the sum is going to be the identity matrix 1 0 0 0 1 0 0 0 1 which is usually written as the identity matrix in 3 dimensions it is an operator ok. Therefore, you see that the same in the bracket notation will be x ket followed by the x bra plus the y ket followed by the y bra which is this operator the sum of the 3 operators z followed by the z bra state the ket state bra state the sum of this will be given as the identity matrix 3 by 3. It is also known as the spectral resolution of identity in mathematics, but we will not worry too much about those speciallingo here what is important for us is that the operators x x y y z z when they sum up they give you the identity operator and therefore, this as a whole does not do anything to anything that it operates on. So, this is the way of decomposing any identity operator or resolving identity. So, let me sum up this part of the lecture that any operator a I will leave this for you to verify any operator a is therefore, given in the basis functions of i and j by this form namely the vector ket i the operator a i j component and the bra vector j please verify that this is true this for both 2 dimensions and 1 dimension and 3 dimensions. We shall follow this in the next lecture with the introduction of the concept known as the linear vector space with a little bit of formal mathematics, but not as rigorously as it should be given, but sufficiently rigorous for us to use these ideas and generalize the concept of vectors to n dimensional and infinite dimensional systems and conclude that part and follow this up with the matrix algebra. Until then, thank you very much.