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The bra-ket notation generally consists of 'ket', i.e. a vector, and a 'bra', i.e. some linear map that maps a vector to a number in the complex plane.

Now, using this bra-ket notation we can compute the inner product of some operator, say $\hat{H}$, so $\langle\psi|\hat{H}|\psi\rangle$ defines the eigenvalue of some hermitian operator $\hat{H}$. This is also called the expectation value of $\hat{H}$ and describes the probability of measuring this operator given the state $\psi$. I hope this is correctly understood.

We can also derive the inner product $\langle\phi|\psi\rangle$. I must admit that I am a little confused about this representation although it makes sence mathematically. Does this mean the probability of being in the state $\phi$ given the state $\psi$? I hope someone can clarify.

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    $\begingroup$ $\left<\psi\mid\hat H\mid\psi\right>$ does NOT define an eigenvalue. Would it help you if you convert $\left<\phi\mid\psi\right>$ to $\left<\phi\mid\hat{𝟙}\mid\psi\right>$? $\endgroup$ Jul 5, 2023 at 17:06
  • $\begingroup$ Multiplying a vector by the unit matrix yields a vector with all the diagonal components. How does this help me realize the physical meaning of $\langle\phi|\psi\rangle$? $\endgroup$ Jul 5, 2023 at 17:10
  • $\begingroup$ because by your question statement it seemed like you think you understood the bra-op-ket notation, yet not the bra-ket notation. Just worth asking explicitly. $\endgroup$ Jul 5, 2023 at 17:22

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Now, using this bra-ket notation we can compute the inner product of some operator, say $\hat{H}$, so $\langle\psi|\hat{H}|\psi\rangle$ defines the eigenvalue of some hermitian operator $\hat{H}$.

The inner product is a thing between two vectors - "the inner product of some operator" is not a meaningful phrase. If $|\psi\rangle$ is a normalized eigenvector of $\hat H$ with eigenvalue $\lambda$, then it's true that $\langle \psi|\hat H|\psi\rangle = \lambda$, but the definition of an eigenvector/eigenvalue pair is that $\hat H|\psi\rangle = \lambda|\psi\rangle$.

This is also called the expectation value of $\hat{H}$ and describes the probability of measuring this operator given the state $\psi$.

$\langle \psi|\hat H|\psi\rangle$ is referred to as the expectation value (or expected value) of $\hat H$ (corresponding to the normalized state vector $|\psi\rangle$). The interpretation of this number is that if you take a large number of identical systems all prepared in the state $|\psi\rangle$ and measured $\hat H$ in each of them, you would expect the mean value of all of those results to be $\langle \psi|\hat H|\psi\rangle$.

We can also derive the inner product $\langle\phi|\psi\rangle$. I must admit that I am a little confused about this representation although it makes sence mathematically. Does this mean the probability of being in the state $\phi$ given the state $\psi$?

There is no immediate physical interpretation of the inner product between two vectors - it is a quantity which shows up in all kinds of different contexts, and essentially measures the "overlap" between $\psi$ and $\phi$. It is analogous to the ordinary dot product between vectors in $\mathbb R^3$.

If $\psi$ is a normalized state vector representing the state of the system and $\phi$ is a normalized eigenvector of some observable $\hat A$ with (non-degenerate) eigenvalue $\lambda$, then $|\langle \phi|\psi\rangle|^2$ is the probability of measuring $\hat A$ to take the value $\lambda$. So that is one context in which the expression could arise. But trying to assign a single physical meaning to the inner product is like trying to assign a single physical meaning to the dot product between vectors in $\mathbb R^3$.

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    $\begingroup$ Thank you very much for the answer. It was very informative and satisfying. $\endgroup$ Jul 5, 2023 at 19:16
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If $H$ is a Hermitian operator, then $\langle \psi|H|\psi\rangle$ represents the expectation value (essentially "average measurement result") when the observable associated with $H$ is measured. Suppose $H = \pmatrix{0 \ 1 \\ 1 \ 0}$ and $|\psi\rangle = \pmatrix{1 \\ 0}$. If a measurement of the observable associated with $H$ is done on the state $\psi$, the expectation value is 0 ($\langle\psi|H|\psi\rangle = \pmatrix{1 \ 0}\pmatrix{0 \\ 1} = 0$). This isn't in any sense an eigenvalue of $H$, the two eigenvalues of $H$ are 1 and -1, and it is in fact impossible for a measurement of the observable associated with $H$ to result in a value of 0, but this still does have meaning that will be established.

