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I'm currently making my way through quantum mechanics by Leonard Susskind, but have got stuck at this part; writing a vector as the sum of basis vectors.

I get that for an $N$ dimensional space and a particular orthonormal basis of kets labeled $|i\rangle$, where $i$ runs from $1$ to $N$, a vector $|A\rangle$ can be written as a sum of basis kets $|A\rangle = \sum a_i|i\rangle$.

But then I don't get why to work out the components you then take the inner product with a basis bra $\langle j|$ and how you can then be left with $\langle j|A\rangle = a_j$.

Maybe I just don't understand the rules of inner product properly, but any help at understanding would be appreciated.

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    $\begingroup$ Isn't it just that $\langle j | i \rangle = \delta_{ij}$? (See Kronecker delta if you're unfamiliar with symbol $\delta_{ij}$) $\endgroup$ – Alfred Centauri Sep 2 '17 at 13:08
  • $\begingroup$ Just note, if you later hear GR: all the stuff about the metric tensor with indices down and up is about having a basis, which is not orthonormal. The inverse metric transforms from the scalar projections to the basis coefficients. $\endgroup$ – lalala Sep 2 '17 at 17:02
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Write $|A\rangle = \sum_i a_i|i\rangle$: you know you can do this because $\left\{|i\rangle\right\}$ are a basis: this is the definition of a basis. Further, they are orthonormal which means that $\langle i|j\rangle = \delta_{ij}$ (this is what being orthonormal means). So now consider $$\begin{align} \langle j|A\rangle &= \langle j|\left(\sum_i a_i|i\rangle\right)\\ &= \sum_i a_i\langle j|i\rangle\\ &= \sum_i a_i \delta_{ji}\\ &= a_j \end{align}$$

As required. The intermediate steps are just moving the basis vector into the sum.

(The answer by CDCM is better than this one, I think).

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  • $\begingroup$ Thank you this is helpful. I'm completely new to this, have only done the scalar product at school so have a few slight things I just haven't realised, so I'm assuming the usual rules follow that you always have to take the inner product on both sides of the equation? (Sorry if this is obvious). And I just don't understand how you are finished with the aj, logically I don't understand this jump but before this everything is perfectly clear. Thanks again. I understand CDCM has shown this but I'm a little confused as they are bras and kets which are different I thought, so they can be mixed? $\endgroup$ – May J Sep 3 '17 at 16:51
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When you say that the basis is orthonormal, mathematically you are saying that $$\langle j | i \rangle = \delta_{ij},$$ where $\delta_{ij}$ is the Kronecker delta, which is equal to $1$ if $i=j$, and $0$ otherwise. Now with this knowledge, let's see what happens to your state vector $|A\rangle = \sum a_i|i\rangle$. $$\langle j | A \rangle = \sum a_i \langle j |i\rangle = ~...+ ~ a_{j-1}\langle j|j-1 \rangle + a_j\langle j | j \rangle + a_{j+1}\langle j|j+1 \rangle + ~... $$ where I've isolated the part of the sum near $j$. Now let's use our orthonomality expression: $$\langle j | A \rangle = \sum a_i \delta_{ij}= ~...+ ~ a_{j-1}\delta_{j (j-1)} + a_j\delta_{jj} + a_{j+1}\delta_{j(j+1)} + ~... $$ Now we can see that $\delta_{ij}$ will be zero for all terms except for the $jj$ term. Hence our sum becomes: $$\langle j| A \rangle = a_1 \times 0 + a_2 \times 0 + ... + ~a_j \times 1 + a_{j+1} \times 0 +~ ... = a_j$$ This post is messy but I'm trying to show you how premultiplying by $\langle j |$ actually goes and picks out the relevant term.

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Maybe it helps to compare the complex inner product to the regular vector dot product. Take for example the vector $\vec v=(5,3,2)$ ; the components in the cartesian basis can be found by taking the dot product with the cartesian basis vectors $\hat x_1, \hat x_2$ and $\hat x_3$ $$v_1=\hat x_1\cdot \vec v=(1,0,0)\ \cdot\ (5,3,2)=5$$ Similarly $v_2=\hat x_2\cdot \vec v=3$ and $v_3=\hat x_3\cdot\vec v=2$

What happens mathematically is that the dot product distributes such that \begin{align}\hat x_1\cdot\vec v&=\hat x_1\cdot (5\hat x_1+3\hat x_2+2\hat x_3)\\&=5\hat x_1\cdot\hat x_1+3\hat x_1\cdot\hat x_2+2\hat x_1\cdot\hat x_3\end{align} Since this is an orthonormal basis, we can make use of the fact that $\hat x_i\cdot \hat x_j=\delta_{ij}$. So everything but $\hat x_1\cdot \hat x_1$ becomes zero and we are left with 5. For this argument we could have used any three vectors that formed an orthonormal basis.

To transform this to the language of bra's and kets you just replace $\vec v$ by $|v\rangle$ and $\hat x_i$ by $\langle i|$. The difference is that bra's and kets can be complex-valued and can have infinite dimensions.