When $|\psi\rangle$ undergoes a measurement associated with $H$'s observable, it will collapse to an eigenvector of $H$ and the measurement will result in the value of the associated eigenvector, and the two unit eigenvectors (up to multiplication by a complex unit circle scalar) of $H$ in the above example are $\pmatrix{1/\sqrt{2} \\ 1/\sqrt{2}}$ and $\pmatrix{1/\sqrt{2} \\ -1/\sqrt{2}}$ respectively for 1 and -1. The Born rule states that the probability of $|\psi\rangle$ collapsing to another state $|\phi\rangle$ in this case is given as $\langle\psi|\phi\rangle\langle\phi|\psi\rangle = |\langle\phi|\psi\rangle|^2$ (since reversing the bra ket is equivalent to complex conjugation and $a a^* = |a|^2$). For the eigenvectors of $H$ and with $|\psi\rangle = \pmatrix{1 \\ 0}$, this value is equal to $1/2$ in both cases, so there is a half chance that, upon measuring $H$'s observable, the result will be 1, and a half chance the result will be -1. If this is attempted multiple times you will get an averaged result of 0, even though that is not a possible result of a single measurement, and that is what $\langle\psi|H|\psi\rangle$, the expectation value of a measurement on $|\psi\rangle$ of the observable associated with $H$, physically represents.

Now, $|\phi\rangle\langle\phi|$ is itself a Hermitian operator, and this means the Born rule probability value $\langle\psi|\phi\rangle\langle\phi|\psi\rangle$ can itself be interpreted as the expectation value of a measurement (one assumes all vectors here are normalized). This Hermitian operator has $|\phi\rangle$ as an eigenvector with eigenvalue 1 and all orthogonal states an eigenvalue of 0, meaning essentially a measurement of 1 corresponds to the state being measured as $|\phi\rangle$ and 0 if measured otherwise (but not probing any more information about the state other than it being orthogonal from $|\phi\rangle$). If the probability of measuring a 1 is $p$ and the other option is 0, then the average result will be $p$, so this interpretation still works.

So, $\langle\phi|\psi\rangle$ by itself is not a probability (for one it is complex), but it does, when multiplied by its complex conjugate or has its absolute value squared, correspond to the probability $|\psi\rangle$ will become $|\phi\rangle$ when a measurement that distinguishes $|\phi\rangle$ is performed. Note that in some higher dimensional cases $|\phi\rangle$ may be an eigenvector of an observable's operator but not the only one of that eigenvalue, and in that case a measurement of $|\psi\rangle$ that resulted in $|\phi\rangle$'s eigenvalue would collapse to an even projection of $|\psi\rangle$ onto the corresponding eigenspace rather than to $|\phi\rangle$. For example, if $|\phi_1\rangle$ and $|\phi_2\rangle$ were both eigenvectors of an operator $A$ with the same eigenvalue and all other eigenvectors of that eigenvalue are linear combinations of the two, the probability that value is measured on $|\psi\rangle$ is $\langle \psi|(|\phi_1\rangle\langle \phi_1| + |\phi_2\rangle\langle \phi_2|)|\psi\rangle$, and the state collapsed to will be $|\phi_1\rangle\langle\phi_1|\psi\rangle + |\phi_2\rangle\langle\phi_2|\psi\rangle$ normalized.

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  • $\begingroup$ Thank you very much for your answer. I need some time to let some of these things sink in. I will read the rest when I am ready to learn something new. But again, thank you. $\endgroup$ Jul 5, 2023 at 19:30
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When first learning QM it is useful to write yourself a little linear algebra / quantum physics dictionary.