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  • $\begingroup$ I understand this so far thank you very much. I just cannot get my head around how this causes ∑ai<j|i> to ∑aj ? (sorry for the lack of maths symbols I'm unsure how to insert them onto here, I will learn). Or is this how before we were just left with 5? I think I'm just unsure about the end expression which is |A> = ∑|i><i|A> and how that is derived? $\endgroup$ – May J Sep 3 '17 at 16:38
  • $\begingroup$ $\sum_i a_i<j|i>$ is indeed a_j for the same reason the expression above yields 5: <j|i> = 1 if i=j and <j|i> = 0 if j $\neq$ i. The last expression basically says this: any vector/ket, call it |A>, can be written as a sum of a number of basis vectors <i| multiplied by the component of |A> along that basis vector (using the scalar product). The component of |A> along <i| can be found by the inner product <i|A>. $\endgroup$ – AccidentalTaylorExpansion Sep 4 '17 at 21:33
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You might have also meant a related question: why is $\langle m | n\rangle$ equal to zero if $m\ne n$? Like in 3D it's obvious that we just choose three vectors to point in perpendicular directions and their dot products with each other are zero, but can we still do this in this weird quantum space?

Generally these basis states $|m\rangle$ were carefully derived because they have some relationship with a quantum operator that we could call, say, the "unperturbed Hamiltonian operator" $\hat H_0$, such that $\hat H_0 |m\rangle = E_m |m\rangle,$ where $E_m$ is some real energy of the state. Now our quantum algebra demands that $\big(\langle \phi|\chi\rangle\big)^* = \langle \chi|\phi\rangle$ for any vectors $|\chi\rangle$ and $|\psi\rangle$, where the $*$ is complex conjugation $a + i b \mapsto a - i b;$ if there is an operator in the middle this becomes the only slightly more complicated

$$\big(\langle \phi|\hat A|\chi\rangle\big)^* = \langle \chi|\hat A^\dagger |\phi\rangle.$$ I mention this but actually the condition that all the $E_m$ be real goes hand in hand with a property called Hermitian-ness, which states that $\hat H_0 = \hat H_0^\dagger$, so for right now you do not need to think about this conjugate transpose operation $\dagger$ very hard.

Looking at $\langle m | \hat H_0 |n \rangle$ we can therefore see that this must simultaneously be two different numbers: $$E_n \langle m |n \rangle = \langle m | \hat H_0 |n \rangle = \big(\langle n | \hat H_0 |m \rangle\big)^* = \big(E_m~\langle n |m \rangle\big)^* = E_m \langle m | n\rangle.$$ That the first of these equals the last of these can also be written as: $$\big(E_m - E_n\big)~\langle m|n\rangle = 0.$$ Now this is just a multiplication of two complex numbers being zero, and complex multiplication does still have the property that $|z_1~z_2| = |z_1|~|z_2|$ which means that if $z_1~z_2 = 0$ then either $z_1=0$ or $z_2 = 0$ or both.

So this leads to two big possibilities. First, there's the usual case: the energies are different and then $\langle m | n\rangle = 0$. In a very real sense this is the only case we really need to think about, as in physics things tend to be a bit "noisy" and so if $m\ne n$ then there is some noise which stops $E_m$ from being equal to $E_n.$ This insight is not unique to physics; Tadashi Tokieda talks about using it in pure mathematics at length in his topology and geometry lectures that he's put for free online.

But you can be poked mathematically by $E_m = E_n$ while $m \ne n$ and this is called a "degeneracy" in the unperturbed Hamiltonian, and there is a way to deal with it. You see, this is an indicator that you did not pick (we'll say) $|m\rangle$ right when you solved for these functions. In fact these equations are saying that you should have picked $$|m'\rangle \propto |m\rangle - |n\rangle \langle n | m\rangle$$Note that this needs to be renormalized so that $\langle m'|m'\rangle = 1$ but that is no bother. If you tried to do this with $|m\rangle = |n\rangle$ you would get just the zero vector which can't be normalized, but any other choice will lead to a new normalizable vector with $E_{m'} = E_m = E_n$, but with the property that $\langle m'|n\rangle = 0$ as desired.

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indeed your problem is with the concept of inner products. Let me drop the bra-ket notation and just work with the standard Euclidean dot product. The bra, kets and inner products in Quantum Mechanics share the properties discussed bellow.

The geometric meaning of the dot product between two vectors $\vec A$ and $\vec B$ is that you project one of the vectors onto the other and then multiply the length of the two colinear straight line segments, $|\vec A|\cos\theta$ and $|\vec B|$, to obtain $|\vec A||\vec B|\cos\theta$.

enter image description here

In particular if you take the dot product between $\vec A$ with a unit vector $\vec j$ then you get $|\vec A|\cos\theta$, i.e., the length of $\vec A$ along the direction of $\vec j$. Now consider an orthonormal basis given by the unit vectors $\vec j$. The projections of $\vec A$ onto the directions $\vec j$, $\vec A\cdot \vec j$, give the components of $\vec A$ in the given basis. Hence $\vec A=\sum_j (\vec A\cdot \vec j)\vec j$.

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