But first, the mathematical setup: one usually has a complex vector space* $V$ (usually a Hilbert space $\mathcal{H}$ or some technical extension) equipped with a hermitean form $\langle{\cdot},\cdot\rangle$. The latter gives us a canonical isomorphism between $V$ and the dual space $V^*$, defined by $x \mapsto \phi_x$ where $\phi_x : y \mapsto \langle x,y \rangle$. By linear operators we mean linear maps $\mathcal{O}:V \rightarrow V$. In QM, many important operators (in particular observables) are self-adjoint. Introductory books written by physicists sometimes use hats such has $\hat{\mathcal{O}}$ to distinguish $\mathcal{O}$ from its classical counterpart. I will forgo this since here the distinction is clear from context.

The beginning of our dictionary is:

  • A ket $|x\rangle$ (the kets make up the physical states) $\leftrightarrow$ The element $x \in V$
  • A bra $\langle x |$ $\leftrightarrow$ The element $\phi_x \in V^*$ as defined above.

Thus $\langle x | y\rangle$ is merely the hermitean product $\phi_x(y) = \langle x , y \rangle$.

For a self adjoint operator $A$, one has $\langle x , A y \rangle = \langle Ax,y\rangle$ and so one might as well write this as an ambivalent $\langle x | A |y \rangle $ in this situation. This is Dirac notation. It only makes sense when $A$ is self-adjoint. With this in mind, we can continue the dictionary:

  • Oberservables $\leftrightarrow$ Self-adjoint operators $A:V \rightarrow V$
  • The expectation value of an observable $A$ given a state $|\psi\rangle$ $\leftrightarrow$ $E_A(\psi) = \langle \psi|A | \psi \rangle / \langle \psi | \psi \rangle$.

There is a slight 'unexpected' notational advantage of Dirac notation, which is helpful in physics. This is that elements of $V$ (which in applications can be some messy tensor product) can be very easily notated. For example, labelling of hydrogen atom states is really clear with Dirac notation; e.g. $|n,j,s\rangle$ is easy to read whereas traditionally mathematicians would probably notate using subscripts $v_{n,j,s}$. You can even use words and symbols in kets, which would look horrible as subscripts.

I will stop here. You should continue to expand this dictionary as you learn more. In particular, try including eigenvalues/eigenvectors. Good books to learn from include those by Sakurai and Messiah.

*For the technically minded, states are really 'rays' in a vector space, or better, elements of the projective vector space $PV$. Doing calculations over $PV$ often amounts to going back out to $V$, but conceptually you should realize that $PV$ is the space you are really working over.

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  • $\begingroup$ Thank you. I think it is a great idea to couple these quantum mechanical concepts to concepts from linear algebra. I do have one question. Is it correctly understood that the expectation value of an observable, say $A$, can be written as either $\langle\psi|A|\psi\rangle$ or $\langle\psi|\psi\rangle$ $\endgroup$ Jul 5, 2023 at 21:40
  • $\begingroup$ It's not really coupling these two concepts together. Linear algebra is the language of quantum mechanics. It seems that you might want a good course in linear algebra (at the level of Lang say). No. $\langle \psi | A | \psi \rangle$ is the object $\langle \psi, A \psi \rangle$ which obviously depends on $A$. This is what is meant by the expectation value of $A$. On the other hand, $\langle \psi | \psi \rangle$ has no $A$-dependence, indeed taking $\psi$ to be normalized by definition means $\langle \psi | \psi \rangle = 1$ trivially. $\endgroup$ Jul 5, 2023 at 21:54
  • $\begingroup$ Thank you. Sakurai explains this concept really well also. I was just confused about the expectation value given as $E_A(\psi)=\langle\psi|A|\psi\rangle/\langle\psi|\psi\rangle$, but maybe the point is to normalize the expectation value? $\endgroup$ Jul 6, 2023 at 15:17
  • $\begingroup$ Exactly. You could of course normalize $|\psi\rangle$ a priori in which case you don't need the denominator.. $\endgroup$ Jul 6, 2023 at 18:28

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